Proof step 1 from proof conttr-thm (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  lim(⌜λ n. βf(σf(n), αf(n))⌝ ) ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ ⊦

Rule argument

Additional information

Proof step 2 from proof conttr-thm (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 3:

  mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ n. # σf(n) = # αf(n) + ⌜m⌝ ⊦ # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜m⌝

• Proof step 4:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ ⊦

Rule argument

Additional information

Used terms: σf, αf, 1

Proof step 3 from proof conttr-thm (previous step) (back to tree)

Lemma twomon-limsize

Proof step 4 from proof conttr-thm (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 5:

  ⊦ mon(σσ) ∧ σσ ≠ [] → (∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ lim(σσ)) ∧ (∀ m. ¬ # lim(σσ) < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ ⌜lim(σσ).take(m)⌝ ⊑ σσ(n)))

• Proof step 6:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), mon(⌜αf⌝ ) ∧ ⌜αf⌝ ≠ [] → (∀ n. ⌜n⌝ < # ⌜αf⌝ → ⌜αf⌝ (n) ⊑ lim(⌜αf⌝ )) ∧ (∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜n⌝ < # ⌜αf⌝ ∧ ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ ⌜αf⌝ (n))), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ ⊦

Rule argument

Additional information

Used terms: ⌜αf⌝

Proof step 5 from proof conttr-thm (previous step) (back to tree)

A spec-lemma

Proof step 6 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 7:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ lim(⌜αf⌝ ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ (n) = nf(n); (s) ⊦ ⌜m⌝ < ∞; (s) ⊦ # ⌜nf⌝ = ∞; (s) ⊦ [] ≠ ⌜nf⌝ ; (s)

Proof step 7 from proof conttr-thm (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 8:

  ⊦ mon(σσ) ∧ σσ ≠ [] → (∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ lim(σσ)) ∧ (∀ m. ¬ # lim(σσ) < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ ⌜lim(σσ).take(m)⌝ ⊑ σσ(n)))

• Proof step 9:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), mon(⌜σf⌝ ) ∧ ⌜σf⌝ ≠ [] → (∀ n. ⌜n⌝ < # ⌜σf⌝ → ⌜σf⌝ (n) ⊑ lim(⌜σf⌝ )) ∧ (∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜n⌝ < # ⌜σf⌝ ∧ ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ ⌜σf⌝ (n))), ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ ⊦

Rule argument

Additional information

Used terms: ⌜σf⌝

Proof step 8 from proof conttr-thm (previous step) (back to tree)

A spec-lemma

Proof step 9 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ (n) = nf(n); (s) ⊦ ⌜m⌝ < ∞; (s) ⊦ # ⌜nf⌝ = ∞; (s) ⊦ [] ≠ ⌜nf⌝ ; (s)

Proof step 10 from proof conttr-thm (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 11:

  ⊦ mon(σσ) ∧ σσ ≠ [] → (∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ lim(σσ)) ∧ (∀ m. ¬ # lim(σσ) < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ ⌜lim(σσ).take(m)⌝ ⊑ σσ(n)))

• Proof step 12:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), mon(⌜λ n. βf(σf(n), αf(n))⌝ ) ∧ ⌜λ n. βf(σf(n), αf(n))⌝ ≠ [] → (∀ n. ⌜n⌝ < # ⌜λ n. βf(σf(n), αf(n))⌝ → ⌜λ n. βf(σf(n), αf(n))⌝ (n) ⊑ lim(⌜λ n. βf(σf(n), αf(n))⌝ )) ∧ (∀ m. ¬ # lim(⌜λ n. βf(σf(n), αf(n))⌝ ) < ⌜m⌝ → (∃ n. ⌜n⌝ < # ⌜λ n. βf(σf(n), αf(n))⌝ ∧ ⌜lim(⌜λ n. βf(σf(n), αf(n))⌝ ).take(m)⌝ ⊑ ⌜λ n. βf(σf(n), αf(n))⌝ (n))), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ ⊦

Rule argument

Additional information

Used terms: ⌜λ n. βf(σf(n), αf(n))⌝

Proof step 11 from proof conttr-thm (previous step) (back to tree)

A spec-lemma

Proof step 12 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), mon(⌜λ n. βf(σf(n), αf(n))⌝ ) → (∀ n. βf(σf(n), αf(n)) ⊑ lim(⌜λ n. βf(σf(n), αf(n))⌝ )) ∧ (∀ m. ¬ # lim(⌜λ n. βf(σf(n), αf(n))⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜λ n. βf(σf(n), αf(n))⌝ ).take(m)⌝ ⊑ βf(σf(n), αf(n)))), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ (n) = nf(n); (s) ⊦ ⌜m⌝ < ∞; (s) ⊦ # ⌜nf⌝ = ∞; (s) ⊦ [] ≠ ⌜nf⌝ ; (s)

Proof step 13 from proof conttr-thm (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 14:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ¬ mon(⌜λ n. βf(σf(n), αf(n))⌝ ), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ ⊦

• Proof step 33:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ m. ¬ # lim(⌜λ n. βf(σf(n), αf(n))⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜λ n. βf(σf(n), αf(n))⌝ ).take(m)⌝ ⊑ βf(σf(n), αf(n))), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ lim(⌜λ n. βf(σf(n), αf(n))⌝ ) ⊦

Rule argument

Additional information

Proof step 14 from proof conttr-thm (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 15:

  ⊦ mon(σσ) ↔ (∀ n. ⌜n⌝ + ⌜1⌝ < # σσ → σσ(n) ⊑ σσ(n + 1))

• Proof step 16:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∃ n. ¬(⌜n⌝ + ⌜1⌝ < # ⌜λ n. βf(σf(n), αf(n))⌝ → ⌜λ n. βf(σf(n), αf(n))⌝ (n) ⊑ ⌜λ n. βf(σf(n), αf(n))⌝ (n + 1)) ⊦

Rule argument

Additional information

Used terms: ⌜λ n. βf(σf(n), αf(n))⌝

Proof step 15 from proof conttr-thm (previous step) (back to tree)

A spec-lemma

Proof step 16 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 17:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ¬ βf(σf(n), αf(n)) ⊑ βf(σf(n + 1), αf(n + 1)), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ (n) = nf(n); (s) add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜n⌝ ≠ ∞; (s) ⊦ ⌜m⌝ + n∞ = ∞ ↔ n∞ = ∞; (s) ⊦ n∞ < ∞ ↔ n∞ ≠ ∞; (s) ⊦ # ⌜nf⌝ = ∞; (s)

Proof step 17 from proof conttr-thm (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 18:

  ⊦ mon(σσ) ↔ (∀ n. ⌜n⌝ + ⌜1⌝ < # σσ → σσ(n) ⊑ σσ(n + 1))

• Proof step 19:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), cont(βf), mon(⌜σf⌝ ), ¬ βf(σf(n), αf(n)) ⊑ βf(σf(n + 1), αf(n + 1)), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜n⌝ + ⌜1⌝ < # ⌜αf⌝ → ⌜αf⌝ (n) ⊑ ⌜αf⌝ (n + 1) ⊦

Rule argument

Additional information

Used terms: ⌜αf⌝

Proof step 18 from proof conttr-thm (previous step) (back to tree)

A spec-lemma

Proof step 19 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), cont(βf), mon(⌜σf⌝ ), ¬ βf(σf(n), αf(n)) ⊑ βf(σf(n + 1), αf(n + 1)), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n0. αf(n0) ⊑ αf(n0 + 1) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ (n) = nf(n); (s) add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜n⌝ ≠ ∞; (s) ⊦ ⌜m⌝ + n∞ = ∞ ↔ n∞ = ∞; (s) ⊦ n∞ < ∞ ↔ n∞ ≠ ∞; (s) ⊦ # ⌜nf⌝ = ∞; (s)

Proof step 20 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 21:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), cont(βf), mon(⌜σf⌝ ), αf(n) ⊑ αf(n + 1), ¬ βf(σf(n), αf(n)) ⊑ βf(σf(n + 1), αf(n + 1)), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ) ⊦

Rule argument

Additional information

Used terms: n

Proof step 21 from proof conttr-thm (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 22:

  ⊦ cont(βf) ↔ mon(βf) ∧ pbd(βf)

• Proof step 23:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), mon(⌜σf⌝ ), mon(βf), pbd(βf), αf(n) ⊑ αf(n + 1), ¬ βf(σf(n), αf(n)) ⊑ βf(σf(n + 1), αf(n + 1)), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ ⊦

Rule argument

Additional information

Used terms: βf

Proof step 22 from proof conttr-thm (previous step) (back to tree)

Lemma conttr-def

Proof step 23 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), mon(⌜σf⌝ ), pbd(βf), αf(n) ⊑ αf(n + 1), ¬ βf(σf(n), αf(n)) ⊑ βf(σf(n + 1), αf(n + 1)), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used simplifier rules: montr-def: ⊦ mon(βf) ↔ (∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2));

Proof step 24 from proof conttr-thm (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 25:

  ⊦ mon(σσ) ↔ (∀ n. ⌜n⌝ + ⌜1⌝ < # σσ → σσ(n) ⊑ σσ(n + 1))

• Proof step 26:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), pbd(βf), αf(n) ⊑ αf(n + 1), ¬ βf(σf(n), αf(n)) ⊑ βf(σf(n + 1), αf(n + 1)), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. ⌜n⌝ + ⌜1⌝ < # ⌜σf⌝ → ⌜σf⌝ (n) ⊑ ⌜σf⌝ (n + 1), ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used terms: ⌜σf⌝

Proof step 25 from proof conttr-thm (previous step) (back to tree)

A spec-lemma

Proof step 26 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 27:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), pbd(βf), αf(n) ⊑ αf(n + 1), ¬ βf(σf(n), αf(n)) ⊑ βf(σf(n + 1), αf(n + 1)), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n0. σf(n0) ⊑ σf(n0 + 1), ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ (n) = nf(n); (s) add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜n⌝ ≠ ∞; (s) ⊦ ⌜m⌝ + n∞ = ∞ ↔ n∞ = ∞; (s) ⊦ n∞ < ∞ ↔ n∞ ≠ ∞; (s) ⊦ # ⌜nf⌝ = ∞; (s)

Proof step 27 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 28:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), pbd(βf), αf(n) ⊑ αf(n + 1), σf(n) ⊑ σf(n + 1), ¬ βf(σf(n), αf(n)) ⊑ βf(σf(n + 1), αf(n + 1)), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used terms: n

Proof step 28 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 29:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), pbd(βf), αf(n) ⊑ αf(n + 1), σf(n) ⊑ σf(n + 1), ¬ βf(σf(n), αf(n)) ⊑ βf(σf(n + 1), αf(n + 1)), # σf(n) = # αf(n) + ⌜1⌝ ∧ # σf(n + 1) = # αf(n + 1) + ⌜1⌝ ∧ σf(n) ⊑ σf(n + 1) ∧ αf(n) ⊑ αf(n + 1) → βf(σf(n), αf(n)) ⊑ βf(σf(n + 1), αf(n + 1)), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ) ⊦

Rule argument

Additional information

Used terms: σf(n), αf(n), σf(n + 1), αf(n + 1)

Proof step 29 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 30:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), pbd(βf), αf(n) ⊑ αf(n + 1), σf(n) ⊑ σf(n + 1), ¬ βf(σf(n), αf(n)) ⊑ βf(σf(n + 1), αf(n + 1)), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ ⊦ # σf(n) = # αf(n) + ⌜1⌝ ∧ # σf(n + 1) = # αf(n + 1) + ⌜1⌝

Rule argument

Additional information

Proof step 30 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 31:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , # σf(n + 1) = # αf(n + 1) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), pbd(βf), αf(n) ⊑ αf(n + 1), σf(n) ⊑ σf(n + 1), ¬ βf(σf(n), αf(n)) ⊑ βf(σf(n + 1), αf(n + 1)), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ) ⊦ # σf(n) = # αf(n) + ⌜1⌝ ∧ # σf(n + 1) = # αf(n + 1) + ⌜1⌝

Rule argument

Additional information

Used terms: n + 1

Proof step 31 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 32:

  # σf(n + 1) = # αf(n + 1) + ⌜1⌝ , # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , # σf(n) ≠ # αf(n) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), pbd(βf), αf(n) ⊑ αf(n + 1), σf(n) ⊑ σf(n + 1), ¬ βf(σf(n), αf(n)) ⊑ βf(σf(n + 1), αf(n + 1)), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ ⊦

Rule argument

Additional information

Proof step 32 from proof conttr-thm (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n

Proof step 33 from proof conttr-thm (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 34:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ¬ # lim(⌜λ n. βf(σf(n), αf(n))⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜λ n. βf(σf(n), αf(n))⌝ ).take(m)⌝ ⊑ βf(σf(n), αf(n))), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ lim(⌜λ n. βf(σf(n), αf(n))⌝ ), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ) ⊦

• Proof step 162:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ¬ # lim(⌜λ n. βf(σf(n), αf(n))⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜λ n. βf(σf(n), αf(n))⌝ ).take(m)⌝ ⊑ βf(σf(n), αf(n))), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ lim(⌜λ n. βf(σf(n), αf(n))⌝ ), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ) ⊦

Rule argument

Additional information

Used simplifier rules: Stream[A] freely generated by ⌝, ⌝;

Proof step 34 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 35:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , ⌜bl⌝ ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ m. ¬ # bl < m → (∃ n. ⌜bl.take(m)⌝ ⊑ βf(σf(n), αf(n))), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ .take(n) = x.take(n); (s) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s)

Proof step 35 from proof conttr-thm (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 36:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , ⌜bl⌝ ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ m. ¬ # bl < m → (∃ n. ⌜bl.take(m)⌝ ⊑ βf(σf(n), αf(n))), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ ⊦

• Proof step 77:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , ⌜bl⌝ ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ m. ¬ # bl < m → (∃ n. ⌜bl.take(m)⌝ ⊑ βf(σf(n), αf(n))), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: Stream[A] freely generated by ⌝, ⌝;

Proof step 36 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 37:

  # lim(⌜σf⌝ ) = ⌜# al⌝ + ⌜1⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , ⌜bl⌝ ≠ βf(lim(⌜σf⌝ ), ⌜al⌝ ), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ¬ # al < m → (∃ n. ⌜al.take(m)⌝ ⊑ αf(n)), ∀ m. ¬ # bl < m → (∃ n. ⌜bl.take(m)⌝ ⊑ βf(σf(n), αf(n))), ∀ m. ¬ # al + 1 < m → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ lim(⌜σf⌝ ) ⊦

Rule argument

Additional information

Used simplifier rules: add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; (s) ⊦ ⌜x⌝ .take(n) = x.take(n); (s) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s)

Proof step 37 from proof conttr-thm (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 38:

  # lim(⌜σf⌝ ) = ⌜# al⌝ + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , ⌜bl⌝ ≠ βf(lim(⌜σf⌝ ), ⌜al⌝ ), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ m. ¬ # al + 1 < m → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ m. ¬ # bl < m → (∃ n. ⌜bl.take(m)⌝ ⊑ βf(σf(n), αf(n))), ∀ m. ¬ # al < m → (∃ n. ⌜al.take(m)⌝ ⊑ αf(n)), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: Stream[A] freely generated by ⌝, ⌝; ⊦ # ⌜nf⌝ = ∞; (s) ⊦ ⌜m⌝ + n∞ = ∞ ↔ n∞ = ∞; (s) ⊦ ⌜n⌝ ≠ ∞; (s) add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf

Proof step 38 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 39:

  # al + 1 = # sl, lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , ⌜bl⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ m. ¬ # bl < m → (∃ n. ⌜bl.take(m)⌝ ⊑ βf(σf(n), αf(n))), ∀ m. ¬ # al < m → (∃ n. ⌜al.take(m)⌝ ⊑ αf(n)), ∀ m. ¬ # sl < m → (∃ n. ⌜sl.take(m)⌝ ⊑ σf(n)), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ .take(n) = x.take(n); (s) add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜n⌝ + ⌜m⌝ = ⌜n0⌝ ↔ n + m = n0; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s)

Proof step 39 from proof conttr-thm (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 40:

  # sl = # al + 1, lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , ⌜bl⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ¬ # sl < m → (∃ n. ⌜sl.take(m)⌝ ⊑ σf(n)), ∀ m. ¬ # al < m → (∃ n. ⌜al.take(m)⌝ ⊑ αf(n)), ∀ m. ¬ # bl < m → (∃ n. ⌜bl.take(m)⌝ ⊑ βf(σf(n), αf(n))), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Used formulas: # sl = # al + 1

Proof step 40 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 41:

  # sl = # al + 1, lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , ⌜bl⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ m. ¬ # bl < m → (∃ n. ⌜bl.take(m)⌝ ⊑ βf(σf(n), αf(n))), ∀ m. ¬ # al < m → (∃ n. ⌜al.take(m)⌝ ⊑ αf(n)), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ¬ # sl < # sl → (∃ n. ⌜sl.take(# sl)⌝ ⊑ σf(n)) ⊦

Rule argument

Additional information

Used terms: # sl

Proof step 41 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 42:

  # sl = # al + 1, lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , ⌜bl⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ⌜sl⌝ ⊑ σf(n), ∀ m. ¬ # al < m → (∃ n. ⌜al.take(m)⌝ ⊑ αf(n)), ∀ m. ¬ # bl < m → (∃ n. ⌜bl.take(m)⌝ ⊑ βf(σf(n), αf(n))), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x = n → x.take(n) = x; (s) ⊦ ¬ n < n; (s)

Proof step 42 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 43:

  # sl = # al + 1, lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , ⌜bl⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ⌜sl⌝ ⊑ σf(n), ∀ m. ¬ # al < m → (∃ n. ⌜al.take(m)⌝ ⊑ αf(n)), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ¬ # bl < # bl → (∃ n. ⌜bl.take(# bl)⌝ ⊑ βf(σf(n), αf(n))) ⊦

Rule argument

Additional information

Used terms: # bl

Proof step 43 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 44:

  # sl = # al + 1, lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , ⌜bl⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), ⌜bl⌝ ⊑ βf(σf(n0), αf(n0)), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ⌜sl⌝ ⊑ σf(n), ∀ m. ¬ # al < m → (∃ n. ⌜al.take(m)⌝ ⊑ αf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x = n → x.take(n) = x; (s) ⊦ ¬ n < n; (s)

Proof step 44 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 45:

  # sl = # al + 1, lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , ⌜bl⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), ⌜bl⌝ ⊑ βf(σf(n0), αf(n0)), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ⌜sl⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ¬ # al < # al → (∃ n. ⌜al.take(# al)⌝ ⊑ αf(n)) ⊦

Rule argument

Additional information

Used terms: # al

Proof step 45 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 46:

  # sl = # al + 1, lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , ⌜bl⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), ⌜al⌝ ⊑ αf(n1), ⌜bl⌝ ⊑ βf(σf(n0), αf(n0)), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ⌜sl⌝ ⊑ σf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x = n → x.take(n) = x; (s) ⊦ ¬ n < n; (s)

Proof step 46 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 47:

  # sl = # al + 1, lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , ⌜bl⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), ⌜al⌝ ⊑ αf(n1), ⌜bl⌝ ⊑ βf(σf(n0), αf(n0)), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ⌜sl⌝ ⊑ σf(n), σf(n) ⊑ ⌜sl⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used terms: n

Proof step 47 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 48:

  # sl = # al + 1, lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , ⌜sl⌝ = σf(n), ⌜bl⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), ⌜al⌝ ⊑ αf(n1), ⌜bl⌝ ⊑ βf(σf(n0), αf(n0)), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x0 = # x1 → (x0 ⊑ x1 ↔ x0 = x1); (s) ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s) ⊦ σ1 ⊑ σ2 → (σ2 ⊑ σ1 ↔ σ1 = σ2); (as)

Proof step 48 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 49:

  # sl = # al + 1, lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , ⌜sl⌝ = σf(n), ⌜bl⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), ⌜al⌝ ⊑ αf(n1), ⌜bl⌝ ⊑ βf(σf(n0), αf(n0)), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), αf(n) ⊑ ⌜al⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Used terms: n

Proof step 49 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 50:

  # sl = # al + 1, lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , ⌜sl⌝ = σf(n), ⌜bl⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), ⌜al⌝ ⊑ αf(n1), ⌜bl⌝ ⊑ βf(σf(n0), αf(n0)), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), αf(n) ⊑ ⌜al⌝ , αf(n) ⊑ αf(n1), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: leq-trans: ⊦ σ ⊑ σ1 ∧ σ1 ⊑ σ2 → σ ⊑ σ2; (forward) from specification stream

Proof step 50 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 51:

  # sl = # al + 1, # σf(n) = # αf(n) + ⌜1⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , ⌜sl⌝ = σf(n), ⌜bl⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), ⌜al⌝ ⊑ αf(n1), ⌜bl⌝ ⊑ βf(σf(n0), αf(n0)), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), αf(n) ⊑ αf(n1), αf(n) ⊑ ⌜al⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Used terms: n

Proof step 51 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 52:

  # sl = # al + 1, ⌜# sl⌝ = # αf(n) + ⌜1⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , ⌜sl⌝ = σf(n), ⌜bl⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), ⌜al⌝ ⊑ αf(n1), ⌜bl⌝ ⊑ βf(σf(n0), αf(n0)), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), αf(n) ⊑ αf(n1), αf(n) ⊑ ⌜al⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s)

Proof step 52 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 53:

  ⌜# sl⌝ = # αf(n) + ⌜1⌝ , # sl = # al + 1, # σf(n1) = # αf(n1) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , ⌜sl⌝ = σf(n), ⌜bl⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), ⌜al⌝ ⊑ αf(n1), ⌜bl⌝ ⊑ βf(σf(n0), αf(n0)), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), αf(n) ⊑ ⌜al⌝ , αf(n) ⊑ αf(n1), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used terms: n1

Proof step 53 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 54:

  # σf(n1) = # αf(n1) + ⌜1⌝ , # sl = # al + 1, ⌜# sl⌝ = # αf(n) + ⌜1⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , ⌜sl⌝ = σf(n), ⌜bl⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), ⌜al⌝ ⊑ αf(n1), ⌜bl⌝ ⊑ βf(σf(n0), αf(n0)), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), αf(n) ⊑ αf(n1), αf(n) ⊑ ⌜al⌝ , αf(n1) ⊑ ⌜al⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Used terms: n1

Proof step 54 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 55:

  # sl = # al + 1, ⌜# sl⌝ = # αf(n) + ⌜1⌝ , # σf(n1) = ⌜# al⌝ + ⌜1⌝ , ⌜al⌝ = αf(n1), lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , ⌜sl⌝ = σf(n), ⌜bl⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), ⌜bl⌝ ⊑ βf(σf(n0), αf(n0)), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), αf(n) ⊑ ⌜al⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x0 = # x1 → (x0 ⊑ x1 ↔ x0 = x1); (s) ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s) leq-trans: ⊦ σ ⊑ σ1 ∧ σ1 ⊑ σ2 → σ ⊑ σ2; (forward) from specification stream ⊦ σ1 ⊑ σ2 → (σ2 ⊑ σ1 ↔ σ1 = σ2); (as)

Proof step 55 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 56:

  # σf(n1) = ⌜# al⌝ + ⌜1⌝ , ⌜# sl⌝ = # αf(n) + ⌜1⌝ , # sl = # al + 1, ⌜al⌝ = αf(n1), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , ⌜sl⌝ = σf(n), ⌜bl⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), ⌜bl⌝ ⊑ βf(σf(n0), αf(n0)), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), αf(n) ⊑ ⌜al⌝ , σf(n1) ⊑ ⌜sl⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used terms: n1

Proof step 56 from proof conttr-thm (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 57:

  # sl = # al + 1, ⌜# sl⌝ = # αf(n) + ⌜1⌝ , # σf(n1) = ⌜# al⌝ + ⌜1⌝ , ⌜al⌝ = αf(n1), lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , ⌜sl⌝ = σf(n), σf(n1) = ⌜sl⌝ , ⌜bl⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), ⌜bl⌝ ⊑ βf(σf(n0), αf(n0)), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), αf(n) ⊑ ⌜al⌝ , σf(n1) ⊑ ⌜sl⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

• Proof step 75:

  # sl = # al + 1, ⌜# sl⌝ = # αf(n) + ⌜1⌝ , # σf(n1) = ⌜# al⌝ + ⌜1⌝ , ⌜al⌝ = αf(n1), lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , ⌜sl⌝ = σf(n), ⌜bl⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), σf(n1) ≠ ⌜sl⌝ , ⌜bl⌝ ⊑ βf(σf(n0), αf(n0)), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), αf(n) ⊑ ⌜al⌝ , σf(n1) ⊑ ⌜sl⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Used formulas: σf(n1) = ⌜sl⌝

Proof step 57 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 58:

  # sl = # al + 1, ⌜# sl⌝ = # αf(n) + ⌜1⌝ , ⌜al⌝ = αf(n1), lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , ⌜sl⌝ = σf(n), σf(n1) = ⌜sl⌝ , ⌜bl⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), ⌜bl⌝ ⊑ βf(σf(n0), αf(n0)), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), αf(n) ⊑ ⌜al⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x0 = # x1 → (x0 ⊑ x1 ↔ x0 = x1); (s) ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s) add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜n⌝ + ⌜m⌝ = ⌜n0⌝ ↔ n + m = n0; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s)

Proof step 58 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 59:

  ⌜# sl⌝ = # αf(n) + ⌜1⌝ , # sl = # al + 1, ⌜al⌝ = αf(n1), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , ⌜sl⌝ = σf(n), σf(n1) = ⌜sl⌝ , ⌜bl⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), ⌜bl⌝ ⊑ βf(σf(n0), αf(n0)), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), αf(n) ⊑ ⌜al⌝ , βf(σf(n1), αf(n1)) ⊑ ⌜bl⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used terms: n1

Proof step 59 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 60:

  # sl = # al + 1, ⌜# sl⌝ = # αf(n) + ⌜1⌝ , ⌜al⌝ = αf(n1), lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , ⌜sl⌝ = σf(n), σf(n1) = ⌜sl⌝ , ⌜bl⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), ⌜bl⌝ ⊑ βf(σf(n0), αf(n0)), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), αf(n) ⊑ ⌜al⌝ , βf(⌜sl⌝ , ⌜al⌝ ) ⊑ ⌜bl⌝ , βf(⌜sl⌝ , ⌜al⌝ ) ⊑ βf(σf(n0), αf(n0)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: leq-trans: ⊦ σ ⊑ σ1 ∧ σ1 ⊑ σ2 → σ ⊑ σ2; (forward) from specification stream

Proof step 60 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 61:

  ⌜# sl⌝ = # αf(n) + ⌜1⌝ , # sl = # al + 1, ⌜al⌝ = αf(n1), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , ⌜sl⌝ = σf(n), σf(n1) = ⌜sl⌝ , ⌜bl⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), ⌜bl⌝ ⊑ βf(σf(n0), αf(n0)), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), αf(n) ⊑ ⌜al⌝ , βf(⌜sl⌝ , ⌜al⌝ ) ⊑ βf(σf(n0), αf(n0)), βf(⌜sl⌝ , ⌜al⌝ ) ⊑ ⌜bl⌝ , βf(σf(n0), αf(n0)) ⊑ ⌜bl⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used terms: n0

Proof step 61 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 62:

  ⌜# sl⌝ = # αf(n) + ⌜1⌝ , # sl = # al + 1, ⌜al⌝ = αf(n1), ⌜bl⌝ = βf(σf(n0), αf(n0)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , ⌜sl⌝ = σf(n), σf(n1) = ⌜sl⌝ , ⌜bl⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), αf(n) ⊑ ⌜al⌝ , βf(⌜sl⌝ , ⌜al⌝ ) ⊑ ⌜bl⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x0 = # x1 → (x0 ⊑ x1 ↔ x0 = x1); (s) ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s) leq-trans: ⊦ σ ⊑ σ1 ∧ σ1 ⊑ σ2 → σ ⊑ σ2; (forward) from specification stream ⊦ σ1 ⊑ σ2 → (σ2 ⊑ σ1 ↔ σ1 = σ2); (as)

Proof step 62 from proof conttr-thm (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 63:

  ⊦ σ = σ0 ↔ σ ⊑ σ0 ∧ σ0 ⊑ σ

• Proof step 64:

  # sl = # al + 1, ⌜# sl⌝ = # αf(n) + ⌜1⌝ , ⌜al⌝ = αf(n1), ⌜bl⌝ = βf(σf(n0), αf(n0)), lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , ⌜sl⌝ = σf(n), σf(n1) = ⌜sl⌝ , cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), αf(n) ⊑ ⌜al⌝ , βf(⌜sl⌝ , ⌜al⌝ ) ⊑ ⌜bl⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ , ¬ ⌜bl⌝ ⊑ βf(⌜sl⌝ , ⌜al⌝ ) ∨ ¬ βf(⌜sl⌝ , ⌜al⌝ ) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used terms: ⌜bl⌝ , βf(⌜sl⌝ , ⌜al⌝ )

Proof step 63 from proof conttr-thm (previous step) (back to tree)

A spec-lemma

Proof step 64 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 65:

  ⌜# sl⌝ = # αf(n) + ⌜1⌝ , # sl = # al + 1, ⌜al⌝ = αf(n1), ⌜bl⌝ = βf(σf(n0), αf(n0)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , ⌜sl⌝ = σf(n), σf(n1) = ⌜sl⌝ , cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), αf(n) ⊑ ⌜al⌝ , βf(⌜sl⌝ , ⌜al⌝ ) ⊑ ⌜bl⌝ , ¬ ⌜bl⌝ ⊑ βf(⌜sl⌝ , ⌜al⌝ ), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Proof step 65 from proof conttr-thm (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 66:

  ⊦ cont(βf) ↔ mon(βf) ∧ pbd(βf)

• Proof step 67:

  # sl = # al + 1, ⌜# sl⌝ = # αf(n) + ⌜1⌝ , ⌜al⌝ = αf(n1), ⌜bl⌝ = βf(σf(n0), αf(n0)), lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , ⌜sl⌝ = σf(n), σf(n1) = ⌜sl⌝ , mon(⌜αf⌝ ), mon(⌜σf⌝ ), mon(βf), pbd(βf), αf(n) ⊑ ⌜al⌝ , βf(⌜sl⌝ , ⌜al⌝ ) ⊑ ⌜bl⌝ , ¬ ⌜bl⌝ ⊑ βf(⌜sl⌝ , ⌜al⌝ ), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Used terms: βf

Proof step 66 from proof conttr-thm (previous step) (back to tree)

Lemma conttr-def

Proof step 67 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 68:

  ⌜# sl⌝ = # αf(n) + ⌜1⌝ , # sl = # al + 1, ⌜al⌝ = αf(n1), ⌜bl⌝ = βf(σf(n0), αf(n0)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , ⌜sl⌝ = σf(n), σf(n1) = ⌜sl⌝ , mon(⌜σf⌝ ), mon(⌜αf⌝ ), pbd(βf), αf(n) ⊑ ⌜al⌝ , βf(⌜sl⌝ , ⌜al⌝ ) ⊑ ⌜bl⌝ , ¬ ⌜bl⌝ ⊑ βf(⌜sl⌝ , ⌜al⌝ ), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used simplifier rules: montr-def: ⊦ mon(βf) ↔ (∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2));

Proof step 68 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 69:

  # sl = # al + 1, ⌜# sl⌝ = # αf(n) + ⌜1⌝ , ⌜al⌝ = αf(n1), ⌜bl⌝ = βf(σf(n0), αf(n0)), lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , ⌜sl⌝ = σf(n), σf(n1) = ⌜sl⌝ , mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), αf(n) ⊑ ⌜al⌝ , βf(⌜sl⌝ , ⌜al⌝ ) ⊑ ⌜bl⌝ , ¬ ⌜bl⌝ ⊑ βf(⌜sl⌝ , ⌜al⌝ ), # σf(n0) = # αf(n0) + ⌜1⌝ ∧ # σf(n1) = # αf(n1) + ⌜1⌝ ∧ σf(n0) ⊑ σf(n1) ∧ αf(n0) ⊑ αf(n1) → βf(σf(n0), αf(n0)) ⊑ βf(σf(n1), αf(n1)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used terms: σf(n0), αf(n0), σf(n1), αf(n1)

Proof step 69 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 70:

  # sl = # al + 1, ⌜# sl⌝ = # αf(n) + ⌜1⌝ , ⌜al⌝ = αf(n1), ⌜bl⌝ = βf(σf(n0), αf(n0)), lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , ⌜sl⌝ = σf(n), σf(n1) = ⌜sl⌝ , mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), αf(n) ⊑ ⌜al⌝ , βf(⌜sl⌝ , ⌜al⌝ ) ⊑ ⌜bl⌝ , ¬ ⌜bl⌝ ⊑ βf(⌜sl⌝ , ⌜al⌝ ), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦ # σf(n0) = # αf(n0) + ⌜1⌝ ∧ σf(n0) ⊑ ⌜sl⌝ ∧ αf(n0) ⊑ ⌜al⌝

Rule argument

Additional information

Used simplifier rules: add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜n⌝ + ⌜m⌝ = ⌜n0⌝ ↔ n + m = n0; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s)

Proof step 70 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 71:

  ⌜# sl⌝ = # αf(n) + ⌜1⌝ , # sl = # al + 1, ⌜al⌝ = αf(n1), ⌜bl⌝ = βf(σf(n0), αf(n0)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , ⌜sl⌝ = σf(n), σf(n1) = ⌜sl⌝ , mon(⌜σf⌝ ), mon(⌜αf⌝ ), pbd(βf), αf(n) ⊑ ⌜al⌝ , αf(n0) ⊑ ⌜al⌝ , βf(⌜sl⌝ , ⌜al⌝ ) ⊑ ⌜bl⌝ , ¬ ⌜bl⌝ ⊑ βf(⌜sl⌝ , ⌜al⌝ ), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦ # σf(n0) = # αf(n0) + ⌜1⌝ ∧ σf(n0) ⊑ ⌜sl⌝ ∧ αf(n0) ⊑ ⌜al⌝

Rule argument

Additional information

Used terms: n0

Proof step 71 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 72:

  # sl = # al + 1, ⌜# sl⌝ = # αf(n) + ⌜1⌝ , ⌜al⌝ = αf(n1), ⌜bl⌝ = βf(σf(n0), αf(n0)), lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , ⌜sl⌝ = σf(n), σf(n1) = ⌜sl⌝ , mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), αf(n0) ⊑ ⌜al⌝ , αf(n) ⊑ ⌜al⌝ , βf(⌜sl⌝ , ⌜al⌝ ) ⊑ ⌜bl⌝ , ¬ ⌜bl⌝ ⊑ βf(⌜sl⌝ , ⌜al⌝ ), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦ # σf(n0) = # αf(n0) + ⌜1⌝ ∧ σf(n0) ⊑ ⌜sl⌝

Rule argument

Additional information

Proof step 72 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 73:

  ⌜# sl⌝ = # αf(n) + ⌜1⌝ , # sl = # al + 1, ⌜al⌝ = αf(n1), ⌜bl⌝ = βf(σf(n0), αf(n0)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , ⌜sl⌝ = σf(n), σf(n1) = ⌜sl⌝ , mon(⌜σf⌝ ), mon(⌜αf⌝ ), pbd(βf), αf(n) ⊑ ⌜al⌝ , αf(n0) ⊑ ⌜al⌝ , βf(⌜sl⌝ , ⌜al⌝ ) ⊑ ⌜bl⌝ , σf(n0) ⊑ ⌜sl⌝ , ¬ ⌜bl⌝ ⊑ βf(⌜sl⌝ , ⌜al⌝ ), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦ # σf(n0) = # αf(n0) + ⌜1⌝ ∧ σf(n0) ⊑ ⌜sl⌝

Rule argument

Additional information

Used terms: n0

Proof step 73 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 74:

  # sl = # al + 1, ⌜# sl⌝ = # αf(n) + ⌜1⌝ , ⌜al⌝ = αf(n1), ⌜bl⌝ = βf(σf(n0), αf(n0)), lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , ⌜sl⌝ = σf(n), σf(n1) = ⌜sl⌝ , # σf(n0) ≠ # αf(n0) + ⌜1⌝ , mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), αf(n0) ⊑ ⌜al⌝ , αf(n) ⊑ ⌜al⌝ , βf(⌜sl⌝ , ⌜al⌝ ) ⊑ ⌜bl⌝ , σf(n0) ⊑ ⌜sl⌝ , ¬ ⌜bl⌝ ⊑ βf(⌜sl⌝ , ⌜al⌝ ), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Proof step 74 from proof conttr-thm (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n0

Proof step 75 from proof conttr-thm (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 76:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

Rule argument

Additional information

Used simplifier rules: ⊦ σ0 + σ = ⌜z⌝ ↔ finp(σ0) ∧ finp(σ) ∧ σ0.l + σ.l = z; (as) ⊦ finp(σ) ↔ (∃ x. ⌜x⌝ = σ); (as) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s) ⊦ ⌜n⌝ + ⌜m⌝ = ⌜n0⌝ ↔ n + m = n0; (s) add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) ⊦ #(x + y) = # x + # y; (s) ⊦ ⌜x⌝ + ⌜y⌝ = ⌜z⌝ ↔ x + y = z; (s) ⊦ x = x + y ↔ y = []; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ ⌜x⌝ .l = x; (s) ⊦ ⌜x + y⌝ = ⌜x⌝ + ⌜y⌝ ; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ m = m + n ↔ n = 0; (s) ⊦ # x = 0 ↔ x = []; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: σf(n1), ⌜sl⌝

Proof step 76 from proof conttr-thm (previous step) (back to tree)

A spec-lemma

Proof step 77 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 78:

  # lim(⌜σf⌝ ) = ∞, lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , ⌜bl⌝ ≠ βf(lim(⌜σf⌝ ), ⌜af⌝ ), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ¬ # bl < m → (∃ n. ⌜bl.take(m)⌝ ⊑ βf(σf(n), αf(n))), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ lim(⌜σf⌝ ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ ∞ < n∞; (s) ⊦ ⌜nf⌝ .take(n) = nf.take(n); (s) ⊦ # ⌜nf⌝ = ∞; (s) add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜m⌝ + ∞ = ∞; (s)

Proof step 78 from proof conttr-thm (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 79:

  # lim(⌜σf⌝ ) = ∞, lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , ⌜bl⌝ ≠ βf(lim(⌜σf⌝ ), ⌜af⌝ ), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n), ∀ m. ¬ # bl < m → (∃ n. ⌜bl.take(m)⌝ ⊑ βf(σf(n), αf(n))), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: Stream[A] freely generated by ⌝, ⌝; ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s) ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 79 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 80:

  lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , ⌜bl⌝ ≠ βf(⌜sf⌝ , ⌜af⌝ ), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ m. ¬ # bl < m → (∃ n. ⌜bl.take(m)⌝ ⊑ βf(σf(n), αf(n))), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ .take(n) = nf.take(n); (s) ⊦ # ⌜nf⌝ = ∞; (s)

Proof step 80 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 81:

  lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , ⌜bl⌝ ≠ βf(⌜sf⌝ , ⌜af⌝ ), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ¬ # bl < # bl → (∃ n. ⌜bl.take(# bl)⌝ ⊑ βf(σf(n), αf(n))) ⊦

Rule argument

Additional information

Used terms: # bl

Proof step 81 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 82:

  lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , ⌜bl⌝ ≠ βf(⌜sf⌝ , ⌜af⌝ ), ⌜bl⌝ ⊑ βf(σf(n), αf(n)), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x = n → x.take(n) = x; (s) ⊦ ¬ n < n; (s)

Proof step 82 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 83:

  lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , ⌜bl⌝ ≠ βf(⌜sf⌝ , ⌜af⌝ ), ⌜bl⌝ ⊑ βf(σf(n), αf(n)), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ ⊦

Rule argument

Additional information

Used terms: n

Proof step 83 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 84:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , ⌜bl⌝ ≠ βf(⌜sf⌝ , ⌜af⌝ ), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x0 = # x1 → (x0 ⊑ x1 ↔ x0 = x1); (s) ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s) ⊦ σ1 ⊑ σ2 → (σ2 ⊑ σ1 ↔ σ1 = σ2); (as)

Proof step 84 from proof conttr-thm (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 85:

  ⊦ cont(βf) ↔ mon(βf) ∧ pbd(βf)

• Proof step 86:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , ⌜bl⌝ ≠ βf(⌜sf⌝ , ⌜af⌝ ), mon(⌜σf⌝ ), mon(⌜αf⌝ ), mon(βf), pbd(βf), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used terms: βf

Proof step 85 from proof conttr-thm (previous step) (back to tree)

Lemma conttr-def

Proof step 86 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 87:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , ⌜bl⌝ ≠ βf(⌜sf⌝ , ⌜af⌝ ), mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used simplifier rules: montr-def: ⊦ mon(βf) ↔ (∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2));

Proof step 87 from proof conttr-thm (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 88:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , ⌜bl⌝ ≠ βf(⌜sf⌝ , ⌜af⌝ ), mon(⌜σf⌝ ), mon(⌜αf⌝ ), pbd(βf), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

• Proof step 130:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , ⌜bl⌝ ≠ βf(⌜sf⌝ , ⌜af⌝ ), mon(⌜σf⌝ ), mon(⌜αf⌝ ), pbd(βf), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used simplifier rules: Stream[A] freely generated by ⌝, ⌝;

Proof step 88 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 89:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, mon(⌜σf⌝ ), mon(⌜αf⌝ ), pbd(βf), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ = ⌜x0⌝ ↔ x = x0; (s)

Proof step 89 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 90:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), # σf(n) = # αf(n) + ⌜1⌝ ∧ # ⌜sf⌝ = # ⌜af⌝ + ⌜1⌝ ∧ σf(n) ⊑ ⌜sf⌝ ∧ αf(n) ⊑ ⌜af⌝ → βf(σf(n), αf(n)) ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used terms: σf(n), αf(n), ⌜sf⌝ , ⌜af⌝

Proof step 90 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 91:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), # σf(n) = # αf(n) + ⌜1⌝ ∧ σf(n) ⊑ ⌜sf⌝ ∧ αf(n) ⊑ ⌜af⌝ → bl ⊑ bl0, ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s) add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜m⌝ + ∞ = ∞; (s) ⊦ # ⌜nf⌝ = ∞; (s)

Proof step 91 from proof conttr-thm (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 92:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, mon(⌜σf⌝ ), mon(⌜αf⌝ ), pbd(βf), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦ # σf(n) = # αf(n) + ⌜1⌝ ∧ σf(n) ⊑ ⌜sf⌝ ∧ αf(n) ⊑ ⌜af⌝

• Proof step 96:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, bl ⊑ bl0, mon(⌜σf⌝ ), mon(⌜αf⌝ ), pbd(βf), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Proof step 92 from proof conttr-thm (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 93:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , # σf(n) ≠ # αf(n) + ⌜1⌝ , bl ≠ bl0, mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

• Proof step 94:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), ¬ σf(n) ⊑ ⌜sf⌝ , ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

• Proof step 95:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), ¬ αf(n) ⊑ ⌜af⌝ , ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Proof step 93 from proof conttr-thm (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n

Proof step 94 from proof conttr-thm (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n

Proof step 95 from proof conttr-thm (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n

Proof step 96 from proof conttr-thm (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 97:

  ⊦ x ⊑ y → (y ⊑ x ↔ x = y)

• Proof step 98:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, bl ⊑ bl0, mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), ¬ bl0 ⊑ bl, ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used terms: bl, bl0

Proof step 97 from proof conttr-thm (previous step) (back to tree)

A spec-lemma

Proof step 98 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 99:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, bl ⊑ bl0, mon(⌜σf⌝ ), mon(⌜αf⌝ ), pbd(βf), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ x ⊑ y → (y ⊑ x ↔ x = y); (s)

Proof step 99 from proof conttr-thm (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 100:

  ⊦ pbd(βf) ↔ (∀ sf, af, bl. (∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝ )

• Proof step 101:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, bl ⊑ bl0, mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ sf, af, bl. (∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used terms: βf

Proof step 100 from proof conttr-thm (previous step) (back to tree)

Lemma pbdtr-def

Proof step 101 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 102:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, bl ⊑ bl0, mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , (∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used terms: sf, af, bl

Proof step 102 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 103:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, bl ⊑ bl0, mon(⌜σf⌝ ), mon(⌜αf⌝ ), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ x ⊑ y → (y ⊑ x ↔ x = y); (s) ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s)

Proof step 103 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 104:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, bl ⊑ bl0, mon(⌜αf⌝ ), mon(⌜σf⌝ ), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∃ n. ⌜af.take(n0)⌝ ⊑ αf(n), ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used terms: n0

Proof step 104 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 105:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, ⌜af.take(n0)⌝ ⊑ αf(n1), bl ⊑ bl0, mon(⌜σf⌝ ), mon(⌜αf⌝ ), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Proof step 105 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 106:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, ⌜af.take(n0)⌝ ⊑ αf(n1), bl ⊑ bl0, mon(⌜αf⌝ ), mon(⌜σf⌝ ), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∃ n. ⌜sf.take(n0 + 1)⌝ ⊑ σf(n), ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used terms: n0 + 1

Proof step 106 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 107:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, ⌜af.take(n0)⌝ ⊑ αf(n1), bl ⊑ bl0, mon(⌜σf⌝ ), mon(⌜αf⌝ ), ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Proof step 107 from proof conttr-thm (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 108:

  mon(σσ), m ≤ n, ⌜n⌝ < # σσ ⊦ σσ(m) ⊑ σσ(n)

• Proof step 109:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, ⌜af.take(n0)⌝ ⊑ αf(n1), bl ⊑ bl0, mon(⌜αf⌝ ), mon(⌜σf⌝ ), ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦ n2 ≤ n1 ∧ ⌜n1⌝ < # ⌜σf⌝

• Proof step 124:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, ⌜af.take(n0)⌝ ⊑ αf(n1), bl ⊑ bl0, mon(⌜αf⌝ ), mon(⌜σf⌝ ), ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), ⌜σf⌝ (n2) ⊑ ⌜σf⌝ (n1), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used terms: ⌜σf⌝ , n2, n1

Proof step 108 from proof conttr-thm (previous step) (back to tree)

A spec-lemma

Proof step 109 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 110:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, ⌜af.take(n0)⌝ ⊑ αf(n1), bl ⊑ bl0, mon(⌜αf⌝ ), mon(⌜σf⌝ ), n1 < n2, ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ ⌜m⌝ < ∞; (s) ⊦ # ⌜nf⌝ = ∞; (s)

Proof step 110 from proof conttr-thm (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 111:

  mon(σσ), m ≤ n, ⌜n⌝ < # σσ ⊦ σσ(m) ⊑ σσ(n)

• Proof step 112:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, ⌜af.take(n0)⌝ ⊑ αf(n1), bl ⊑ bl0, mon(⌜σf⌝ ), mon(⌜αf⌝ ), n1 < n2, ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), ⌜σf⌝ (n1) ⊑ ⌜σf⌝ (n2), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s) ⊦ # ⌜nf⌝ = ∞; (s) ⊦ ⌜m⌝ < ∞; (s) Used terms: ⌜σf⌝ , n1, n2

Proof step 111 from proof conttr-thm (previous step) (back to tree)

A spec-lemma

Proof step 112 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 113:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, ⌜af.take(n0)⌝ ⊑ αf(n1), bl ⊑ bl0, mon(⌜αf⌝ ), mon(⌜σf⌝ ), n1 < n2, ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), σf(n1) ⊑ σf(n2), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ (n) = nf(n); (s)

Proof step 113 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 114:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, ⌜af.take(n0)⌝ ⊑ αf(n1), bl ⊑ bl0, mon(⌜σf⌝ ), mon(⌜αf⌝ ), n1 < n2, ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), σf(n1) ⊑ σf(n2), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , # ⌜sf.take(n0 + 1)⌝ = # ⌜af.take(n0)⌝ + ⌜1⌝ ∧ # σf(n2) = # αf(n2) + ⌜1⌝ ∧ ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2) ∧ ⌜af.take(n0)⌝ ⊑ αf(n2) → βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ βf(σf(n2), αf(n2)), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used terms: ⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ , σf(n2), αf(n2)

Proof step 114 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 115:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, ⌜af.take(n0)⌝ ⊑ αf(n1), bl ⊑ bl0, mon(⌜αf⌝ ), mon(⌜σf⌝ ), n1 < n2, ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), σf(n1) ⊑ σf(n2), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , # σf(n2) = # αf(n2) + ⌜1⌝ ∧ ⌜af.take(n0)⌝ ⊑ αf(n2) → βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ βf(σf(n2), αf(n2)), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used simplifier rules: add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜n⌝ = ⌜n0⌝ ↔ n = n0; (s) ⊦ n∞ + ⌜0⌝ = n∞; (s) ⊦ m ≤ n → (⌜(* n)⌝ + n∞1 = ⌜(* m)⌝ + n∞2 ↔ ⌜n - m⌝ + n∞1 = n∞2); (s) ⊦ # ⌜nf.take(n)⌝ = ⌜n⌝ ; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)

Proof step 115 from proof conttr-thm (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 116:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, ⌜af.take(n0)⌝ ⊑ αf(n1), bl ⊑ bl0, mon(⌜σf⌝ ), mon(⌜αf⌝ ), n1 < n2, ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), σf(n1) ⊑ σf(n2), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦ # σf(n2) = # αf(n2) + ⌜1⌝ ∧ ⌜af.take(n0)⌝ ⊑ αf(n2)

• Proof step 123:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, ⌜af.take(n0)⌝ ⊑ αf(n1), bl ⊑ bl0, mon(⌜σf⌝ ), mon(⌜αf⌝ ), n1 < n2, ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ βf(σf(n2), αf(n2)), σf(n1) ⊑ σf(n2), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Proof step 116 from proof conttr-thm (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 117:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , # σf(n2) ≠ # αf(n2) + ⌜1⌝ , bl ≠ bl0, ⌜af.take(n0)⌝ ⊑ αf(n1), bl ⊑ bl0, mon(⌜αf⌝ ), mon(⌜σf⌝ ), n1 < n2, ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), σf(n1) ⊑ σf(n2), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

• Proof step 118:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, ⌜af.take(n0)⌝ ⊑ αf(n1), bl ⊑ bl0, mon(⌜αf⌝ ), mon(⌜σf⌝ ), n1 < n2, ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), σf(n1) ⊑ σf(n2), ¬ ⌜af.take(n0)⌝ ⊑ αf(n2), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Proof step 117 from proof conttr-thm (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n2

Proof step 118 from proof conttr-thm (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 119:

  ⊦ σ ⊑ σ1 ∧ σ1 ⊑ σ2 → σ ⊑ σ2

• Proof step 120:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, ⌜af.take(n0)⌝ ⊑ αf(n1), bl ⊑ bl0, mon(⌜σf⌝ ), mon(⌜αf⌝ ), n1 < n2, ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), σf(n1) ⊑ σf(n2), ¬ ⌜af.take(n0)⌝ ⊑ αf(n2), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦ ⌜af.take(n0)⌝ ⊑ αf(n1) ∧ αf(n1) ⊑ αf(n2)

Rule argument

Additional information

Used terms: αf(n1), ⌜af.take(n0)⌝ , αf(n2)

Proof step 119 from proof conttr-thm (previous step) (back to tree)

A spec-lemma

Proof step 120 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 121:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, ⌜af.take(n0)⌝ ⊑ αf(n1), bl ⊑ bl0, mon(⌜αf⌝ ), mon(⌜σf⌝ ), n1 < n2, ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), σf(n1) ⊑ σf(n2), ¬ ⌜af.take(n0)⌝ ⊑ αf(n2), ¬ αf(n1) ⊑ αf(n2), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Proof step 121 from proof conttr-thm (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 122:

  mon(σσ), m ≤ n, ⌜n⌝ < # σσ ⊦ σσ(m) ⊑ σσ(n)

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ (n) = nf(n); (s) ⊦ m < n → m ≤ n; (s) ⊦ # ⌜nf⌝ = ∞; (s) ⊦ ⌜m⌝ < ∞; (s) Used terms: ⌜αf⌝ , n1, n2

Proof step 122 from proof conttr-thm (previous step) (back to tree)

A spec-lemma

Proof step 123 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: leq-trans: ⊦ σ ⊑ σ1 ∧ σ1 ⊑ σ2 → σ ⊑ σ2; (forward) from specification stream Used terms: n2

Proof step 124 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 125:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, ⌜af.take(n0)⌝ ⊑ αf(n1), bl ⊑ bl0, mon(⌜σf⌝ ), mon(⌜αf⌝ ), ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), ⌜sf.take(n0 + 1)⌝ ⊑ σf(n1), σf(n2) ⊑ σf(n1), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used simplifier rules: leq-trans: ⊦ σ ⊑ σ1 ∧ σ1 ⊑ σ2 → σ ⊑ σ2; (forward) from specification stream ⊦ ⌜nf⌝ (n) = nf(n); (s)

Proof step 125 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 126:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, ⌜af.take(n0)⌝ ⊑ αf(n1), bl ⊑ bl0, mon(⌜αf⌝ ), mon(⌜σf⌝ ), ⌜sf.take(n0 + 1)⌝ ⊑ σf(n1), ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), σf(n2) ⊑ σf(n1), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , # ⌜sf.take(n0 + 1)⌝ = # ⌜af.take(n0)⌝ + ⌜1⌝ ∧ # σf(n1) = # αf(n1) + ⌜1⌝ ∧ ⌜sf.take(n0 + 1)⌝ ⊑ σf(n1) ∧ ⌜af.take(n0)⌝ ⊑ αf(n1) → βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ βf(σf(n1), αf(n1)), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ ⊦

Rule argument

Additional information

Used terms: ⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ , σf(n1), αf(n1)

Proof step 126 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 127:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, ⌜af.take(n0)⌝ ⊑ αf(n1), bl ⊑ bl0, mon(⌜σf⌝ ), mon(⌜αf⌝ ), ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), ⌜sf.take(n0 + 1)⌝ ⊑ σf(n1), σf(n2) ⊑ σf(n1), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , # σf(n1) = # αf(n1) + ⌜1⌝ → βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ βf(σf(n1), αf(n1)), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜n⌝ = ⌜n0⌝ ↔ n = n0; (s) ⊦ n∞ + ⌜0⌝ = n∞; (s) ⊦ m ≤ n → (⌜(* n)⌝ + n∞1 = ⌜(* m)⌝ + n∞2 ↔ ⌜n - m⌝ + n∞1 = n∞2); (s) ⊦ # ⌜nf.take(n)⌝ = ⌜n⌝ ; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)

Proof step 127 from proof conttr-thm (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 128:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , # σf(n1) ≠ # αf(n1) + ⌜1⌝ , bl ≠ bl0, ⌜af.take(n0)⌝ ⊑ αf(n1), bl ⊑ bl0, mon(⌜αf⌝ ), mon(⌜σf⌝ ), ⌜sf.take(n0 + 1)⌝ ⊑ σf(n1), ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), σf(n2) ⊑ σf(n1), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ ⊦

• Proof step 129:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl ≠ bl0, ⌜af.take(n0)⌝ ⊑ αf(n1), bl ⊑ bl0, mon(⌜αf⌝ ), mon(⌜σf⌝ ), ⌜sf.take(n0 + 1)⌝ ⊑ σf(n1), ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ βf(σf(n1), αf(n1)), σf(n2) ⊑ σf(n1), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ ⊦

Rule argument

Additional information

Proof step 128 from proof conttr-thm (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n1

Proof step 129 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: leq-trans: ⊦ σ ⊑ σ1 ∧ σ1 ⊑ σ2 → σ ⊑ σ2; (forward) from specification stream Used terms: n1

Proof step 130 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 131:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , mon(⌜σf⌝ ), mon(⌜αf⌝ ), pbd(βf), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ ≠ ⌜nf⌝ ; (s)

Proof step 131 from proof conttr-thm (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 132:

  ⊦ pbd(βf) ↔ (∀ sf, af, bl. (∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝ )

• Proof step 133:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ sf, af, bl. (∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used terms: βf

Proof step 132 from proof conttr-thm (previous step) (back to tree)

Lemma pbdtr-def

Proof step 133 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 134:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , (∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used terms: sf, af, bl

Proof step 134 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 135:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , mon(⌜σf⌝ ), mon(⌜αf⌝ ), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ ⌜nf⌝ ⊑ ⌜x⌝ ; (s)

Proof step 135 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 136:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , mon(⌜αf⌝ ), mon(⌜σf⌝ ), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∃ n. ⌜af.take(n0)⌝ ⊑ αf(n), ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used terms: n0

Proof step 136 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 137:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , ⌜af.take(n0)⌝ ⊑ αf(n1), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Proof step 137 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 138:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , ⌜af.take(n0)⌝ ⊑ αf(n1), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∃ n. ⌜sf.take(n0 + 1)⌝ ⊑ σf(n), ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used terms: n0 + 1

Proof step 138 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 139:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , ⌜af.take(n0)⌝ ⊑ αf(n1), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Proof step 139 from proof conttr-thm (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 140:

  mon(σσ), m ≤ n, ⌜n⌝ < # σσ ⊦ σσ(m) ⊑ σσ(n)

• Proof step 141:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , ⌜af.take(n0)⌝ ⊑ αf(n1), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦ n1 ≤ n2 ∧ ⌜n2⌝ < # ⌜αf⌝

• Proof step 156:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , ⌜af.take(n0)⌝ ⊑ αf(n1), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), ⌜αf⌝ (n1) ⊑ ⌜αf⌝ (n2), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used terms: ⌜αf⌝ , n1, n2

Proof step 140 from proof conttr-thm (previous step) (back to tree)

A spec-lemma

Proof step 141 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 142:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , ⌜af.take(n0)⌝ ⊑ αf(n1), mon(⌜αf⌝ ), mon(⌜σf⌝ ), n2 < n1, ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ ⌜m⌝ < ∞; (s) ⊦ # ⌜nf⌝ = ∞; (s)

Proof step 142 from proof conttr-thm (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 143:

  mon(σσ), m ≤ n, ⌜n⌝ < # σσ ⊦ σσ(m) ⊑ σσ(n)

• Proof step 144:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , ⌜af.take(n0)⌝ ⊑ αf(n1), mon(⌜σf⌝ ), mon(⌜αf⌝ ), n2 < n1, ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), ⌜αf⌝ (n2) ⊑ ⌜αf⌝ (n1), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s) ⊦ # ⌜nf⌝ = ∞; (s) ⊦ ⌜m⌝ < ∞; (s) Used terms: ⌜αf⌝ , n2, n1

Proof step 143 from proof conttr-thm (previous step) (back to tree)

A spec-lemma

Proof step 144 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 145:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , ⌜af.take(n0)⌝ ⊑ αf(n1), mon(⌜αf⌝ ), mon(⌜σf⌝ ), n2 < n1, ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), αf(n2) ⊑ αf(n1), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ (n) = nf(n); (s)

Proof step 145 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 146:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , ⌜af.take(n0)⌝ ⊑ αf(n1), mon(⌜σf⌝ ), mon(⌜αf⌝ ), n2 < n1, ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), αf(n2) ⊑ αf(n1), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , # ⌜sf.take(n0 + 1)⌝ = # ⌜af.take(n0)⌝ + ⌜1⌝ ∧ # σf(n1) = # αf(n1) + ⌜1⌝ ∧ ⌜sf.take(n0 + 1)⌝ ⊑ σf(n1) ∧ ⌜af.take(n0)⌝ ⊑ αf(n1) → βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ βf(σf(n1), αf(n1)), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used terms: ⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ , σf(n1), αf(n1)

Proof step 146 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 147:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , ⌜af.take(n0)⌝ ⊑ αf(n1), mon(⌜αf⌝ ), mon(⌜σf⌝ ), n2 < n1, ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), αf(n2) ⊑ αf(n1), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , # σf(n1) = # αf(n1) + ⌜1⌝ ∧ ⌜sf.take(n0 + 1)⌝ ⊑ σf(n1) → βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ βf(σf(n1), αf(n1)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ ⊦

Rule argument

Additional information

Used simplifier rules: add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜n⌝ = ⌜n0⌝ ↔ n = n0; (s) ⊦ n∞ + ⌜0⌝ = n∞; (s) ⊦ m ≤ n → (⌜(* n)⌝ + n∞1 = ⌜(* m)⌝ + n∞2 ↔ ⌜n - m⌝ + n∞1 = n∞2); (s) ⊦ # ⌜nf.take(n)⌝ = ⌜n⌝ ; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)

Proof step 147 from proof conttr-thm (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 148:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , ⌜af.take(n0)⌝ ⊑ αf(n1), mon(⌜σf⌝ ), mon(⌜αf⌝ ), n2 < n1, ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), αf(n2) ⊑ αf(n1), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ ⊦ # σf(n1) = # αf(n1) + ⌜1⌝ ∧ ⌜sf.take(n0 + 1)⌝ ⊑ σf(n1)

• Proof step 155:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , ⌜af.take(n0)⌝ ⊑ αf(n1), mon(⌜σf⌝ ), mon(⌜αf⌝ ), n2 < n1, ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), αf(n2) ⊑ αf(n1), βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ βf(σf(n1), αf(n1)), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Proof step 148 from proof conttr-thm (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 149:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , # σf(n1) ≠ # αf(n1) + ⌜1⌝ , ⌜af.take(n0)⌝ ⊑ αf(n1), mon(⌜αf⌝ ), mon(⌜σf⌝ ), n2 < n1, ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), αf(n2) ⊑ αf(n1), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ ⊦

• Proof step 150:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , ⌜af.take(n0)⌝ ⊑ αf(n1), mon(⌜αf⌝ ), mon(⌜σf⌝ ), n2 < n1, ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), αf(n2) ⊑ αf(n1), ¬ ⌜sf.take(n0 + 1)⌝ ⊑ σf(n1), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ ⊦

Rule argument

Additional information

Proof step 149 from proof conttr-thm (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n1

Proof step 150 from proof conttr-thm (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 151:

  ⊦ σ ⊑ σ1 ∧ σ1 ⊑ σ2 → σ ⊑ σ2

• Proof step 152:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , ⌜af.take(n0)⌝ ⊑ αf(n1), mon(⌜σf⌝ ), mon(⌜αf⌝ ), n2 < n1, ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), αf(n2) ⊑ αf(n1), ¬ ⌜sf.take(n0 + 1)⌝ ⊑ σf(n1), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ ⊦ ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2) ∧ σf(n2) ⊑ σf(n1)

Rule argument

Additional information

Used terms: σf(n2), ⌜sf.take(n0 + 1)⌝ , σf(n1)

Proof step 151 from proof conttr-thm (previous step) (back to tree)

A spec-lemma

Proof step 152 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 153:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , ⌜af.take(n0)⌝ ⊑ αf(n1), mon(⌜αf⌝ ), mon(⌜σf⌝ ), n2 < n1, ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), αf(n2) ⊑ αf(n1), ¬ ⌜sf.take(n0 + 1)⌝ ⊑ σf(n1), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ¬ σf(n2) ⊑ σf(n1), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ ⊦

Rule argument

Additional information

Proof step 153 from proof conttr-thm (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 154:

  mon(σσ), m ≤ n, ⌜n⌝ < # σσ ⊦ σσ(m) ⊑ σσ(n)

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ (n) = nf(n); (s) ⊦ m < n → m ≤ n; (s) ⊦ # ⌜nf⌝ = ∞; (s) ⊦ ⌜m⌝ < ∞; (s) Used terms: ⌜σf⌝ , n2, n1

Proof step 154 from proof conttr-thm (previous step) (back to tree)

A spec-lemma

Proof step 155 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: leq-trans: ⊦ σ ⊑ σ1 ∧ σ1 ⊑ σ2 → σ ⊑ σ2; (forward) from specification stream Used terms: n1

Proof step 156 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 157:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , ⌜af.take(n0)⌝ ⊑ αf(n1), ⌜af.take(n0)⌝ ⊑ αf(n2), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), αf(n1) ⊑ αf(n2), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used simplifier rules: leq-trans: ⊦ σ ⊑ σ1 ∧ σ1 ⊑ σ2 → σ ⊑ σ2; (forward) from specification stream ⊦ ⌜nf⌝ (n) = nf(n); (s)

Proof step 157 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 158:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , ⌜af.take(n0)⌝ ⊑ αf(n2), ⌜af.take(n0)⌝ ⊑ αf(n1), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), αf(n1) ⊑ αf(n2), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , # ⌜sf.take(n0 + 1)⌝ = # ⌜af.take(n0)⌝ + ⌜1⌝ ∧ # σf(n2) = # αf(n2) + ⌜1⌝ ∧ ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2) ∧ ⌜af.take(n0)⌝ ⊑ αf(n2) → βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ βf(σf(n2), αf(n2)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ ⊦

Rule argument

Additional information

Used terms: ⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ , σf(n2), αf(n2)

Proof step 158 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 159:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , ⌜af.take(n0)⌝ ⊑ αf(n1), ⌜af.take(n0)⌝ ⊑ αf(n2), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), αf(n1) ⊑ αf(n2), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , # σf(n2) = # αf(n2) + ⌜1⌝ → βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ βf(σf(n2), αf(n2)), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜n⌝ = ⌜n0⌝ ↔ n = n0; (s) ⊦ n∞ + ⌜0⌝ = n∞; (s) ⊦ m ≤ n → (⌜(* n)⌝ + n∞1 = ⌜(* m)⌝ + n∞2 ↔ ⌜n - m⌝ + n∞1 = n∞2); (s) ⊦ # ⌜nf.take(n)⌝ = ⌜n⌝ ; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)

Proof step 159 from proof conttr-thm (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 160:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , # σf(n2) ≠ # αf(n2) + ⌜1⌝ , ⌜af.take(n0)⌝ ⊑ αf(n2), ⌜af.take(n0)⌝ ⊑ αf(n1), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), αf(n1) ⊑ αf(n2), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ ⊦

• Proof step 161:

  ⌜bl⌝ = βf(σf(n), αf(n)), lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bl⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , ⌜af.take(n0)⌝ ⊑ αf(n2), ⌜af.take(n0)⌝ ⊑ αf(n1), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ⌜sf.take(n0 + 1)⌝ ⊑ σf(n2), αf(n1) ⊑ αf(n2), βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ βf(σf(n2), αf(n2)), ¬ βf(⌜sf.take(n0 + 1)⌝ , ⌜af.take(n0)⌝ ) ⊑ ⌜bl⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ ⊦

Rule argument

Additional information

Proof step 160 from proof conttr-thm (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n2

Proof step 161 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: leq-trans: ⊦ σ ⊑ σ1 ∧ σ1 ⊑ σ2 → σ ⊑ σ2; (forward) from specification stream Used terms: n2

Proof step 162 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 163:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , ⌜bf⌝ ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ .take(n) = nf.take(n); (s) ⊦ ¬ ∞ < n∞; (s) ⊦ # ⌜nf⌝ = ∞; (s)

Proof step 163 from proof conttr-thm (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 164:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , ⌜bf⌝ ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ ⊦

• Proof step 212:

  # lim(⌜σf⌝ ) = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , ⌜bf⌝ ≠ βf(lim(⌜σf⌝ ), lim(⌜αf⌝ )), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ).take(m)⌝ ⊑ σf(n)), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Used simplifier rules: Stream[A] freely generated by ⌝, ⌝;

Proof step 164 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 165:

  ⌜# sl⌝ = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , ⌜bf⌝ ≠ βf(⌜sl⌝ , lim(⌜αf⌝ )), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ m. ¬ # sl < m → (∃ n. ⌜sl.take(m)⌝ ⊑ σf(n)), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ .take(n) = x.take(n); (s) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s)

Proof step 165 from proof conttr-thm (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 166:

  ⌜# sl⌝ = # lim(⌜αf⌝ ) + ⌜1⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , ⌜bf⌝ ≠ βf(⌜sl⌝ , lim(⌜αf⌝ )), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ¬ # sl < m → (∃ n. ⌜sl.take(m)⌝ ⊑ σf(n)), ∀ m. ¬ # lim(⌜αf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n)), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: Stream[A] freely generated by ⌝, ⌝; ⊦ # ⌜nf⌝ = ∞; (s) ⊦ ⌜m⌝ + ∞ = ∞; (s) add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 166 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 167:

  # al + 1 = # sl, lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , ⌜bf⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ¬ # sl < m → (∃ n. ⌜sl.take(m)⌝ ⊑ σf(n)), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ m. ¬ # al < m → (∃ n. ⌜al.take(m)⌝ ⊑ αf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ .take(n) = x.take(n); (s) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s) add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜n⌝ + ⌜m⌝ = ⌜n0⌝ ↔ n + m = n0; (s)

Proof step 167 from proof conttr-thm (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 168:

  # al + 1 = # sl, lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bl⌝ , ⌜bf⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ m. ¬ # al < m → (∃ n. ⌜al.take(m)⌝ ⊑ αf(n)), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ m. ¬ # sl < m → (∃ n. ⌜sl.take(m)⌝ ⊑ σf(n)), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ ⊦

• Proof step 187:

  # al + 1 = # sl, lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bf0⌝ , ⌜bf⌝ ≠ βf(⌜sl⌝ , ⌜al⌝ ), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ m. ¬ # al < m → (∃ n. ⌜al.take(m)⌝ ⊑ αf(n)), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ m. ¬ # sl < m → (∃ n. ⌜sl.take(m)⌝ ⊑ σf(n)), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Used simplifier rules: Stream[A] freely generated by ⌝, ⌝;

Proof step 168 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 169:

  # al + 1 = # sl, lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bl⌝ , cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ m. ¬ # al < m → (∃ n. ⌜al.take(m)⌝ ⊑ αf(n)), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ m. ¬ # sl < m → (∃ n. ⌜sl.take(m)⌝ ⊑ σf(n)), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ ≠ ⌜nf⌝ ; (s)

Proof step 169 from proof conttr-thm (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 170:

  ⊦ cont(βf) ↔ mon(βf) ∧ pbd(βf)

• Proof step 171:

  # al + 1 = # sl, lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bl⌝ , mon(⌜αf⌝ ), mon(⌜σf⌝ ), mon(βf), pbd(βf), ∀ m. ¬ # sl < m → (∃ n. ⌜sl.take(m)⌝ ⊑ σf(n)), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ m. ¬ # al < m → (∃ n. ⌜al.take(m)⌝ ⊑ αf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Used terms: βf

Proof step 170 from proof conttr-thm (previous step) (back to tree)

Lemma conttr-def

Proof step 171 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 172:

  # al + 1 = # sl, lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bl⌝ , mon(⌜σf⌝ ), mon(⌜αf⌝ ), pbd(βf), ∀ m. ¬ # al < m → (∃ n. ⌜al.take(m)⌝ ⊑ αf(n)), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ m. ¬ # sl < m → (∃ n. ⌜sl.take(m)⌝ ⊑ σf(n)), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used simplifier rules: montr-def: ⊦ mon(βf) ↔ (∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2));

Proof step 172 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 173:

  # al + 1 = # sl, lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bl⌝ , mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), ∀ m. ¬ # sl < m → (∃ n. ⌜sl.take(m)⌝ ⊑ σf(n)), ∀ m. ¬ # al < m → (∃ n. ⌜al.take(m)⌝ ⊑ αf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ , ∃ n. ⌜bf.take(# bl + 1)⌝ ⊑ βf(σf(n), αf(n)), ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used terms: # bl + 1

Proof step 173 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 174:

  # al + 1 = # sl, lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bl⌝ , ⌜bf.take(# bl + 1)⌝ ⊑ βf(σf(n), αf(n)), mon(⌜σf⌝ ), mon(⌜αf⌝ ), pbd(βf), ∀ m. ¬ # al < m → (∃ n. ⌜al.take(m)⌝ ⊑ αf(n)), ∀ m. ¬ # sl < m → (∃ n. ⌜sl.take(m)⌝ ⊑ σf(n)), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Proof step 174 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 175:

  # al + 1 = # sl, lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bl⌝ , ⌜bf.take(# bl + 1)⌝ ⊑ βf(σf(n), αf(n)), mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), # σf(n) = # αf(n) + ⌜1⌝ ∧ # ⌜sl⌝ = # ⌜al⌝ + ⌜1⌝ ∧ σf(n) ⊑ ⌜sl⌝ ∧ αf(n) ⊑ ⌜al⌝ → βf(σf(n), αf(n)) ⊑ βf(⌜sl⌝ , ⌜al⌝ ), ∀ m. ¬ # sl < m → (∃ n. ⌜sl.take(m)⌝ ⊑ σf(n)), ∀ m. ¬ # al < m → (∃ n. ⌜al.take(m)⌝ ⊑ αf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Used terms: σf(n), αf(n), ⌜sl⌝ , ⌜al⌝

Proof step 175 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 176:

  # al + 1 = # sl, lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bl⌝ , ⌜bf.take(# bl + 1)⌝ ⊑ βf(σf(n), αf(n)), mon(⌜σf⌝ ), mon(⌜αf⌝ ), pbd(βf), # σf(n) = # αf(n) + ⌜1⌝ ∧ σf(n) ⊑ ⌜sl⌝ ∧ αf(n) ⊑ ⌜al⌝ → βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ m. ¬ # al < m → (∃ n. ⌜al.take(m)⌝ ⊑ αf(n)), ∀ m. ¬ # sl < m → (∃ n. ⌜sl.take(m)⌝ ⊑ σf(n)), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Used simplifier rules: add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜n⌝ + ⌜m⌝ = ⌜n0⌝ ↔ n + m = n0; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s)

Proof step 176 from proof conttr-thm (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 177:

  # al + 1 = # sl, lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bl⌝ , ⌜bf.take(# bl + 1)⌝ ⊑ βf(σf(n), αf(n)), mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), ∀ m. ¬ # sl < m → (∃ n. ⌜sl.take(m)⌝ ⊑ σf(n)), ∀ m. ¬ # al < m → (∃ n. ⌜al.take(m)⌝ ⊑ αf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦ # σf(n) = # αf(n) + ⌜1⌝ ∧ σf(n) ⊑ ⌜sl⌝ ∧ αf(n) ⊑ ⌜al⌝

• Proof step 181:

  # al + 1 = # sl, lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bl⌝ , ⌜bf.take(# bl + 1)⌝ ⊑ βf(σf(n), αf(n)), mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ m. ¬ # sl < m → (∃ n. ⌜sl.take(m)⌝ ⊑ σf(n)), ∀ m. ¬ # al < m → (∃ n. ⌜al.take(m)⌝ ⊑ αf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Proof step 177 from proof conttr-thm (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 178:

  # al + 1 = # sl, lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bl⌝ , # σf(n) ≠ # αf(n) + ⌜1⌝ , ⌜bf.take(# bl + 1)⌝ ⊑ βf(σf(n), αf(n)), mon(⌜σf⌝ ), mon(⌜αf⌝ ), pbd(βf), ∀ m. ¬ # al < m → (∃ n. ⌜al.take(m)⌝ ⊑ αf(n)), ∀ m. ¬ # sl < m → (∃ n. ⌜sl.take(m)⌝ ⊑ σf(n)), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ ⊦

• Proof step 179:

  # al + 1 = # sl, lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bl⌝ , ⌜bf.take(# bl + 1)⌝ ⊑ βf(σf(n), αf(n)), mon(⌜σf⌝ ), mon(⌜αf⌝ ), pbd(βf), ¬ σf(n) ⊑ ⌜sl⌝ , ∀ m. ¬ # al < m → (∃ n. ⌜al.take(m)⌝ ⊑ αf(n)), ∀ m. ¬ # sl < m → (∃ n. ⌜sl.take(m)⌝ ⊑ σf(n)), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ ⊦

• Proof step 180:

  # al + 1 = # sl, lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bl⌝ , ⌜bf.take(# bl + 1)⌝ ⊑ βf(σf(n), αf(n)), mon(⌜σf⌝ ), mon(⌜αf⌝ ), pbd(βf), ¬ αf(n) ⊑ ⌜al⌝ , ∀ m. ¬ # al < m → (∃ n. ⌜al.take(m)⌝ ⊑ αf(n)), ∀ m. ¬ # sl < m → (∃ n. ⌜sl.take(m)⌝ ⊑ σf(n)), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Proof step 178 from proof conttr-thm (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n

Proof step 179 from proof conttr-thm (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n

Proof step 180 from proof conttr-thm (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n

Proof step 181 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 182:

  # al + 1 = # sl, lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bl⌝ , ⌜bf.take(# bl + 1)⌝ ⊑ βf(σf(n), αf(n)), bf.take(# bl + 1) ⊑ bl, mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ m. ¬ # sl < m → (∃ n. ⌜sl.take(m)⌝ ⊑ σf(n)), ∀ m. ¬ # al < m → (∃ n. ⌜al.take(m)⌝ ⊑ αf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s) leq-trans: ⊦ σ ⊑ σ1 ∧ σ1 ⊑ σ2 → σ ⊑ σ2; (forward) from specification stream

Proof step 182 from proof conttr-thm (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 183:

  ⊦ x ⊑ y ↔ (∃ z. x + z = y)

• Proof step 184:

  # al + 1 = # sl, lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bl⌝ , ⌜bf.take(# bl + 1)⌝ ⊑ βf(σf(n), αf(n)), mon(⌜σf⌝ ), mon(⌜αf⌝ ), pbd(βf), βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ m. ¬ # al < m → (∃ n. ⌜al.take(m)⌝ ⊑ αf(n)), ∀ m. ¬ # sl < m → (∃ n. ⌜sl.take(m)⌝ ⊑ σf(n)), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∃ z. bf.take(# bl + 1) + z = bl ⊦

Rule argument

Additional information

Used terms: bf.take(# bl + 1), bl

Proof step 183 from proof conttr-thm (previous step) (back to tree)

A spec-lemma

Proof step 184 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 185:

  # al + 1 = # sl, bf.take(# bl + 1) + bl0 = bl, lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bl⌝ , ⌜bf.take(# bl + 1)⌝ ⊑ βf(σf(n), αf(n)), mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , ∀ m. ¬ # sl < m → (∃ n. ⌜sl.take(m)⌝ ⊑ σf(n)), ∀ m. ¬ # al < m → (∃ n. ⌜al.take(m)⌝ ⊑ αf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Proof step 185 from proof conttr-thm (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 186:

  ⊦ x = y ↔ # x = # y ∧ (∀ n. n < # x → x(n) = y(n))

Rule argument

Additional information

Used simplifier rules: ⊦ #(x + y) = # x + # y; (s) ⊦ # nf.take(n) = n; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ m = m + n ↔ n = 0; (s) ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: bf.take(# bl + 1) + bl0, bl

Proof step 186 from proof conttr-thm (previous step) (back to tree)

A spec-lemma

Proof step 187 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 188:

  # al + 1 = # sl, lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bf0⌝ , bf ≠ bf0, cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ m. ¬ # al < m → (∃ n. ⌜al.take(m)⌝ ⊑ αf(n)), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ m. ¬ # sl < m → (∃ n. ⌜sl.take(m)⌝ ⊑ σf(n)), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ = ⌜nf0⌝ ↔ nf = nf0; (s)

Proof step 188 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 189:

  # al + 1 = # sl, lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bf0⌝ , bf ≠ bf0, cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ¬ # sl < m → (∃ n. ⌜sl.take(m)⌝ ⊑ σf(n)), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ , ¬ # al < # al → (∃ n. ⌜al.take(# al)⌝ ⊑ αf(n)) ⊦

Rule argument

Additional information

Used terms: # al

Proof step 189 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 190:

  # al + 1 = # sl, lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bf0⌝ , bf ≠ bf0, ⌜al⌝ ⊑ αf(n), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ m. ¬ # sl < m → (∃ n. ⌜sl.take(m)⌝ ⊑ σf(n)), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x = n → x.take(n) = x; (s) ⊦ ¬ n < n; (s)

Proof step 190 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 191:

  # al + 1 = # sl, lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bf0⌝ , bf ≠ bf0, ⌜al⌝ ⊑ αf(n), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), αf(n) ⊑ ⌜al⌝ , ∀ m. ¬ # sl < m → (∃ n. ⌜sl.take(m)⌝ ⊑ σf(n)), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Used terms: n

Proof step 191 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 192:

  # al + 1 = # sl, ⌜al⌝ = αf(n), lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bf0⌝ , bf ≠ bf0, cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ¬ # sl < m → (∃ n. ⌜sl.take(m)⌝ ⊑ σf(n)), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x0 = # x1 → (x0 ⊑ x1 ↔ x0 = x1); (s) ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s) ⊦ σ1 ⊑ σ2 → (σ2 ⊑ σ1 ↔ σ1 = σ2); (as)

Proof step 192 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 193:

  # al + 1 = # sl, ⌜al⌝ = αf(n), lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bf0⌝ , bf ≠ bf0, cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ¬ # sl < # sl → (∃ n. ⌜sl.take(# sl)⌝ ⊑ σf(n)) ⊦

Rule argument

Additional information

Used terms: # sl

Proof step 193 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 194:

  # al + 1 = # sl, ⌜al⌝ = αf(n), lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bf0⌝ , bf ≠ bf0, cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ⌜sl⌝ ⊑ σf(n0), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x = n → x.take(n) = x; (s) ⊦ ¬ n < n; (s)

Proof step 194 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 195:

  # al + 1 = # sl, ⌜al⌝ = αf(n), lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bf0⌝ , bf ≠ bf0, cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ⌜sl⌝ ⊑ σf(n0), σf(n0) ⊑ ⌜sl⌝ , ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Used terms: n0

Proof step 195 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 196:

  # al + 1 = # sl, ⌜al⌝ = αf(n), lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , ⌜sl⌝ = σf(n0), βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bf0⌝ , bf ≠ bf0, cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x0 = # x1 → (x0 ⊑ x1 ↔ x0 = x1); (s) ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s) ⊦ σ1 ⊑ σ2 → (σ2 ⊑ σ1 ↔ σ1 = σ2); (as)

Proof step 196 from proof conttr-thm (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 197:

  mon(σσ), m ≤ n, ⌜n⌝ < # σσ ⊦ σσ(m) ⊑ σσ(n)

• Proof step 198:

  # al + 1 = # sl, ⌜al⌝ = αf(n), lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , ⌜sl⌝ = σf(n0), βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bf0⌝ , bf ≠ bf0, cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦ n0 ≤ n ∧ ⌜n⌝ < # ⌜σf⌝

• Proof step 208:

  # al + 1 = # sl, ⌜al⌝ = αf(n), lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , ⌜sl⌝ = σf(n0), βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bf0⌝ , bf ≠ bf0, cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ⌜σf⌝ (n0) ⊑ ⌜σf⌝ (n), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Used terms: ⌜σf⌝ , n0, n

Proof step 197 from proof conttr-thm (previous step) (back to tree)

A spec-lemma

Proof step 198 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 199:

  # al + 1 = # sl, ⌜al⌝ = αf(n), lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , ⌜sl⌝ = σf(n0), βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bf0⌝ , bf ≠ bf0, cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), n < n0, ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ ⌜m⌝ < ∞; (s) ⊦ # ⌜nf⌝ = ∞; (s)

Proof step 199 from proof conttr-thm (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 200:

  mon(σσ), m ≤ n, ⌜n⌝ < # σσ ⊦ σσ(m) ⊑ σσ(n)

• Proof step 201:

  # al + 1 = # sl, ⌜al⌝ = αf(n), lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , ⌜sl⌝ = σf(n0), βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bf0⌝ , bf ≠ bf0, cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), n < n0, ⌜σf⌝ (n) ⊑ ⌜σf⌝ (n0), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s) ⊦ # ⌜nf⌝ = ∞; (s) ⊦ ⌜m⌝ < ∞; (s) Used terms: ⌜σf⌝ , n, n0

Proof step 200 from proof conttr-thm (previous step) (back to tree)

A spec-lemma

Proof step 201 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 202:

  # al + 1 = # sl, ⌜al⌝ = αf(n), lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , ⌜sl⌝ = σf(n0), βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bf0⌝ , bf ≠ bf0, cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), n < n0, σf(n) ⊑ ⌜sl⌝ , ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ (n) = nf(n); (s)

Proof step 202 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 203:

  # al + 1 = # sl, ⌜al⌝ = αf(n), lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , ⌜sl⌝ = σf(n0), βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bf0⌝ , bf ≠ bf0, cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), n < n0, βf(σf(n0), αf(n0)) ⊑ ⌜bf⌝ , σf(n) ⊑ ⌜sl⌝ , ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ ⊦

Rule argument

Additional information

Used terms: n0

Proof step 203 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 204:

  # al + 1 = # sl, ⌜al⌝ = αf(n), lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , ⌜sl⌝ = σf(n0), βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bf0⌝ , bf ≠ bf0, cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), n < n0, βf(⌜sl⌝ , αf(n0)) ⊑ ⌜bf⌝ , σf(n) ⊑ ⌜sl⌝ , ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Proof step 204 from proof conttr-thm (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 205:

  mon(σσ), m ≤ n, ⌜n⌝ < # σσ ⊦ σσ(m) ⊑ σσ(n)

• Proof step 206:

  # al + 1 = # sl, ⌜al⌝ = αf(n), lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , ⌜sl⌝ = σf(n0), βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bf0⌝ , bf ≠ bf0, cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), n < n0, ⌜αf⌝ (n) ⊑ ⌜αf⌝ (n0), βf(⌜sl⌝ , αf(n0)) ⊑ ⌜bf⌝ , σf(n) ⊑ ⌜sl⌝ , ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s) ⊦ # ⌜nf⌝ = ∞; (s) ⊦ ⌜m⌝ < ∞; (s) Used terms: ⌜αf⌝ , n, n0

Proof step 205 from proof conttr-thm (previous step) (back to tree)

A spec-lemma

Proof step 206 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 207:

  # al + 1 = # sl, ⌜al⌝ = αf(n), lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , ⌜sl⌝ = σf(n0), βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bf0⌝ , bf ≠ bf0, ⌜al⌝ ⊑ αf(n0), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), n < n0, βf(⌜sl⌝ , αf(n0)) ⊑ ⌜bf⌝ , σf(n) ⊑ ⌜sl⌝ , ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. σf(n) ⊑ ⌜sl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ (n) = nf(n); (s)

Proof step 207 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ σ1 ⊑ σ2 → (σ2 ⊑ σ1 ↔ σ1 = σ2); (as) ⊦ ⌜nf⌝ ⊑ ⌜nf0⌝ ↔ nf = nf0; (s) Used terms: n0

Proof step 208 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 209:

  # al + 1 = # sl, ⌜al⌝ = αf(n), lim(⌜αf⌝ ) = ⌜al⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sl⌝ , ⌜sl⌝ = σf(n0), βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bf0⌝ , bf ≠ bf0, cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ⌜sl⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ n. σf(n) ⊑ ⌜sl⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ (n) = nf(n); (s)

Proof step 209 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 210:

  # al + 1 = # sl, ⌜al⌝ = αf(n), lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , ⌜sl⌝ = σf(n0), βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bf0⌝ , bf ≠ bf0, cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ⌜sl⌝ ⊑ σf(n), σf(n) ⊑ ⌜sl⌝ , ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ ⊦

Rule argument

Additional information

Used terms: n

Proof step 210 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 211:

  # al + 1 = # sl, ⌜al⌝ = αf(n), lim(⌜σf⌝ ) = ⌜sl⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜αf⌝ ) = ⌜al⌝ , ⌜sl⌝ = σf(n), ⌜sl⌝ = σf(n0), βf(⌜sl⌝ , ⌜al⌝ ) = ⌜bf0⌝ , bf ≠ bf0, cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜al⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x0 = # x1 → (x0 ⊑ x1 ↔ x0 = x1); (s) ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s) ⊦ σ1 ⊑ σ2 → (σ2 ⊑ σ1 ↔ σ1 = σ2); (as)

Proof step 211 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ ⊑ ⌜nf0⌝ ↔ nf = nf0; (s) Used terms: n

Proof step 212 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 213:

  # lim(⌜αf⌝ ) = ∞, lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , ⌜bf⌝ ≠ βf(⌜sf⌝ , lim(⌜αf⌝ )), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ m. ∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. σf(n) ⊑ ⌜sf⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ ∞ < n∞; (s) ⊦ ⌜nf⌝ .take(n) = nf.take(n); (s) ⊦ # ⌜nf⌝ = ∞; (s) add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜m⌝ + n∞ = ∞ ↔ n∞ = ∞; (s)

Proof step 213 from proof conttr-thm (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 214:

  # lim(⌜αf⌝ ) = ∞, lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , ⌜bf⌝ ≠ βf(⌜sf⌝ , lim(⌜αf⌝ )), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ m. ∃ n. ⌜lim(⌜αf⌝ ).take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ lim(⌜αf⌝ ), ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Used simplifier rules: Stream[A] freely generated by ⌝, ⌝; ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s) ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 214 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 215:

  lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , ⌜bf⌝ ≠ βf(⌜sf⌝ , ⌜af⌝ ), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ .take(n) = nf.take(n); (s) ⊦ # ⌜nf⌝ = ∞; (s)

Proof step 215 from proof conttr-thm (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 216:

  lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl⌝ , ⌜bf⌝ ≠ βf(⌜sf⌝ , ⌜af⌝ ), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ ⊦

• Proof step 232:

  lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf0⌝ , ⌜bf⌝ ≠ βf(⌜sf⌝ , ⌜af⌝ ), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ ⊦

Rule argument

Additional information

Used simplifier rules: Stream[A] freely generated by ⌝, ⌝;

Proof step 216 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 217:

  lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl⌝ , cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ ≠ ⌜nf⌝ ; (s)

Proof step 217 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 218:

  lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl⌝ , cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∃ n. ⌜bf.take(# bl + 1)⌝ ⊑ βf(σf(n), αf(n)) ⊦

Rule argument

Additional information

Used terms: # bl + 1

Proof step 218 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 219:

  lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl⌝ , ⌜bf.take(# bl + 1)⌝ ⊑ βf(σf(n), αf(n)), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ ⊦

Rule argument

Additional information

Proof step 219 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 220:

  lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl⌝ , ⌜bf.take(# bl + 1)⌝ ⊑ βf(σf(n), αf(n)), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), σf(n) ⊑ ⌜sf⌝ , ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Used terms: n

Proof step 220 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 221:

  lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl⌝ , ⌜bf.take(# bl + 1)⌝ ⊑ βf(σf(n), αf(n)), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), αf(n) ⊑ ⌜af⌝ , σf(n) ⊑ ⌜sf⌝ , ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ ⊦

Rule argument

Additional information

Used terms: n

Proof step 221 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 222:

  # σf(n) = # αf(n) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl⌝ , ⌜bf.take(# bl + 1)⌝ ⊑ βf(σf(n), αf(n)), cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), αf(n) ⊑ ⌜af⌝ , σf(n) ⊑ ⌜sf⌝ , ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Used terms: n

Proof step 222 from proof conttr-thm (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 223:

  ⊦ cont(βf) ↔ mon(βf) ∧ pbd(βf)

• Proof step 224:

  # σf(n) = # αf(n) + ⌜1⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl⌝ , ⌜bf.take(# bl + 1)⌝ ⊑ βf(σf(n), αf(n)), mon(⌜αf⌝ ), mon(⌜σf⌝ ), mon(βf), pbd(βf), αf(n) ⊑ ⌜af⌝ , σf(n) ⊑ ⌜sf⌝ , ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Used terms: βf

Proof step 223 from proof conttr-thm (previous step) (back to tree)

Lemma conttr-def

Proof step 224 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 225:

  # σf(n) = # αf(n) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl⌝ , ⌜bf.take(# bl + 1)⌝ ⊑ βf(σf(n), αf(n)), mon(⌜σf⌝ ), mon(⌜αf⌝ ), pbd(βf), αf(n) ⊑ ⌜af⌝ , σf(n) ⊑ ⌜sf⌝ , ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used simplifier rules: montr-def: ⊦ mon(βf) ↔ (∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2));

Proof step 225 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 226:

  # σf(n) = # αf(n) + ⌜1⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl⌝ , ⌜bf.take(# bl + 1)⌝ ⊑ βf(σf(n), αf(n)), mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), αf(n) ⊑ ⌜af⌝ , σf(n) ⊑ ⌜sf⌝ , # σf(n) = # αf(n) + ⌜1⌝ ∧ # ⌜sf⌝ = # ⌜af⌝ + ⌜1⌝ ∧ σf(n) ⊑ ⌜sf⌝ ∧ αf(n) ⊑ ⌜af⌝ → βf(σf(n), αf(n)) ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Used terms: σf(n), αf(n), ⌜sf⌝ , ⌜af⌝

Proof step 226 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 227:

  # σf(n) = # αf(n) + ⌜1⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl⌝ , ⌜bf.take(# bl + 1)⌝ ⊑ βf(σf(n), αf(n)), bf.take(# bl + 1) ⊑ bl, mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), αf(n) ⊑ ⌜af⌝ , βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , σf(n) ⊑ ⌜sf⌝ , ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s) leq-trans: ⊦ σ ⊑ σ1 ∧ σ1 ⊑ σ2 → σ ⊑ σ2; (forward) from specification stream add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜m⌝ + ∞ = ∞; (s) ⊦ # ⌜nf⌝ = ∞; (s)

Proof step 227 from proof conttr-thm (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 228:

  ⊦ x ⊑ y ↔ (∃ z. x + z = y)

• Proof step 229:

  # σf(n) = # αf(n) + ⌜1⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl⌝ , ⌜bf.take(# bl + 1)⌝ ⊑ βf(σf(n), αf(n)), mon(⌜σf⌝ ), mon(⌜αf⌝ ), pbd(βf), αf(n) ⊑ ⌜af⌝ , βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , σf(n) ⊑ ⌜sf⌝ , ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∃ z. bf.take(# bl + 1) + z = bl ⊦

Rule argument

Additional information

Used terms: bf.take(# bl + 1), bl

Proof step 228 from proof conttr-thm (previous step) (back to tree)

A spec-lemma

Proof step 229 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 230:

  # σf(n) = # αf(n) + ⌜1⌝ , bf.take(# bl + 1) + bl0 = bl, lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl⌝ , ⌜bf.take(# bl + 1)⌝ ⊑ βf(σf(n), αf(n)), mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), αf(n) ⊑ ⌜af⌝ , βf(σf(n), αf(n)) ⊑ ⌜bl⌝ , σf(n) ⊑ ⌜sf⌝ , ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Proof step 230 from proof conttr-thm (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 231:

  ⊦ x = y ↔ # x = # y ∧ (∀ n. n < # x → x(n) = y(n))

Rule argument

Additional information

Used simplifier rules: ⊦ #(x + y) = # x + # y; (s) ⊦ # nf.take(n) = n; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ m = m + n ↔ n = 0; (s) ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: bf.take(# bl + 1) + bl0, bl

Proof step 231 from proof conttr-thm (previous step) (back to tree)

A spec-lemma

Proof step 232 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 233:

  lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf0⌝ , bf ≠ bf0, cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ = ⌜nf0⌝ ↔ nf = nf0; (s)

Proof step 233 from proof conttr-thm (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 234:

  ⊦ bf ≠ bf0 ↔ (∃ n. bf(n) ≠ bf0(n) ∧ (∀ n0. n0 < n → bf(n0) = bf0(n0)))

• Proof step 235:

  lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf0⌝ , cont(βf), mon(⌜σf⌝ ), mon(⌜αf⌝ ), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∃ n. bf(n) ≠ bf0(n) ∧ (∀ n0. n0 < n → bf(n0) = bf0(n0)) ⊦

Rule argument

Additional information

Used terms: bf, bf0

Proof step 234 from proof conttr-thm (previous step) (back to tree)

Lemma fun-neq

Proof step 235 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 236:

  lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf0⌝ , bf(n) ≠ bf0(n), cont(βf), mon(⌜αf⌝ ), mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n0. n0 < n → bf(n0) = bf0(n0) ⊦

Rule argument

Additional information

Proof step 236 from proof conttr-thm (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 237:

  ⊦ cont(βf) ↔ mon(βf) ∧ pbd(βf)

• Proof step 238:

  lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf0⌝ , bf(n) ≠ bf0(n), mon(⌜σf⌝ ), mon(⌜αf⌝ ), mon(βf), pbd(βf), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n0. n0 < n → bf(n0) = bf0(n0) ⊦

Rule argument

Additional information

Used terms: βf

Proof step 237 from proof conttr-thm (previous step) (back to tree)

Lemma conttr-def

Proof step 238 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 239:

  lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf0⌝ , bf(n) ≠ bf0(n), mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜bf.take(m)⌝ ⊑ βf(σf(n), αf(n)), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n0. n0 < n → bf(n0) = bf0(n0), ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used simplifier rules: montr-def: ⊦ mon(βf) ↔ (∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2));

Proof step 239 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 240:

  lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf0⌝ , bf(n) ≠ bf0(n), mon(⌜σf⌝ ), mon(⌜αf⌝ ), pbd(βf), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n0. n0 < n → bf(n0) = bf0(n0), ∃ n0. ⌜bf.take(n + 1)⌝ ⊑ βf(σf(n0), αf(n0)), ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Used terms: n + 1

Proof step 240 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 241:

  lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf0⌝ , bf(n) ≠ bf0(n), ⌜bf.take(n + 1)⌝ ⊑ βf(σf(n0), αf(n0)), mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n0. n0 < n → bf(n0) = bf0(n0), ∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → βf(σ1, α1) ⊑ βf(σ2, α2) ⊦

Rule argument

Additional information

Proof step 241 from proof conttr-thm (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 242:

  lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf0⌝ , bf(n) ≠ bf0(n), ⌜bf.take(n + 1)⌝ ⊑ βf(σf(n0), αf(n0)), mon(⌜σf⌝ ), mon(⌜αf⌝ ), pbd(βf), # σf(n0) = # αf(n0) + ⌜1⌝ ∧ # ⌜sf⌝ = # ⌜af⌝ + ⌜1⌝ ∧ σf(n0) ⊑ ⌜sf⌝ ∧ αf(n0) ⊑ ⌜af⌝ → βf(σf(n0), αf(n0)) ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n0. n0 < n → bf(n0) = bf0(n0) ⊦

Rule argument

Additional information

Used terms: σf(n0), αf(n0), ⌜sf⌝ , ⌜af⌝

Proof step 242 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 243:

  lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf0⌝ , bf(n) ≠ bf0(n), ⌜bf.take(n + 1)⌝ ⊑ βf(σf(n0), αf(n0)), mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), # σf(n0) = # αf(n0) + ⌜1⌝ ∧ σf(n0) ⊑ ⌜sf⌝ ∧ αf(n0) ⊑ ⌜af⌝ → βf(σf(n0), αf(n0)) ⊑ ⌜bf0⌝ , ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n0. n0 < n → bf(n0) = bf0(n0) ⊦

Rule argument

Additional information

Used simplifier rules: add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜m⌝ + ∞ = ∞; (s) ⊦ # ⌜nf⌝ = ∞; (s)

Proof step 243 from proof conttr-thm (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 244:

  lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf0⌝ , bf(n) ≠ bf0(n), ⌜bf.take(n + 1)⌝ ⊑ βf(σf(n0), αf(n0)), mon(⌜σf⌝ ), mon(⌜αf⌝ ), pbd(βf), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n0. n0 < n → bf(n0) = bf0(n0) ⊦ # σf(n0) = # αf(n0) + ⌜1⌝ ∧ σf(n0) ⊑ ⌜sf⌝ ∧ αf(n0) ⊑ ⌜af⌝

• Proof step 248:

  lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf0⌝ , bf(n) ≠ bf0(n), ⌜bf.take(n + 1)⌝ ⊑ βf(σf(n0), αf(n0)), mon(⌜σf⌝ ), mon(⌜αf⌝ ), pbd(βf), βf(σf(n0), αf(n0)) ⊑ ⌜bf0⌝ , ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n0. n0 < n → bf(n0) = bf0(n0) ⊦

Rule argument

Additional information

Proof step 244 from proof conttr-thm (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 245:

  lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf0⌝ , # σf(n0) ≠ # αf(n0) + ⌜1⌝ , bf(n) ≠ bf0(n), ⌜bf.take(n + 1)⌝ ⊑ βf(σf(n0), αf(n0)), mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n0. n0 < n → bf(n0) = bf0(n0) ⊦

• Proof step 246:

  lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf0⌝ , bf(n) ≠ bf0(n), ⌜bf.take(n + 1)⌝ ⊑ βf(σf(n0), αf(n0)), mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), ¬ σf(n0) ⊑ ⌜sf⌝ , ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n0. n0 < n → bf(n0) = bf0(n0) ⊦

• Proof step 247:

  lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf0⌝ , bf(n) ≠ bf0(n), ⌜bf.take(n + 1)⌝ ⊑ βf(σf(n0), αf(n0)), mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), ¬ αf(n0) ⊑ ⌜af⌝ , ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n0. n0 < n → bf(n0) = bf0(n0) ⊦

Rule argument

Additional information

Proof step 245 from proof conttr-thm (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n0

Proof step 246 from proof conttr-thm (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n0

Proof step 247 from proof conttr-thm (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n0

Proof step 248 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 249:

  lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf0⌝ , bf(n) ≠ bf0(n), ⌜bf.take(n + 1)⌝ ⊑ βf(σf(n0), αf(n0)), ⌜bf.take(n + 1)⌝ ⊑ ⌜bf0⌝ , mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), βf(σf(n0), αf(n0)) ⊑ ⌜bf0⌝ , ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n0. n0 < n → bf(n0) = bf0(n0) ⊦

Rule argument

Additional information

Used simplifier rules: leq-trans: ⊦ σ ⊑ σ1 ∧ σ1 ⊑ σ2 → σ ⊑ σ2; (forward) from specification stream

Proof step 249 from proof conttr-thm (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 250:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 251:

  lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf0⌝ , bf(n) ≠ bf0(n), ⌜bf.take(n + 1)⌝ ⊑ βf(σf(n0), αf(n0)), mon(⌜σf⌝ ), mon(⌜αf⌝ ), pbd(βf), βf(σf(n0), αf(n0)) ⊑ ⌜bf0⌝ , ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n0. n0 < n → bf(n0) = bf0(n0), ∃ β. ⌜bf.take(n + 1)⌝ + β = ⌜bf0⌝ ⊦

Rule argument

Additional information

Used terms: ⌜bf.take(n + 1)⌝ , ⌜bf0⌝

Proof step 250 from proof conttr-thm (previous step) (back to tree)

A spec-lemma

Proof step 251 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 252:

  lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜αf⌝ ) = ⌜af⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf.take(n + 1)⌝ + ⌜bf1⌝ , bf(n) ≠ (bf.take(n + 1) + bf1)(n), ⌜bf.take(n + 1)⌝ ⊑ βf(σf(n0), αf(n0)), ⌜bf.take(n + 1)⌝ ⊑ βf(⌜sf⌝ , ⌜af⌝ ), mon(⌜σf⌝ ), mon(⌜αf⌝ ), pbd(βf), βf(σf(n0), αf(n0)) ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n1. n1 < n → bf(n1) = (bf.take(n + 1) + bf1)(n1) ⊦

Rule argument

Additional information

Used simplifier rules: leq-trans: ⊦ σ ⊑ σ1 ∧ σ1 ⊑ σ2 → σ ⊑ σ2; (forward) from specification stream ⊦ ⌜nf⌝ .f = nf; (s) ⊦ ⌜x + nf⌝ = ⌜x⌝ + ⌜nf⌝ ; (s) ⊦ finp(σ) ↔ ¬(∃ nf. ⌜nf⌝ = σ); (ss) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ ⌜x⌝ + σ = ⌜nf⌝ ↔ ¬ finp(σ) ∧ x + σ.f = nf; (as)

Proof step 252 from proof conttr-thm (previous step) (back to tree)

User applied rule subst equation

And obtained the following new goals

• Proof step 253:

  lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , bf(n) ≠ (bf.take(n + 1) + bf1)(n), ⌜bf.take(n + 1)⌝ ⊑ ⌜bf.take(n + 1)⌝ + ⌜bf1⌝ , ⌜bf.take(n + 1)⌝ ⊑ βf(σf(n0), αf(n0)), mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), βf(σf(n0), αf(n0)) ⊑ ⌜bf.take(n + 1)⌝ + ⌜bf1⌝ , ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n1. n1 < n → bf(n1) = (bf.take(n + 1) + bf1)(n1) ⊦

Rule argument

Additional information

Proof step 253 from proof conttr-thm (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 254:

  lim(⌜αf⌝ ) = ⌜af⌝ , lim(⌜σf⌝ ) = ⌜sf⌝ , lim(⌜λ n. βf(σf(n), αf(n))⌝ ) = ⌜bf⌝ , bf(n) ≠ (bf.take(n + 1) + bf1)(n), ⌜bf.take(n + 1)⌝ ⊑ βf(σf(n0), αf(n0)), mon(⌜αf⌝ ), mon(⌜σf⌝ ), pbd(βf), βf(σf(n0), αf(n0)) ⊑ ⌜bf.take(n + 1)⌝ + ⌜bf1⌝ , ∀ m. ∃ n. ⌜af.take(m)⌝ ⊑ αf(n), ∀ m. ∃ n. ⌜sf.take(m)⌝ ⊑ σf(n), ∀ n. βf(σf(n), αf(n)) ⊑ ⌜bf⌝ , ∀ n. # σf(n) = # αf(n) + ⌜1⌝ , ∀ n. αf(n) ⊑ ⌜af⌝ , ∀ n. σf(n) ⊑ ⌜sf⌝ , ∀ n1. n1 < n → bf(n1) = (bf.take(n + 1) + bf1)(n1) ⊦

Rule argument

Additional information

Used simplifier rules: append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ σ1 ⊑ σ1 + σ2; (s) leq-trans: ⊦ σ ⊑ σ1 ∧ σ1 ⊑ σ2 → σ ⊑ σ2; (forward) from specification stream

Proof step 254 from proof conttr-thm (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 255:

  ⊦ x + nf = (λ n. (n < # x ⊃ x(n);nf(n - # x)))

Rule argument

Additional information

Used simplifier rules: ⊦ m < n + 1 ↔ ¬ n < m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic m < n ⊦ nf.take(n)(m) = nf(m); (s) ⊦ # nf.take(n) = n; (s) ⊦ ¬ n < n; (s) Used terms: bf.take(n + 1), bf1

Proof step 255 from proof conttr-thm (previous step) (back to tree)

A spec-lemma