Proof step 1 from proof lim-extractprefix (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ) ⊦

Rule argument

Additional information

Proof step 2 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 3:

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

• Proof step 4:

  lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), mon(⌜σf⌝ ) ∧ ⌜σf⌝ ≠ [] → (∀ n. ⌜n⌝ < # ⌜σf⌝ → ⌜σf⌝ [n] ⊑ lim(⌜σf⌝ )) ∧ (∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜n⌝ < # ⌜σf⌝ ∧ ⌜lim(⌜σf⌝ ) to m⌝ ⊑ ⌜σf⌝ [n])) ⊦

Rule argument

Additional information

Used terms: ⌜σf⌝

Proof step 3 from proof lim-extractprefix (previous step) (back to tree)

Lemma lim-def

Proof step 4 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)) ⊦

Rule argument

Additional information

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

Proof step 5 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 6:

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

• Proof step 7:

  lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), mon(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ∧ ⌜λ n. ⌜x⌝ + σf(n)⌝ ≠ [] → (∀ n. ⌜n⌝ < # ⌜λ n. ⌜x⌝ + σf(n)⌝ → ⌜λ n. ⌜x⌝ + σf(n)⌝ [n] ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )) ∧ (∀ m. ¬ # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) < ⌜m⌝ → (∃ n. ⌜n⌝ < # ⌜λ n. ⌜x⌝ + σf(n)⌝ ∧ ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜λ n. ⌜x⌝ + σf(n)⌝ [n])), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ) ⊦

Rule argument

Additional information

Used terms: ⌜λ n. ⌜x⌝ + σf(n)⌝

Proof step 6 from proof lim-extractprefix (previous step) (back to tree)

Lemma lim-def

Proof step 7 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), mon(⌜λ n. ⌜x⌝ + σf(n)⌝ ) → (∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )) ∧ (∀ m. ¬ # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n))), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)) ⊦

Rule argument

Additional information

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

Proof step 8 from proof lim-extractprefix (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 9:

  lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ¬ mon(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ) ⊦

• Proof step 11:

  lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ¬ # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n)) ⊦

Rule argument

Additional information

Proof step 9 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 10:

  ⊦ mon(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ↔ mon(⌜σf⌝ )

Rule argument

Additional information

Used terms: x, σf

Proof step 10 from proof lim-extractprefix (previous step) (back to tree)

Lemma mon-rmprefix

Proof step 11 from proof lim-extractprefix (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 12:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ∀ m. ¬ # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n)), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)) ⊦

• Proof step 136:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ∀ m. ¬ # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n)), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)) ⊦

Rule argument

Additional information

Used formulas: # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞

Proof step 12 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ ∞ < n∞; (s)

Proof step 13 from proof lim-extractprefix (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 14:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ⊦

• Proof step 115:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) ≠ ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ⊦

Rule argument

Additional information

Used formulas: # lim(⌜σf⌝ ) = ∞

Proof step 14 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ ∞ < n∞; (s)

Proof step 15 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 16:

  ⊦ σ0 = σ1 ↔ # σ0 = # σ1 ∧ (∀ n. ⌜n⌝ < # σ0 → σ0[n] = σ1[n])

• Proof step 17:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ #(⌜x⌝ + lim(⌜σf⌝ )) ∨ (∃ n. ¬(⌜n⌝ < # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) → lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n] = (⌜x⌝ + lim(⌜σf⌝ ))[n])) ⊦

Rule argument

Additional information

Used terms: lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ⌜x⌝ + lim(⌜σf⌝ )

Proof step 16 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 17 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n] ≠ (⌜x⌝ + lim(⌜σf⌝ ))[n], mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ∞; (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) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s) ⊦ #(σ + σ1) = # σ + # σ1; (s)

Proof step 18 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 19:

  ⊦ ⌜m⌝ < # σ → (σ + σ1)[m] = σ[m]

• Proof step 20:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n] ≠ (⌜x⌝ + lim(⌜σf⌝ ))[n], mon(⌜σf⌝ ), ¬ ⌜n⌝ < # ⌜x⌝ , ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

• Proof step 99:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n] ≠ ⌜x⌝ [n], ⌜n⌝ < # ⌜x⌝ , mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

Rule argument

Additional information

Used terms: ⌜x⌝ , lim(⌜σf⌝ ), n

Proof step 19 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 20 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 21:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n] ≠ lim(⌜σf⌝ )[n - # x], mon(⌜σf⌝ ), # x ≤ n, ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ ⌜n⌝ .n = n; (s) ⊦ ¬ ⌜m⌝ < # σ → (σ + σ1)[m] = σ1[m - (# σ).n]; (s)

Proof step 21 from proof lim-extractprefix (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 22:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, n = n0 + # x, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n] ≠ lim(⌜σf⌝ )[n0], mon(⌜σf⌝ ), # x ≤ n, ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

Rule argument

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, # x, n0

Proof step 22 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 23:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] ≠ lim(⌜σf⌝ )[n0], mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n) ⊦

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m ≤ m + n; (s)

Proof step 23 from proof lim-extractprefix (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 24:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] ≠ lim(⌜σf⌝ )[n0], mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∃ n. ⌜lim(⌜σf⌝ ) to n0 + 1⌝ ⊑ σf(n) ⊦

Rule argument

Additional information

Used terms: n0 + 1

Proof step 24 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 25:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] ≠ lim(⌜σf⌝ )[n0], mon(⌜σf⌝ ), ⌜lim(⌜σf⌝ ) to n0 + 1⌝ ⊑ σf(n), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n) ⊦

Rule argument

Additional information

Proof step 25 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 26:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 27:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] ≠ lim(⌜σf⌝ )[n0], mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∃ σ1. ⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1 = σf(n) ⊦

Rule argument

Additional information

Used terms: ⌜lim(⌜σf⌝ ) to n0 + 1⌝ , σf(n)

Proof step 26 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 27 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 28:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, ⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1 = σf(n), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] ≠ lim(⌜σf⌝ )[n0], mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n) ⊦

Rule argument

Additional information

Proof step 28 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 29:

  ⊦ σ0 = σ1 ↔ # σ0 = # σ1 ∧ (∀ n. ⌜n⌝ < # σ0 → σ0[n] = σ1[n])

• Proof step 30:

  #(⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1) = # σf(n), # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] ≠ lim(⌜σf⌝ )[n0], mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n1. ⌜n1⌝ < #(⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1) → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1] ⊦

Rule argument

Additional information

Used terms: ⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1, σf(n)

Proof step 29 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 30 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 31:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] ≠ lim(⌜σf⌝ )[n0], mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1] ⊦

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⌝ < ⌜1⌝ + n∞ ↔ ¬ n∞ < ⌜n⌝ ; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream 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) ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = n; (s) ⊦ ¬ ∞ < n∞; (s) ⊦ # ⌜σ to n⌝ = ⌜#(σ to n)⌝ ; (s) ⊦ #(σ + σ1) = # σ + # σ1; (s)

Proof step 31 from proof lim-extractprefix (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 32:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] ≠ lim(⌜σf⌝ )[n0], mon(⌜σf⌝ ), ¬ ⌜n0⌝ + # σ1 < ⌜n0⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n0] = σf(n)[n0], ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

Rule argument

Additional information

Used terms: n0

Proof step 32 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 33:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] ≠ lim(⌜σf⌝ )[n0], mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1] ⊦

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∞ + n∞1 < n∞1; (s)

Proof step 33 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 34:

  ⊦ ⌜m⌝ < # σ → (σ + σ1)[m] = σ[m]

• Proof step 35:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), ⌜lim(⌜σf⌝ ) to n0 + 1⌝ [n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] ≠ lim(⌜σf⌝ )[n0], ⌜n0⌝ < # ⌜lim(⌜σf⌝ ) to n0 + 1⌝ , mon(⌜σf⌝ ), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ ⌜m⌝ < # σ → (σ + σ1)[m] = σ1[m - (# σ).n]; (s) ⊦ ⌜n⌝ ≠ ∞; (s) n∞ ≠ ∞ ⊦ (⌜m⌝ + n∞).n = m + n∞.n; (s) ⊦ ⌜n⌝ .n = n; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ # ⌜σ to n⌝ = ⌜#(σ to n)⌝ ; (s) ⊦ ¬ ∞ < n∞; (s) ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = n; (s) ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ ⌜n⌝ < ⌜1⌝ + n∞ ↔ ¬ n∞ < ⌜n⌝ ; (s) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ ¬ n < 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 Used terms: ⌜lim(⌜σf⌝ ) to n0 + 1⌝ , σ1, n0

Proof step 34 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 35 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 36:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] ≠ σf(n)[n0], mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1] ⊦

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 < n; (s) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ ⌜n⌝ < ⌜1⌝ + n∞ ↔ ¬ 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) ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = n; (s) ⊦ ¬ ∞ < n∞; (s) ⊦ # ⌜σ to n⌝ = ⌜#(σ to n)⌝ ; (s) m < n, ¬ # σ < ⌜n⌝ ⊦ (σ to n)[m] = σ[m]; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ ⌜x⌝ [n] = x[n]; (s)

Proof step 36 from proof lim-extractprefix (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 37:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] ≠ σf(n)[n0], mon(⌜σf⌝ ), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ ⊑ ⌜x⌝ + σf(n) ⊦

Rule argument

Additional information

Used terms: n0 + 1 + # x

Proof step 37 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 38:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] ≠ σf(n)[n0], mon(⌜σf⌝ ), ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ ⊑ ⌜x⌝ + σf(n1), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1] ⊦

Rule argument

Additional information

Proof step 38 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 39:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 40:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] ≠ σf(n)[n0], mon(⌜σf⌝ ), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∃ σ1. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ1 = ⌜x⌝ + σf(n1) ⊦

Rule argument

Additional information

Used terms: ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ , ⌜x⌝ + σf(n1)

Proof step 39 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 40 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 41:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ = ⌜x⌝ + σf(n1), (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] ≠ σf(n)[n0], mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1] ⊦

Rule argument

Additional information

Proof step 41 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 42:

  ⊦ σ0 = σ1 ↔ # σ0 = # σ1 ∧ (∀ n. ⌜n⌝ < # σ0 → σ0[n] = σ1[n])

• Proof step 43:

  #(⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ) = #(⌜x⌝ + σf(n1)), # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] ≠ σf(n)[n0], mon(⌜σf⌝ ), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜n⌝ < #(⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ) → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n] = (⌜x⌝ + σf(n1))[n] ⊦

Rule argument

Additional information

Used terms: ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ, ⌜x⌝ + σf(n1)

Proof step 42 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 43 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 44:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] ≠ σf(n)[n0], mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2] ⊦

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⌝ < ⌜1⌝ + n∞ ↔ ¬ n∞ < ⌜n⌝ ; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream 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) ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = n; (s) ⊦ ¬ ∞ < n∞; (s) ⊦ # ⌜σ to n⌝ = ⌜#(σ to n)⌝ ; (s) ⊦ #(σ + σ1) = # σ + # σ1; (s) ⊦ ⌜n⌝ + n∞1 = ⌜n⌝ + n∞2 ↔ n∞1 = n∞2; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s)

Proof step 44 from proof lim-extractprefix (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 45:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] ≠ σf(n)[n0], mon(⌜σf⌝ ), ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n0 + # x⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n0 + # x] = (⌜x⌝ + σf(n1))[n0 + # x], ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2], ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

Rule argument

Additional information

Used terms: n0 + # x

Proof step 45 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 46:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), lim(⌜σf⌝ )[n0] = σf(n)[n0], (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n0 + # x] = (⌜x⌝ + σf(n1))[n0 + # x], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] ≠ σf(n)[n0], mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2] ⊦

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∞ + n∞1 < n∞1; (s) ⊦ ⌜n⌝ + n∞1 < ⌜n⌝ + n∞2 ↔ n∞1 < n∞2; (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 46 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 47:

  ⊦ ⌜m⌝ < # σ → (σ + σ1)[m] = σ[m]

• Proof step 48:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ [n0 + # x] = (⌜x⌝ + σf(n1))[n0 + # x], lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] ≠ σf(n)[n0], ⌜n0 + # x⌝ < # ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ , mon(⌜σf⌝ ), ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2], ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ ⌜m⌝ < # σ → (σ + σ1)[m] = σ1[m - (# σ).n]; (s) ⊦ ⌜m⌝ + n∞ = ∞ ↔ n∞ = ∞; (s) ⊦ ⌜n⌝ ≠ ∞; (s) n∞ ≠ ∞ ⊦ (⌜m⌝ + n∞).n = m + n∞.n; (s) ⊦ ⌜n⌝ .n = n; (s) ⊦ m + n - n + n1 = m - n1; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ # ⌜σ to n⌝ = ⌜#(σ to n)⌝ ; (s) ⊦ ¬ ∞ < n∞; (s) ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = n; (s) ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ ⌜n⌝ + n∞1 < ⌜n⌝ + n∞2 ↔ n∞1 < n∞2; (s) ⊦ ⌜n⌝ < ⌜1⌝ + n∞ ↔ ¬ n∞ < ⌜n⌝ ; (s) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ ¬ n < 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 Used terms: ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ , σ, n0 + # x

Proof step 47 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 48 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 49:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = (⌜x⌝ + σf(n1))[n0 + # x], (⌜x⌝ + σf(n1))[n0 + # x] ≠ σf(n)[n0], mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2] ⊦

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 < n; (s) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ ⌜n⌝ < ⌜1⌝ + n∞ ↔ ¬ n∞ < ⌜n⌝ ; (s) ⊦ ⌜n⌝ + n∞1 < ⌜n⌝ + n∞2 ↔ n∞1 < n∞2; (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) ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = n; (s) ⊦ ¬ ∞ < n∞; (s) ⊦ # ⌜σ to n⌝ = ⌜#(σ to n)⌝ ; (s) m < n, ¬ # σ < ⌜n⌝ ⊦ (σ to n)[m] = σ[m]; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ m + n < m + n0 ↔ n < n0; (s) ⊦ ⌜x⌝ [n] = x[n]; (s)

Proof step 49 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 50:

  ⊦ ¬ ⌜m⌝ < # σ → (σ + σ1)[m] = σ1[m - (# σ).n]

• Proof step 51:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0 + # x - (# ⌜x⌝ ).n], lim(⌜σf⌝ )[n0] = σf(n)[n0], σf(n1)[n0 + # x - (# ⌜x⌝ ).n] ≠ σf(n)[n0], mon(⌜σf⌝ ), ¬ ⌜n0 + # x⌝ < # ⌜x⌝ , ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2], ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < # σ → (σ + σ1)[m] = σ[m]; (s) ⊦ ⌜x⌝ [n] = x[n]; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s) ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; (s) ⊦ ¬ m + n < m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic 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 Used terms: ⌜x⌝ , σf(n1), n0 + # x

Proof step 50 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 51 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 52:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], σf(n1)[n0] ≠ σf(n)[n0], mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2] ⊦

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 c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ ¬ m + n < m; (s) ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s) ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) ⊦ m + n - m = n; (s) ⊦ ⌜n⌝ .n = n; (s)

Proof step 52 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 53:

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

• Proof step 54:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], lim(⌜σf⌝ )[n0] = σf(n)[n0], σf(n1)[n0] ≠ σf(n)[n0], mon(⌜σf⌝ ), ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2], ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦ n1 ≤ n ∧ ⌜n⌝ < # ⌜σf⌝

• Proof step 78:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], lim(⌜σf⌝ )[n0] = σf(n)[n0], σf(n1)[n0] ≠ σf(n)[n0], mon(⌜σf⌝ ), ⌜σf⌝ [n1] ⊑ ⌜σf⌝ [n], ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2], ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

Rule argument

Additional information

Used terms: ⌜σf⌝ , n1, n

Proof step 53 from proof lim-extractprefix (previous step) (back to tree)

Lemma mon-lem

Proof step 54 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 55:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], lim(⌜σf⌝ )[n0] = σf(n)[n0], σf(n1)[n0] ≠ σf(n)[n0], n < n1, mon(⌜σf⌝ ), ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2], ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

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 55 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 56:

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

• Proof step 57:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], σf(n1)[n0] ≠ σf(n)[n0], n < n1, mon(⌜σf⌝ ), ⌜σf⌝ [n] ⊑ ⌜σf⌝ [n1], ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2] ⊦

Rule argument

Additional information

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

Proof step 56 from proof lim-extractprefix (previous step) (back to tree)

Lemma mon-lem

Proof step 57 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 58:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], lim(⌜σf⌝ )[n0] = σf(n)[n0], σf(n1)[n0] ≠ σf(n)[n0], n < n1, mon(⌜σf⌝ ), σf(n) ⊑ σf(n1), ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2], ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

Rule argument

Additional information

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

Proof step 58 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 59:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 60:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], σf(n1)[n0] ≠ σf(n)[n0], n < n1, mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2], ∃ σ1. σf(n) + σ1 = σf(n1) ⊦

Rule argument

Additional information

Used terms: σf(n), σf(n1)

Proof step 59 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 60 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 61:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, σf(n) + σ0 = σf(n1), ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], lim(⌜σf⌝ )[n0] = σf(n)[n0], σf(n1)[n0] ≠ σf(n)[n0], n < n1, mon(⌜σf⌝ ), ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2], ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

Rule argument

Additional information

Proof step 61 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 62:

  ⊦ σ0 = σ1 ↔ # σ0 = # σ1 ∧ (∀ n. ⌜n⌝ < # σ0 → σ0[n] = σ1[n])

• Proof step 63:

  #(σf(n) + σ0) = # σf(n1), # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], σf(n1)[n0] ≠ σf(n)[n0], n < n1, mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2], ∀ n0. ⌜n0⌝ < #(σf(n) + σ0) → (σf(n) + σ0)[n0] = σf(n1)[n0] ⊦

Rule argument

Additional information

Used terms: σf(n) + σ0, σf(n1)

Proof step 62 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 63 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 64:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), # σf(n) + # σ0 = # σf(n1), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], lim(⌜σf⌝ )[n0] = σf(n)[n0], σf(n1)[n0] ≠ σf(n)[n0], n < n1, mon(⌜σf⌝ ), ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2], ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n2. ⌜n2⌝ < # σf(n1) → (σf(n) + σ0)[n2] = σf(n1)[n2] ⊦

Rule argument

Additional information

Used simplifier rules: append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ #(σ + σ1) = # σ + # σ1; (s)

Proof step 64 from proof lim-extractprefix (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 65:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # σf(n) + # σ0 = # σf(n1), (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], σf(n1)[n0] ≠ σf(n)[n0], n < n1, mon(⌜σf⌝ ), ⌜n0⌝ < # σf(n1) → (σf(n) + σ0)[n0] = σf(n1)[n0], ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2] ⊦

Rule argument

Additional information

Used terms: n0

Proof step 65 from proof lim-extractprefix (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 66:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), # σf(n) + # σ0 = # σf(n1), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], lim(⌜σf⌝ )[n0] = σf(n)[n0], σf(n1)[n0] ≠ σf(n)[n0], n < n1, ⌜n0⌝ < # σf(n1), mon(⌜σf⌝ ), ⌜n0⌝ < # σf(n1) → (σf(n) + σ0)[n0] = σf(n1)[n0], ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2], ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

• Proof step 75:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), # σf(n) + # σ0 = # σf(n1), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], lim(⌜σf⌝ )[n0] = σf(n)[n0], σf(n1)[n0] ≠ σf(n)[n0], n < n1, mon(⌜σf⌝ ), ¬ ⌜n0⌝ < # σf(n1), ⌜n0⌝ < # σf(n1) → (σf(n) + σ0)[n0] = σf(n1)[n0], ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2], ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

Rule argument

Additional information

Used formulas: ⌜n0⌝ < # σf(n1)

Proof step 66 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 67:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # σf(n) + # σ0 = # σf(n1), (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], (σf(n) + σ0)[n0] = σf(n1)[n0], σf(n1)[n0] ≠ σf(n)[n0], ⌜n0⌝ < # σf(n1), n < n1, mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2] ⊦

Rule argument

Additional information

Proof step 67 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 68:

  ⊦ ⌜m⌝ < # σ → (σ + σ1)[m] = σ[m]

• Proof step 69:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), # σf(n) + # σ0 = # σf(n1), (σf(n) + σ0)[n0] = σf(n1)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], lim(⌜σf⌝ )[n0] = σf(n)[n0], σf(n1)[n0] ≠ σf(n)[n0], n < n1, ⌜n0⌝ < # σf(n1), mon(⌜σf⌝ ), ¬ ⌜n0⌝ < # σf(n), ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2], ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

Rule argument

Additional information

Used terms: σf(n), σ0, n0

Proof step 68 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 69 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 70:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # σf(n) + # σ0 = # σf(n1), (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], σ0[n0 - (# σf(n)).n] = σf(n1)[n0], σf(n1)[n0] ≠ σf(n)[n0], ⌜n0⌝ < # σf(n1), n < n1, # σf(n) < # σf(n1), mon(⌜σf⌝ ), ¬ ⌜n0⌝ < # σf(n), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2] ⊦

Rule argument

Additional information

Used simplifier rules: lels-trans: ⊦ ¬ n∞1 < n∞ ∧ n∞1 < n∞2 → n∞ < n∞2; (forward) from specification nat-inf append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ ¬ ⌜m⌝ < # σ → (σ + σ1)[m] = σ1[m - (# σ).n]; (s)

Proof step 70 from proof lim-extractprefix (previous step) (back to tree)

User applied rule subst equation

And obtained the following new goals

• Proof step 71:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n) + # σ0, (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), σ0[n0 - (# σf(n)).n] = σf(n1)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], lim(⌜σf⌝ )[n0] = σf(n)[n0], σf(n1)[n0] ≠ σf(n)[n0], # σf(n) < # σf(n) + # σ0, n < n1, ⌜n0⌝ < # σf(n) + # σ0, mon(⌜σf⌝ ), ¬ ⌜n0⌝ < # σf(n), ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2], ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

Rule argument

Additional information

Proof step 71 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 72:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n) + # σ0, (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), σ0[n0 - (# σf(n)).n] = σf(n1)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], lim(⌜σf⌝ )[n0] = σf(n)[n0], # σf(n) ≠ ∞, σf(n1)[n0] ≠ σf(n)[n0], σ0 ≠ [], n < n1, ⌜n0⌝ < # σf(n) + # σ0, mon(⌜σf⌝ ), ¬ ⌜n0⌝ < # σf(n), ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2], ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

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 ⊦ # σ = ⌜0⌝ ↔ σ = []; (s) ⊦ n∞1 < n∞ + n∞1 ↔ n∞ ≠ ⌜0⌝ ∧ n∞1 ≠ ∞; (s) lels-trans: ⊦ ¬ n∞1 < n∞ ∧ n∞1 < n∞2 → n∞ < n∞2; (forward) from specification nat-inf

Proof step 72 from proof lim-extractprefix (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 73:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # σf(n) = ⌜n3⌝ , (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n) + # σ0, lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], σ0[n0 - n3] = σf(n1)[n0], # σf(n) ≠ ∞, σf(n1)[n0] ≠ σf(n)[n0], σ0 ≠ [], ⌜n0⌝ < # σf(n) + # σ0, n < n1, mon(⌜σf⌝ ), ¬ ⌜n0⌝ < # σf(n), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2] ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: # σf(n), n3

Proof step 73 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 74:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # σf(n) = ⌜n3⌝ , (⌜n0⌝ + ⌜1⌝ ) + # σ1 = ⌜n3⌝ , ⌜n0⌝ + ⌜1⌝ + # σ = ⌜n3⌝ + # σ0, lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], σ0[n0 - n3] = σf(n1)[n0], σf(n1)[n0] ≠ σf(n)[n0], σ0 ≠ [], ⌜n0⌝ < ⌜n3⌝ + # σ0, n < n1, mon(⌜σf⌝ ), n3 ≤ n0, ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2] ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < 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 ⊦ # σ = ⌜0⌝ ↔ σ = []; (s) ⊦ ⌜n⌝ < n∞ + ⌜n⌝ ↔ n∞ ≠ ⌜0⌝ ; (s) ⊦ ⌜n⌝ ≠ ∞; (s) lels-trans: ⊦ ¬ n∞1 < n∞ ∧ n∞1 < n∞2 → n∞ < n∞2; (forward) from specification nat-inf

Proof step 74 from proof lim-extractprefix (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ ⌜n⌝ + n∞ = ⌜n⌝ ↔ n∞ = ⌜0⌝ ; (s) ⊦ ⌜n⌝ + n∞ = ⌜0⌝ ↔ n∞ = ⌜0⌝ ∧ n = 0; (ws) ⊦ n < m → (⌜(* n)⌝ = ⌜(* m)⌝ + n∞2 ↔ false); (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 ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n0, n3, n4

Proof step 75 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 76:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # σf(n) + # σ0 = # σf(n1), (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], σf(n1)[n0] ≠ σf(n)[n0], n < n1, mon(⌜σf⌝ ), ¬ ⌜n0⌝ < # σf(n1), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2] ⊦

Rule argument

Additional information

Proof step 76 from proof lim-extractprefix (previous step) (back to tree)

User applied rule subst equation

And obtained the following new goals

• Proof step 77:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), ((⌜n0⌝ + ⌜1⌝ ) + # σ1) + # σ0 = # σf(n1), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], lim(⌜σf⌝ )[n0] = σf(n)[n0], σf(n1)[n0] ≠ σf(n)[n0], n < n1, mon(⌜σf⌝ ), ¬ ⌜n0⌝ < # σf(n1), ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2], ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

Rule argument

Additional information

Proof step 77 from proof lim-extractprefix (previous step) (back to tree)

User applied rule subst equation

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ n → (⌜(* n)⌝ + n∞1 = ⌜(* m)⌝ + n∞2 ↔ ⌜n - m⌝ + n∞1 = n∞2); (s) ⊦ n∞ + ⌜0⌝ = n∞; (s) ⊦ ⌜n⌝ + n∞1 = ⌜n⌝ + n∞2 ↔ n∞1 = n∞2; (s) ⊦ ⌜n⌝ < ⌜1⌝ + n∞ ↔ ¬ n∞ < ⌜n⌝ ; (s) ⊦ ¬ n∞ + n∞1 < n∞1; (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 78 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 79:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], σf(n1)[n0] ≠ σf(n)[n0], mon(⌜σf⌝ ), σf(n1) ⊑ σf(n), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2] ⊦

Rule argument

Additional information

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

Proof step 79 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 80:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 81:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], lim(⌜σf⌝ )[n0] = σf(n)[n0], σf(n1)[n0] ≠ σf(n)[n0], mon(⌜σf⌝ ), ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2], ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∃ σ1. σf(n1) + σ1 = σf(n) ⊦

Rule argument

Additional information

Used terms: σf(n1), σf(n)

Proof step 80 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 81 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 82:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, σf(n1) + σ0 = σf(n), (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], σf(n1)[n0] ≠ σf(n)[n0], mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2] ⊦

Rule argument

Additional information

Proof step 82 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 83:

  ⊦ σ0 = σ1 ↔ # σ0 = # σ1 ∧ (∀ n. ⌜n⌝ < # σ0 → σ0[n] = σ1[n])

• Proof step 84:

  #(σf(n1) + σ0) = # σf(n), # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], lim(⌜σf⌝ )[n0] = σf(n)[n0], σf(n1)[n0] ≠ σf(n)[n0], mon(⌜σf⌝ ), ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2], ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n0. ⌜n0⌝ < #(σf(n1) + σ0) → (σf(n1) + σ0)[n0] = σf(n)[n0] ⊦

Rule argument

Additional information

Used terms: σf(n1) + σ0, σf(n)

Proof step 83 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 84 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 85:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), # σf(n1) + # σ0 = # σf(n), lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], σf(n1)[n0] ≠ σf(n)[n0], mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2], ∀ n2. ⌜n2⌝ < # σf(n) → (σf(n1) + σ0)[n2] = σf(n)[n2] ⊦

Rule argument

Additional information

Used simplifier rules: append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ #(σ + σ1) = # σ + # σ1; (s)

Proof step 85 from proof lim-extractprefix (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 86:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, # σf(n1) + # σ0 = # σf(n), ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], lim(⌜σf⌝ )[n0] = σf(n)[n0], σf(n1)[n0] ≠ σf(n)[n0], mon(⌜σf⌝ ), ⌜n0⌝ < # σf(n) → (σf(n1) + σ0)[n0] = σf(n)[n0], ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2], ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

Rule argument

Additional information

Used terms: n0

Proof step 86 from proof lim-extractprefix (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 87:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), # σf(n1) + # σ0 = # σf(n), lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], σf(n1)[n0] ≠ σf(n)[n0], ⌜n0⌝ < # σf(n), mon(⌜σf⌝ ), ⌜n0⌝ < # σf(n) → (σf(n1) + σ0)[n0] = σf(n)[n0], ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2] ⊦

• Proof step 96:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), # σf(n1) + # σ0 = # σf(n), lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], σf(n1)[n0] ≠ σf(n)[n0], mon(⌜σf⌝ ), ¬ ⌜n0⌝ < # σf(n), ⌜n0⌝ < # σf(n) → (σf(n1) + σ0)[n0] = σf(n)[n0], ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2] ⊦

Rule argument

Additional information

Used formulas: ⌜n0⌝ < # σf(n)

Proof step 87 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 88:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, # σf(n1) + # σ0 = # σf(n), ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], lim(⌜σf⌝ )[n0] = σf(n)[n0], (σf(n1) + σ0)[n0] = σf(n)[n0], σf(n1)[n0] ≠ σf(n)[n0], ⌜n0⌝ < # σf(n), mon(⌜σf⌝ ), ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2], ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

Rule argument

Additional information

Proof step 88 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 89:

  ⊦ ⌜m⌝ < # σ → (σ + σ1)[m] = σ[m]

• Proof step 90:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), # σf(n1) + # σ0 = # σf(n), (σf(n1) + σ0)[n0] = σf(n)[n0], lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], σf(n1)[n0] ≠ σf(n)[n0], ⌜n0⌝ < # σf(n), mon(⌜σf⌝ ), ¬ ⌜n0⌝ < # σf(n1), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2] ⊦

Rule argument

Additional information

Used terms: σf(n1), σ0, n0

Proof step 89 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 90 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 91:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, # σf(n1) + # σ0 = # σf(n), ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], lim(⌜σf⌝ )[n0] = σf(n)[n0], σ0[n0 - (# σf(n1)).n] = σf(n)[n0], σf(n1)[n0] ≠ σf(n)[n0], ⌜n0⌝ < # σf(n), # σf(n1) < # σf(n), mon(⌜σf⌝ ), ¬ ⌜n0⌝ < # σf(n1), ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2], ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

Rule argument

Additional information

Used simplifier rules: lels-trans: ⊦ ¬ n∞1 < n∞ ∧ n∞1 < n∞2 → n∞ < n∞2; (forward) from specification nat-inf append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ ¬ ⌜m⌝ < # σ → (σ + σ1)[m] = σ1[m - (# σ).n]; (s)

Proof step 91 from proof lim-extractprefix (previous step) (back to tree)

User applied rule subst equation

And obtained the following new goals

• Proof step 92:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n1) + # σ0, ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), σ0[n0 - (# σf(n1)).n] = σf(n)[n0], lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], σf(n1)[n0] ≠ σf(n)[n0], # σf(n1) < # σf(n1) + # σ0, ⌜n0⌝ < # σf(n1) + # σ0, mon(⌜σf⌝ ), ¬ ⌜n0⌝ < # σf(n1), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2] ⊦

Rule argument

Additional information

Proof step 92 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 93:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n1) + # σ0, ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), σ0[n0 - (# σf(n1)).n] = σf(n)[n0], lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], # σf(n1) ≠ ∞, σf(n1)[n0] ≠ σf(n)[n0], σ0 ≠ [], ⌜n0⌝ < # σf(n1) + # σ0, mon(⌜σf⌝ ), ¬ ⌜n0⌝ < # σf(n1), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2] ⊦

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 ⊦ # σ = ⌜0⌝ ↔ σ = []; (s) ⊦ n∞1 < n∞ + n∞1 ↔ n∞ ≠ ⌜0⌝ ∧ n∞1 ≠ ∞; (s) lels-trans: ⊦ ¬ n∞1 < n∞ ∧ n∞1 < n∞2 → n∞ < n∞2; (forward) from specification nat-inf

Proof step 93 from proof lim-extractprefix (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 94:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, # σf(n1) = ⌜n3⌝ , ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n1) + # σ0, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], lim(⌜σf⌝ )[n0] = σf(n)[n0], σ0[n0 - n3] = σf(n)[n0], # σf(n1) ≠ ∞, σf(n1)[n0] ≠ σf(n)[n0], σ0 ≠ [], ⌜n0⌝ < # σf(n1) + # σ0, mon(⌜σf⌝ ), ¬ ⌜n0⌝ < # σf(n1), ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2], ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: # σf(n1), n3

Proof step 94 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 95:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, # σf(n1) = ⌜n3⌝ , ⌜n0⌝ + ⌜1⌝ + # σ = ⌜n3⌝ , (⌜n0⌝ + ⌜1⌝ ) + # σ1 = ⌜n3⌝ + # σ0, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], lim(⌜σf⌝ )[n0] = σf(n)[n0], σ0[n0 - n3] = σf(n)[n0], σf(n1)[n0] ≠ σf(n)[n0], σ0 ≠ [], ⌜n0⌝ < ⌜n3⌝ + # σ0, mon(⌜σf⌝ ), n3 ≤ n0, ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2], ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < 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 ⊦ # σ = ⌜0⌝ ↔ σ = []; (s) ⊦ ⌜n⌝ < n∞ + ⌜n⌝ ↔ n∞ ≠ ⌜0⌝ ; (s) ⊦ ⌜n⌝ ≠ ∞; (s) lels-trans: ⊦ ¬ n∞1 < n∞ ∧ n∞1 < n∞2 → n∞ < n∞2; (forward) from specification nat-inf

Proof step 95 from proof lim-extractprefix (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ ⌜n⌝ + n∞ = ⌜n⌝ ↔ n∞ = ⌜0⌝ ; (s) ⊦ ⌜n⌝ + n∞ = ⌜0⌝ ↔ n∞ = ⌜0⌝ ∧ n = 0; (ws) ⊦ n < m → (⌜(* n)⌝ = ⌜(* m)⌝ + n∞2 ↔ false); (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 ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n0, n3, n4

Proof step 96 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 97:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, # σf(n1) + # σ0 = # σf(n), ⌜n0⌝ + ⌜1⌝ + # σ = # σf(n1), (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], lim(⌜σf⌝ )[n0] = σf(n)[n0], σf(n1)[n0] ≠ σf(n)[n0], mon(⌜σf⌝ ), ¬ ⌜n0⌝ < # σf(n), ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2], ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

Rule argument

Additional information

Proof step 97 from proof lim-extractprefix (previous step) (back to tree)

User applied rule subst equation

And obtained the following new goals

• Proof step 98:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, (⌜n0⌝ + ⌜1⌝ ) + # σ1 = # σf(n), (⌜n0⌝ + ⌜1⌝ + # σ) + # σ0 = # σf(n), lim(⌜σf⌝ )[n0] = σf(n)[n0], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n0 + # x] = σf(n1)[n0], σf(n1)[n0] ≠ σf(n)[n0], mon(⌜σf⌝ ), ¬ ⌜n0⌝ < # σf(n), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n1. ¬ ⌜n0⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜σf⌝ ) to n0 + 1⌝ + σ1)[n1] = σf(n)[n1], ∀ n2. ¬ ⌜n0⌝ + ⌜# x⌝ + # σ < ⌜n2⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n0 + 1 + # x⌝ + σ)[n2] = (⌜x⌝ + σf(n1))[n2] ⊦

Rule argument

Additional information

Proof step 98 from proof lim-extractprefix (previous step) (back to tree)

User applied rule subst equation

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ n → (⌜(* n)⌝ + n∞1 = ⌜(* m)⌝ + n∞2 ↔ ⌜n - m⌝ + n∞1 = n∞2); (s) ⊦ n∞ + ⌜0⌝ = n∞; (s) ⊦ ⌜n⌝ + n∞1 = ⌜n⌝ + n∞2 ↔ n∞1 = n∞2; (s) ⊦ ⌜n⌝ < ⌜1⌝ + n∞ ↔ ¬ n∞ < ⌜n⌝ ; (s) ⊦ ¬ n∞ + n∞1 < n∞1; (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 99 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 100:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n] ≠ x[n], x ≠ [], n < # x, mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s) ⊦ ⌜x⌝ [n] = x[n]; (s)

Proof step 100 from proof lim-extractprefix (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 101:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n] ≠ x[n], x ≠ [], n < # x, mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∃ n0. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n + 1 + # x⌝ ⊑ ⌜x⌝ + σf(n0) ⊦

Rule argument

Additional information

Used terms: n + 1 + # x

Proof step 101 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 102:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n] ≠ x[n], x ≠ [], n < # x, mon(⌜σf⌝ ), ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n + 1 + # x⌝ ⊑ ⌜x⌝ + σf(n0), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

Rule argument

Additional information

Proof step 102 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 103:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 104:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n] ≠ x[n], x ≠ [], n < # x, mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∃ σ1. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n + 1 + # x⌝ + σ1 = ⌜x⌝ + σf(n0) ⊦

Rule argument

Additional information

Used terms: ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n + 1 + # x⌝ , ⌜x⌝ + σf(n0)

Proof step 103 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 104 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 105:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n + 1 + # x⌝ + σ1 = ⌜x⌝ + σf(n0), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n] ≠ x[n], x ≠ [], n < # x, mon(⌜σf⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

Rule argument

Additional information

Proof step 105 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 106:

  ⊦ σ0 = σ1 ↔ # σ0 = # σ1 ∧ (∀ n. ⌜n⌝ < # σ0 → σ0[n] = σ1[n])

• Proof step 107:

  #(⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n + 1 + # x⌝ + σ1) = #(⌜x⌝ + σf(n0)), # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n] ≠ x[n], x ≠ [], n < # x, mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n1. ⌜n1⌝ < #(⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n + 1 + # x⌝ + σ1) → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n + 1 + # x⌝ + σ1)[n1] = (⌜x⌝ + σf(n0))[n1] ⊦

Rule argument

Additional information

Used terms: ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n + 1 + # x⌝ + σ1, ⌜x⌝ + σf(n0)

Proof step 106 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 107 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 108:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, ⌜n⌝ + ⌜1⌝ + # σ1 = # σf(n0), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n] ≠ x[n], x ≠ [], n < # x, mon(⌜σf⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n1. ¬ ⌜n⌝ + ⌜# x⌝ + # σ1 < ⌜n1⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n + 1 + # x⌝ + σ1)[n1] = (⌜x⌝ + σf(n0))[n1] ⊦

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⌝ < ⌜1⌝ + n∞ ↔ ¬ n∞ < ⌜n⌝ ; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream 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) ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = n; (s) ⊦ ¬ ∞ < n∞; (s) ⊦ # ⌜σ to n⌝ = ⌜#(σ to n)⌝ ; (s) ⊦ #(σ + σ1) = # σ + # σ1; (s) ⊦ ⌜n⌝ + n∞1 = ⌜n⌝ + n∞2 ↔ n∞1 = n∞2; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s)

Proof step 108 from proof lim-extractprefix (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 109:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, ⌜n⌝ + ⌜1⌝ + # σ1 = # σf(n0), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n] ≠ x[n], x ≠ [], n < # x, mon(⌜σf⌝ ), ¬ ⌜n⌝ + ⌜# x⌝ + # σ1 < ⌜n⌝ → (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n + 1 + # x⌝ + σ1)[n] = (⌜x⌝ + σf(n0))[n], ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ) ⊦

Rule argument

Additional information

Used terms: n

Proof step 109 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 110:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, ⌜n⌝ + ⌜1⌝ + # σ1 = # σf(n0), (⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n + 1 + # x⌝ + σ1)[n] = (⌜x⌝ + σf(n0))[n], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n] ≠ x[n], x ≠ [], n < # x, mon(⌜σf⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

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 < n → (⌜n⌝ + n∞ < ⌜m⌝ ↔ false); (s)

Proof step 110 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 111:

  ⊦ ⌜m⌝ < # σ → (σ + σ1)[m] = σ[m]

• Proof step 112:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, ⌜n⌝ + ⌜1⌝ + # σ1 = # σf(n0), ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n + 1 + # x⌝ [n] = (⌜x⌝ + σf(n0))[n], lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n] ≠ x[n], x ≠ [], n < # x, ⌜n⌝ < # ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n + 1 + # x⌝ , mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ ⌜m⌝ < # σ → (σ + σ1)[m] = σ1[m - (# σ).n]; (s) ⊦ ⌜m⌝ + n∞ = ∞ ↔ n∞ = ∞; (s) ⊦ ⌜n⌝ ≠ ∞; (s) n∞ ≠ ∞ ⊦ (⌜m⌝ + n∞).n = m + n∞.n; (s) ⊦ ⌜n⌝ .n = n; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ # ⌜σ to n⌝ = ⌜#(σ to n)⌝ ; (s) ⊦ ¬ ∞ < n∞; (s) ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = n; (s) ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ m < n → ⌜m⌝ < ⌜n⌝ + n∞0; (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 Used terms: ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n + 1 + # x⌝ , σ1, n

Proof step 111 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 112 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 113:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ∞, ⌜n⌝ + ⌜1⌝ + # σ1 = # σf(n0), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )[n] = (⌜x⌝ + σf(n0))[n], (⌜x⌝ + σf(n0))[n] ≠ x[n], x ≠ [], n < # x, mon(⌜σf⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

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 < n → ⌜m⌝ < ⌜n⌝ + n∞0; (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) ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = n; (s) ⊦ ¬ ∞ < n∞; (s) ⊦ # ⌜σ to n⌝ = ⌜#(σ to n)⌝ ; (s) m < n, ¬ # σ < ⌜n⌝ ⊦ (σ to n)[m] = σ[m]; (s) ⊦ ¬ m + n < m; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ ⌜x⌝ [n] = x[n]; (s)

Proof step 113 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 114:

  ⊦ ⌜m⌝ < # σ → (σ + σ1)[m] = σ[m]

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ [n] = x[n]; (s) ⊦ ¬ ⌜m⌝ < # σ → (σ + σ1)[m] = σ1[m - (# σ).n]; (s) ⊦ ⌜n⌝ .n = n; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m < n → (n ≤ m ↔ false); (s) Used terms: ⌜x⌝ , σf(n0), n

Proof step 114 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 115 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 116:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) ≠ ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n) ⊦

Rule argument

Additional information

Proof step 116 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 117:

  ⊦ n∞ = ∞ ↔ ¬(∃ n. n∞ = ⌜n⌝ )

• Proof step 118:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∃ n. # lim(⌜σf⌝ ) = ⌜n⌝ ⊦

Rule argument

Additional information

Used terms: # lim(⌜σf⌝ )

Proof step 117 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 118 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 119:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ⌜n⌝ , lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)

Proof step 119 from proof lim-extractprefix (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 120:

  # lim(⌜σf⌝ ) = ⌜n⌝ , # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∃ n0. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n + 1 + # x⌝ ⊑ ⌜x⌝ + σf(n0) ⊦

Rule argument

Additional information

Used terms: n + 1 + # x

Proof step 120 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 121:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ⌜n⌝ , lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n + 1 + # x⌝ ⊑ ⌜x⌝ + σf(n0), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)) ⊦

Rule argument

Additional information

Proof step 121 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 122:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 123:

  # lim(⌜σf⌝ ) = ⌜n⌝ , # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∃ σ1. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n + 1 + # x⌝ + σ1 = ⌜x⌝ + σf(n0) ⊦

Rule argument

Additional information

Used terms: ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n + 1 + # x⌝ , ⌜x⌝ + σf(n0)

Proof step 122 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 123 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 124:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ⌜n⌝ , ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n + 1 + # x⌝ + σ1 = ⌜x⌝ + σf(n0), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)) ⊦

Rule argument

Additional information

Proof step 124 from proof lim-extractprefix (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 125:

  # lim(⌜σf⌝ ) = ⌜n⌝ , # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n + 1 + # x⌝ + σ1 = ⌜x⌝ + σf(n0), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), σf(n0) ⊑ lim(⌜σf⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n) ⊦

Rule argument

Additional information

Used terms: n0

Proof step 125 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 126:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 127:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # lim(⌜σf⌝ ) = ⌜n⌝ , ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n + 1 + # x⌝ + σ1 = ⌜x⌝ + σf(n0), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)), ∃ σ1. σf(n0) + σ1 = lim(⌜σf⌝ ) ⊦

Rule argument

Additional information

Used terms: σf(n0), lim(⌜σf⌝ )

Proof step 126 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 127 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 128:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n + 1 + # x⌝ + σ1 = ⌜x⌝ + σf(n0), σf(n0) + σ = lim(⌜σf⌝ ), # σf(n0) + # σ = ⌜n⌝ , lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + σf(n0) + σ, mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ σf(n0) + σ, ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜σf(n0) + σ to m⌝ ⊑ σf(n)) ⊦

Rule argument

Additional information

Used simplifier rules: append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ #(σ + σ1) = # σ + # σ1; (s)

Proof step 128 from proof lim-extractprefix (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 129:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # σf(n0) + # σ = ⌜n⌝ , σf(n0) + σ = lim(⌜σf⌝ ), ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n + 1 + # x⌝ + σ1 = ⌜x⌝ + σf(n0), σ = ⌜x0⌝ , lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + σf(n0) + σ, mon(⌜σf⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜σf(n0) + σ to m⌝ ⊑ σf(n)), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ σf(n0) + σ, ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n) ⊦

Rule argument

Additional information

Used simplifier rules: stream('a) freely generated by ⌜ ⌝, ⌜ ⌝;; ⊦ # ⌜nf⌝ = ∞; (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 ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 129 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 130:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n + 1 + # x⌝ + σ1 = ⌜x⌝ + σf(n0), σf(n0) + ⌜x0⌝ = lim(⌜σf⌝ ), # σf(n0) + ⌜# x0⌝ = ⌜n⌝ , lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + σf(n0) + ⌜x0⌝ , mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ σf(n0) + ⌜x0⌝ , ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜σf(n0) + ⌜x0⌝ to m⌝ ⊑ σf(n)) ⊦

Rule argument

Additional information

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

Proof step 130 from proof lim-extractprefix (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 131:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, # σf(n0) + ⌜# x0⌝ = ⌜n⌝ , σf(n0) + ⌜x0⌝ = lim(⌜σf⌝ ), ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n + 1 + # x⌝ + σ1 = ⌜x⌝ + σf(n0), ⌜x1⌝ = σf(n0), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + σf(n0) + ⌜x0⌝ , mon(⌜σf⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜σf(n0) + ⌜x0⌝ to m⌝ ⊑ σf(n)), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ σf(n0) + ⌜x0⌝ , ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n) ⊦

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 131 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 132:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to (# x1 + # x0) + 1 + # x⌝ + σ1 = ⌜x⌝ + ⌜x1⌝ , ⌜x1⌝ + ⌜x0⌝ = lim(⌜σf⌝ ), ⌜x1⌝ = σf(n0), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + ⌜x1⌝ + ⌜x0⌝ , mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ ⌜x1⌝ + ⌜x0⌝ , ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ¬ # x1 + # x0 < m → (∃ n. ⌜⌜x1⌝ + ⌜x0⌝ to m⌝ ⊑ σf(n)) ⊦

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 132 from proof lim-extractprefix (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 133:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, ⌜x1⌝ + ⌜x0⌝ = lim(⌜σf⌝ ), ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to (# x1 + # x0) + 1 + # x⌝ + σ1 = ⌜x⌝ + ⌜x1⌝ , σ1 = ⌜x2⌝ , ⌜x1⌝ = σf(n0), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + ⌜x1⌝ + ⌜x0⌝ , mon(⌜σf⌝ ), ∀ m. ¬ # x1 + # x0 < m → (∃ n. ⌜⌜x1⌝ + ⌜x0⌝ to m⌝ ⊑ σf(n)), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ ⌜x1⌝ + ⌜x0⌝ , ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n) ⊦

Rule argument

Additional information

Used simplifier rules: stream('a) freely generated by ⌜ ⌝, ⌜ ⌝;; ⊦ σ0 + ⌜x⌝ = σ + ⌜nf0⌝ ↔ σ0 = σ + ⌜nf0⌝ ; (s) ⊦ ⌜x⌝ ≠ σ + ⌜nf0⌝ ; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream

Proof step 133 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 134:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ∞, ⌜x1⌝ + ⌜x0⌝ = lim(⌜σf⌝ ), (lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to (# x1 + # x0) + 1 + # x) + x2 = x + x1, ⌜x1⌝ = σf(n0), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + ⌜x1⌝ + ⌜x0⌝ , mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n), ∀ n. σf(n) ⊑ ⌜x1⌝ + ⌜x0⌝ , ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ¬ # x1 + # x0 < m → (∃ n. ⌜⌜x1⌝ + ⌜x0⌝ to m⌝ ⊑ σf(n)) ⊦

Rule argument

Additional information

Used simplifier rules: append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ ⌜x⌝ + ⌜y⌝ = ⌜x0⌝ + ⌜z⌝ ↔ x + y = x0 + z; (s)

Proof step 134 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 135:

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

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) ⊦ ¬ ∞ < n∞; (s) ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = n; (s) ⊦ #(x + y) = # x + # y; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ m + n = m + n0 ↔ n = n0; (s) ⊦ m = m + n ↔ n = 0; (s) ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws) ⊦ # x = 0 ↔ x = []; (s) ⊦ m + 1 ≠ 0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: (lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to (# x1 + # x0) + 1 + # x) + x2, x + x1

Proof step 135 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 136 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 137:

  ⊦ n∞ = ∞ ↔ ¬(∃ n. n∞ = ⌜n⌝ )

• Proof step 138:

  lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ¬ # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n)), ∃ n. # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ⌜n⌝ ⊦

Rule argument

Additional information

Used terms: # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )

Proof step 137 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 138 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 139:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ⌜n⌝ , lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n)) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)

Proof step 139 from proof lim-extractprefix (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 140:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ⌜n⌝ , # lim(⌜σf⌝ ) = ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n)), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)) ⊦

• Proof step 156:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ⌜n⌝ , # lim(⌜σf⌝ ) ≠ ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n)), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)) ⊦

Rule argument

Additional information

Used formulas: # lim(⌜σf⌝ ) = ∞

Proof step 140 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 141:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ⌜n⌝ , # lim(⌜σf⌝ ) = ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n)), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ ∞ < n∞; (s)

Proof step 141 from proof lim-extractprefix (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 142:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ⌜n⌝ , lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n)), ∃ n0. ⌜lim(⌜σf⌝ ) to n + 1 + # x⌝ ⊑ σf(n0) ⊦

Rule argument

Additional information

Used terms: n + 1 + # x

Proof step 142 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 143:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ⌜n⌝ , # lim(⌜σf⌝ ) = ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ⌜lim(⌜σf⌝ ) to n + 1 + # x⌝ ⊑ σf(n0), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n)), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n) ⊦

Rule argument

Additional information

Proof step 143 from proof lim-extractprefix (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 144:

  # lim(⌜σf⌝ ) = ∞, # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ⌜n⌝ , lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ⌜lim(⌜σf⌝ ) to n + 1 + # x⌝ ⊑ σf(n0), ⌜x⌝ + σf(n0) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n)) ⊦

Rule argument

Additional information

Used terms: n0

Proof step 144 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 145:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 146:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ⌜n⌝ , # lim(⌜σf⌝ ) = ∞, lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ⌜lim(⌜σf⌝ ) to n + 1 + # x⌝ ⊑ σf(n0), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n)), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∃ σ1. (⌜x⌝ + σf(n0)) + σ1 = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ⊦

Rule argument

Additional information

Used terms: ⌜x⌝ + σf(n0), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )

Proof step 145 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 146 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 147:

  # lim(⌜σf⌝ ) = ∞, (⌜x⌝ + σf(n0)) + σ1 = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ⌜# x⌝ + # σf(n0) + # σ1 = ⌜n⌝ , σf(n0) + σ1 ≠ lim(⌜σf⌝ ), mon(⌜σf⌝ ), ⌜lim(⌜σf⌝ ) to n + 1 + # x⌝ ⊑ σf(n0), ∀ m. ¬ n < m → (∃ n. ⌜(⌜x⌝ + σf(n0)) + σ1 to m⌝ ⊑ ⌜x⌝ + σf(n)), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n1. σf(n1) ⊑ σf(n0) + σ1 ⊦

Rule argument

Additional information

Used simplifier rules: append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ ⌜x⌝ + σ1 ⊑ ⌜x⌝ + σ2 ↔ σ1 ⊑ σ2; (s) ⊦ # x = # y → (⌜x⌝ + σ = ⌜y⌝ + σ0 ↔ x = y ∧ σ = σ0); (s) ⊦ #(σ + σ1) = # σ + # σ1; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s)

Proof step 147 from proof lim-extractprefix (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 148:

  # lim(⌜σf⌝ ) = ∞, ⌜# x⌝ + # σf(n0) + # σ1 = ⌜n⌝ , (⌜x⌝ + σf(n0)) + σ1 = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), σ1 = ⌜x0⌝ , σf(n0) + σ1 ≠ lim(⌜σf⌝ ), mon(⌜σf⌝ ), ⌜lim(⌜σf⌝ ) to n + 1 + # x⌝ ⊑ σf(n0), ∀ n1. σf(n1) ⊑ σf(n0) + σ1, ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜(⌜x⌝ + σf(n0)) + σ1 to m⌝ ⊑ ⌜x⌝ + σf(n)) ⊦

Rule argument

Additional information

Used simplifier rules: stream('a) freely generated by ⌜ ⌝, ⌜ ⌝;; ⊦ # ⌜nf⌝ = ∞; (s) ⊦ ⌜m⌝ + ∞ = ∞; (s) ⊦ 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)

Proof step 148 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 149:

  # lim(⌜σf⌝ ) = ∞, (⌜x⌝ + σf(n0)) + ⌜x0⌝ = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ⌜# x⌝ + # σf(n0) + ⌜# x0⌝ = ⌜n⌝ , σf(n0) + ⌜x0⌝ ≠ lim(⌜σf⌝ ), mon(⌜σf⌝ ), ⌜lim(⌜σf⌝ ) to n + 1 + # x⌝ ⊑ σf(n0), ∀ m. ¬ n < m → (∃ n. ⌜(⌜x⌝ + σf(n0)) + ⌜x0⌝ to m⌝ ⊑ ⌜x⌝ + σf(n)), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n1. σf(n1) ⊑ σf(n0) + ⌜x0⌝ ⊦

Rule argument

Additional information

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

Proof step 149 from proof lim-extractprefix (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 150:

  # lim(⌜σf⌝ ) = ∞, ⌜# x⌝ + # σf(n0) + ⌜# x0⌝ = ⌜n⌝ , (⌜x⌝ + σf(n0)) + ⌜x0⌝ = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ⌜x1⌝ = σf(n0), σf(n0) + ⌜x0⌝ ≠ lim(⌜σf⌝ ), mon(⌜σf⌝ ), ⌜lim(⌜σf⌝ ) to n + 1 + # x⌝ ⊑ σf(n0), ∀ n1. σf(n1) ⊑ σf(n0) + ⌜x0⌝ , ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜(⌜x⌝ + σf(n0)) + ⌜x0⌝ to m⌝ ⊑ ⌜x⌝ + σf(n)) ⊦

Rule argument

Additional information

Used simplifier rules: stream('a) freely generated by ⌜ ⌝, ⌜ ⌝;; ⊦ # ⌜nf⌝ = ∞; (s) ⊦ ⌜m⌝ + ∞ = ∞; (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)

Proof step 150 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 151:

  # lim(⌜σf⌝ ) = ∞, (⌜x⌝ + ⌜x1⌝ ) + ⌜x0⌝ = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ⌜x1⌝ = σf(n0), ⌜x1⌝ + ⌜x0⌝ ≠ lim(⌜σf⌝ ), mon(⌜σf⌝ ), ⌜lim(⌜σf⌝ ) to (# x + # x1 + # x0) + 1 + # x⌝ ⊑ ⌜x1⌝ , ∀ m. ¬ # x + # x1 + # x0 < m → (∃ n. ⌜(⌜x⌝ + ⌜x1⌝ ) + ⌜x0⌝ to m⌝ ⊑ ⌜x⌝ + σf(n)), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n1. σf(n1) ⊑ ⌜x1⌝ + ⌜x0⌝ ⊦

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 ⊦ ⌜n1⌝ + ⌜n⌝ + ⌜m⌝ = ⌜n0⌝ ↔ n1 + n + m = n0; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s)

Proof step 151 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 152:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 153:

  # lim(⌜σf⌝ ) = ∞, (⌜x⌝ + ⌜x1⌝ ) + ⌜x0⌝ = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ⌜x1⌝ = σf(n0), ⌜x1⌝ + ⌜x0⌝ ≠ lim(⌜σf⌝ ), mon(⌜σf⌝ ), ∀ n1. σf(n1) ⊑ ⌜x1⌝ + ⌜x0⌝ , ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ¬ # x + # x1 + # x0 < m → (∃ n. ⌜(⌜x⌝ + ⌜x1⌝ ) + ⌜x0⌝ to m⌝ ⊑ ⌜x⌝ + σf(n)), ∃ σ1. ⌜lim(⌜σf⌝ ) to (# x + # x1 + # x0) + 1 + # x⌝ + σ1 = ⌜x1⌝ ⊦

Rule argument

Additional information

Used terms: ⌜lim(⌜σf⌝ ) to (# x + # x1 + # x0) + 1 + # x⌝ , ⌜x1⌝

Proof step 152 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 153 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 154:

  # lim(⌜σf⌝ ) = ∞, ⌜lim(⌜σf⌝ ) to (# x + # x1 + # x0) + 1 + # x⌝ + σ1 = ⌜x1⌝ , (⌜x⌝ + ⌜x1⌝ ) + ⌜x0⌝ = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ⌜x1⌝ = σf(n0), ⌜x1⌝ + ⌜x0⌝ ≠ lim(⌜σf⌝ ), mon(⌜σf⌝ ), ∀ m. ¬ # x + # x1 + # x0 < m → (∃ n. ⌜(⌜x⌝ + ⌜x1⌝ ) + ⌜x0⌝ to m⌝ ⊑ ⌜x⌝ + σf(n)), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ m. ∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n), ∀ n1. σf(n1) ⊑ ⌜x1⌝ + ⌜x0⌝ ⊦

Rule argument

Additional information

Proof step 154 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 155:

  ⊦ σ0 = σ1 ↔ # σ0 = # σ1 ∧ (∀ n. ⌜n⌝ < # σ0 → σ0[n] = σ1[n])

Rule argument

Additional information

Used simplifier rules: ⊦ #(σ + σ1) = # σ + # σ1; (s) ⊦ # ⌜σ to n⌝ = ⌜#(σ to n)⌝ ; (s) ⊦ ¬ ∞ < n∞; (s) ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = n; (s) ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s) ⊦ ⌜n⌝ + n∞ = ⌜n⌝ ↔ n∞ = ⌜0⌝ ; (s) ⊦ ⌜n⌝ + n∞ = ⌜0⌝ ↔ n∞ = ⌜0⌝ ∧ n = 0; (ws) ⊦ n < m → (⌜(* n)⌝ = ⌜(* m)⌝ + n∞2 ↔ false); (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 Used terms: ⌜lim(⌜σf⌝ ) to (# x + # x1 + # x0) + 1 + # x⌝ + σ1, ⌜x1⌝

Proof step 155 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 156 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 157:

  ⊦ n∞ = ∞ ↔ ¬(∃ n. n∞ = ⌜n⌝ )

• Proof step 158:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ⌜n⌝ , lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ∀ m. ¬ # lim(⌜σf⌝ ) < ⌜m⌝ → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n)), ∃ n. # lim(⌜σf⌝ ) = ⌜n⌝ ⊦

Rule argument

Additional information

Used terms: # lim(⌜σf⌝ )

Proof step 157 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 158 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 159:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ⌜n⌝ , # lim(⌜σf⌝ ) = ⌜n0⌝ , lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n)), ∀ m. ¬ n0 < m → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)

Proof step 159 from proof lim-extractprefix (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 160:

  # lim(⌜σf⌝ ) = ⌜n0⌝ , # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ⌜n⌝ , lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ¬ n0 < n0 → (∃ n. ⌜lim(⌜σf⌝ ) to n0⌝ ⊑ σf(n)), ∀ m. ¬ n0 < m → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n)), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ) ⊦

Rule argument

Additional information

Used terms: n0

Proof step 160 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 161:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ⌜n⌝ , # lim(⌜σf⌝ ) = ⌜n0⌝ , lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ⌜lim(⌜σf⌝ ).l⌝ ⊑ σf(n1), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n)), ∀ m. ¬ n0 < m → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # σ = ⌜n⌝ → σ to n = σ.l; (s) ⊦ ¬ n < n; (s)

Proof step 161 from proof lim-extractprefix (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 162:

  # lim(⌜σf⌝ ) = ⌜n0⌝ , # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ⌜n⌝ , lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), ⌜lim(⌜σf⌝ ).l⌝ ⊑ σf(n1), σf(n1) ⊑ lim(⌜σf⌝ ), ∀ m. ¬ n0 < m → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n)), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ) ⊦

Rule argument

Additional information

Used terms: n1

Proof step 162 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 163:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ⌜n⌝ , # lim(⌜σf⌝ ) = ⌜n0⌝ , lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), mon(⌜σf⌝ ), σf(n1) ⊑ lim(⌜σf⌝ ), ⌜lim(⌜σf⌝ ).l⌝ ⊑ σf(n1), ⌜lim(⌜σf⌝ ).l⌝ ⊑ lim(⌜σf⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n)), ∀ m. ¬ n0 < m → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)) ⊦

Rule argument

Additional information

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

Proof step 163 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 164:

  ⊦ finp(σ) → ⌜σ.l⌝ = σ

• Proof step 165:

  # lim(⌜σf⌝ ) = ⌜n0⌝ , # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ⌜n⌝ , lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + lim(⌜σf⌝ ), finp(lim(⌜σf⌝ )), mon(⌜σf⌝ ), lim(⌜σf⌝ ) ⊑ lim(⌜σf⌝ ), lim(⌜σf⌝ ) ⊑ σf(n1), σf(n1) ⊑ lim(⌜σf⌝ ), ∀ m. ¬ n0 < m → (∃ n. ⌜lim(⌜σf⌝ ) to m⌝ ⊑ σf(n)), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n)), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ lim(⌜σf⌝ ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ finp(σ) ↔ ¬(∃ nf. ⌜nf⌝ = σ); (ss) leq-trans: ⊦ σ ⊑ σ1 ∧ σ1 ⊑ σ2 → σ ⊑ σ2; (forward) from specification stream ⊦ # ⌜nf⌝ = ∞; (s) ⊦ ⌜n⌝ ≠ ∞; (s) Used terms: lim(⌜σf⌝ )

Proof step 164 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 165 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 166:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ⌜n⌝ , ⌜x0⌝ = σf(n1), ⌜x0⌝ = lim(⌜σf⌝ ), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + ⌜x0⌝ , mon(⌜σf⌝ ), ∀ m. ¬ # x0 < m → (∃ n2. ⌜x0 to m⌝ ⊑ σf(n2)), ∀ n. σf(n) ⊑ ⌜x0⌝ , ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n)) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ to n = x to n; (s) ⊦ σ ⊑ σ; (s) ⊦ ⌜n⌝ = ⌜n0⌝ ↔ n = n0; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s) ⊦ finp(σ) ↔ (∃ x. ⌜x⌝ = σ); (as) ⊦ σ1 ⊑ σ2 → (σ2 ⊑ σ1 ↔ σ1 = σ2); (s)

Proof step 166 from proof lim-extractprefix (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 167:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ⌜n⌝ , ⌜x0⌝ = lim(⌜σf⌝ ), ⌜x0⌝ = σf(n1), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + ⌜x0⌝ , mon(⌜σf⌝ ), ¬ n < n → (∃ n0. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to n⌝ ⊑ ⌜x⌝ + σf(n0)), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n)), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ ⌜x0⌝ , ∀ m. ¬ # x0 < m → (∃ n2. ⌜x0 to m⌝ ⊑ σf(n2)) ⊦

Rule argument

Additional information

Used terms: n

Proof step 167 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 168:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ⌜n⌝ , ⌜x0⌝ = σf(n1), ⌜x0⌝ = lim(⌜σf⌝ ), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + ⌜x0⌝ , mon(⌜σf⌝ ), ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ).l⌝ ⊑ ⌜x⌝ + σf(n0), ∀ m. ¬ # x0 < m → (∃ n2. ⌜x0 to m⌝ ⊑ σf(n2)), ∀ n. σf(n) ⊑ ⌜x0⌝ , ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n)) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # σ = ⌜n⌝ → σ to n = σ.l; (s) ⊦ ¬ n < n; (s)

Proof step 168 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 169:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 170:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ⌜n⌝ , ⌜x0⌝ = lim(⌜σf⌝ ), ⌜x0⌝ = σf(n1), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + ⌜x0⌝ , mon(⌜σf⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n)), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ ⌜x0⌝ , ∀ m. ¬ # x0 < m → (∃ n2. ⌜x0 to m⌝ ⊑ σf(n2)), ∃ σ1. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ).l⌝ + σ1 = ⌜x⌝ + σf(n0) ⊦

Rule argument

Additional information

Used terms: ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ).l⌝ , ⌜x⌝ + σf(n0)

Proof step 169 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 170 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 171:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ⌜n⌝ , ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ).l⌝ + σ1 = ⌜x⌝ + σf(n0), ⌜x0⌝ = σf(n1), ⌜x0⌝ = lim(⌜σf⌝ ), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + ⌜x0⌝ , mon(⌜σf⌝ ), ∀ m. ¬ # x0 < m → (∃ n2. ⌜x0 to m⌝ ⊑ σf(n2)), ∀ n. σf(n) ⊑ ⌜x0⌝ , ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n)) ⊦

Rule argument

Additional information

Proof step 171 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 172:

  ⊦ finp(σ) → ⌜σ.l⌝ = σ

• Proof step 173:

  # lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) = ⌜n⌝ , lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) + σ1 = ⌜x⌝ + σf(n0), ⌜x0⌝ = lim(⌜σf⌝ ), ⌜x0⌝ = σf(n1), lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) ≠ ⌜x⌝ + ⌜x0⌝ , finp(lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )), mon(⌜σf⌝ ), ∀ m. ¬ n < m → (∃ n. ⌜lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ) to m⌝ ⊑ ⌜x⌝ + σf(n)), ∀ n. ⌜x⌝ + σf(n) ⊑ lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ∀ n. σf(n) ⊑ ⌜x0⌝ , ∀ m. ¬ # x0 < m → (∃ n2. ⌜x0 to m⌝ ⊑ σf(n2)) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ finp(σ) ↔ ¬(∃ nf. ⌜nf⌝ = σ); (ss) ⊦ # ⌜nf⌝ = ∞; (s) ⊦ ⌜n⌝ ≠ ∞; (s) Used terms: lim(⌜λ n. ⌜x⌝ + σf(n)⌝ )

Proof step 172 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 173 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 174:

  ⌜x1⌝ + σ1 = ⌜x⌝ + σf(n0), ⌜x1⌝ = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ⌜x0⌝ = σf(n1), ⌜x0⌝ = lim(⌜σf⌝ ), x + x0 ≠ x1, mon(⌜σf⌝ ), ∀ m. ¬ # x1 < m → (∃ n2. ⌜x1 to m⌝ ⊑ ⌜x⌝ + σf(n2)), ∀ m. ¬ # x0 < m → (∃ n2. ⌜x0 to m⌝ ⊑ σf(n2)), ∀ n. σf(n) ⊑ ⌜x0⌝ , ∀ n. ⌜x⌝ + σf(n) ⊑ ⌜x1⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ to n = x to n; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ ⌜x⌝ + ⌜y⌝ = ⌜z⌝ ↔ x + y = z; (s) ⊦ ⌜n⌝ = ⌜n0⌝ ↔ n = n0; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s) ⊦ finp(σ) ↔ (∃ x. ⌜x⌝ = σ); (as)

Proof step 174 from proof lim-extractprefix (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 175:

  ⌜x1⌝ + σ1 = ⌜x⌝ + σf(n0), σ1 = ⌜x2⌝ , ⌜x0⌝ = lim(⌜σf⌝ ), ⌜x0⌝ = σf(n1), ⌜x1⌝ = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), x + x0 ≠ x1, mon(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ ⌜x1⌝ , ∀ n. σf(n) ⊑ ⌜x0⌝ , ∀ m. ¬ # x0 < m → (∃ n2. ⌜x0 to m⌝ ⊑ σf(n2)), ∀ m. ¬ # x1 < m → (∃ n2. ⌜x1 to m⌝ ⊑ ⌜x⌝ + σf(n2)) ⊦

• Proof step 190:

  ⌜x1⌝ + σ1 = ⌜x⌝ + σf(n0), σ1 = ⌜nf⌝ , ⌜x0⌝ = lim(⌜σf⌝ ), ⌜x0⌝ = σf(n1), ⌜x1⌝ = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), x + x0 ≠ x1, mon(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ ⌜x1⌝ , ∀ n. σf(n) ⊑ ⌜x0⌝ , ∀ m. ¬ # x0 < m → (∃ n2. ⌜x0 to m⌝ ⊑ σf(n2)), ∀ m. ¬ # x1 < m → (∃ n2. ⌜x1 to m⌝ ⊑ ⌜x⌝ + σf(n2)) ⊦

Rule argument

Additional information

Used simplifier rules: stream('a) freely generated by ⌜ ⌝, ⌜ ⌝;;

Proof step 175 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 176:

  ⌜x1⌝ + ⌜x2⌝ = ⌜x⌝ + σf(n0), ⌜x1⌝ = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ⌜x0⌝ = σf(n1), ⌜x0⌝ = lim(⌜σf⌝ ), x + x0 ≠ x1, mon(⌜σf⌝ ), ∀ m. ¬ # x1 < m → (∃ n2. ⌜x1 to m⌝ ⊑ ⌜x⌝ + σf(n2)), ∀ m. ¬ # x0 < m → (∃ n2. ⌜x0 to m⌝ ⊑ σf(n2)), ∀ n. σf(n) ⊑ ⌜x0⌝ , ∀ n. ⌜x⌝ + σf(n) ⊑ ⌜x1⌝ ⊦

Rule argument

Additional information

Proof step 176 from proof lim-extractprefix (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 177:

  ⌜x1⌝ + ⌜x2⌝ = ⌜x⌝ + σf(n0), ⌜x0⌝ = lim(⌜σf⌝ ), ⌜x0⌝ = σf(n1), ⌜x1⌝ = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), x + x0 ≠ x1, mon(⌜σf⌝ ), ⌜x⌝ + σf(n1) ⊑ ⌜x1⌝ , ∀ n. ⌜x⌝ + σf(n) ⊑ ⌜x1⌝ , ∀ n. σf(n) ⊑ ⌜x0⌝ , ∀ m. ¬ # x0 < m → (∃ n2. ⌜x0 to m⌝ ⊑ σf(n2)), ∀ m. ¬ # x1 < m → (∃ n2. ⌜x1 to m⌝ ⊑ ⌜x⌝ + σf(n2)) ⊦

Rule argument

Additional information

Used terms: n1

Proof step 177 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 178:

  ⌜x1⌝ + ⌜x2⌝ = ⌜x⌝ + σf(n0), ⌜x1⌝ = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ⌜x0⌝ = σf(n1), ⌜x0⌝ = lim(⌜σf⌝ ), x + x0 ≠ x1, mon(⌜σf⌝ ), ⌜x⌝ + ⌜x0⌝ ⊑ ⌜x1⌝ , ∀ m. ¬ # x1 < m → (∃ n2. ⌜x1 to m⌝ ⊑ ⌜x⌝ + σf(n2)), ∀ m. ¬ # x0 < m → (∃ n2. ⌜x0 to m⌝ ⊑ σf(n2)), ∀ n. σf(n) ⊑ ⌜x0⌝ , ∀ n. ⌜x⌝ + σf(n) ⊑ ⌜x1⌝ ⊦

Rule argument

Additional information

Proof step 178 from proof lim-extractprefix (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 179:

  ⌜x1⌝ + ⌜x2⌝ = ⌜x⌝ + σf(n0), ⌜x0⌝ = lim(⌜σf⌝ ), ⌜x0⌝ = σf(n1), ⌜x1⌝ = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), x + x0 ≠ x1, mon(⌜σf⌝ ), ⌜x⌝ + ⌜x0⌝ ⊑ ⌜x1⌝ , ¬ # x1 < # x1 → (∃ n2. ⌜x1 to # x1⌝ ⊑ ⌜x⌝ + σf(n2)), ∀ n. ⌜x⌝ + σf(n) ⊑ ⌜x1⌝ , ∀ n. σf(n) ⊑ ⌜x0⌝ , ∀ m. ¬ # x0 < m → (∃ n2. ⌜x0 to m⌝ ⊑ σf(n2)), ∀ m. ¬ # x1 < m → (∃ n2. ⌜x1 to m⌝ ⊑ ⌜x⌝ + σf(n2)) ⊦

Rule argument

Additional information

Used terms: # x1

Proof step 179 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 180:

  ⌜x1⌝ + ⌜x2⌝ = ⌜x⌝ + σf(n0), ⌜x0⌝ = lim(⌜σf⌝ ), ⌜x0⌝ = σf(n1), ⌜x1⌝ = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), x + x0 ≠ x1, mon(⌜σf⌝ ), ⌜x1⌝ ⊑ ⌜x⌝ + σf(n2), ⌜x⌝ + ⌜x0⌝ ⊑ ⌜x1⌝ , ⌜x0⌝ ⊑ σf(n2), ∀ n. ⌜x⌝ + σf(n) ⊑ ⌜x1⌝ , ∀ n. σf(n) ⊑ ⌜x0⌝ , ∀ m. ¬ # x0 < m → (∃ n2. ⌜x0 to m⌝ ⊑ σf(n2)), ∀ m. ¬ # x1 < m → (∃ n2. ⌜x1 to m⌝ ⊑ ⌜x⌝ + σf(n2)) ⊦

Rule argument

Additional information

Used simplifier rules: append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ ⌜x⌝ + σ1 ⊑ ⌜x⌝ + σ2 ↔ σ1 ⊑ σ2; (s) leq-trans: ⊦ σ ⊑ σ1 ∧ σ1 ⊑ σ2 → σ ⊑ σ2; (forward) from specification stream ⊦ # x = n → x to n = x; (s) ⊦ ¬ n < n; (s)

Proof step 180 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 181:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 182:

  ⌜x1⌝ + ⌜x2⌝ = ⌜x⌝ + σf(n0), ⌜x1⌝ = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ⌜x0⌝ = σf(n1), ⌜x0⌝ = lim(⌜σf⌝ ), x + x0 ≠ x1, mon(⌜σf⌝ ), ⌜x⌝ + ⌜x0⌝ ⊑ ⌜x1⌝ , ⌜x1⌝ ⊑ ⌜x⌝ + σf(n2), ∀ m. ¬ # x1 < m → (∃ n2. ⌜x1 to m⌝ ⊑ ⌜x⌝ + σf(n2)), ∀ m. ¬ # x0 < m → (∃ n2. ⌜x0 to m⌝ ⊑ σf(n2)), ∀ n. σf(n) ⊑ ⌜x0⌝ , ∀ n. ⌜x⌝ + σf(n) ⊑ ⌜x1⌝ , ∃ σ1. ⌜x0⌝ + σ1 = σf(n2) ⊦

Rule argument

Additional information

Used terms: ⌜x0⌝ , σf(n2)

Proof step 181 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 182 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 183:

  ⌜x0⌝ + σ1 = σf(n2), ⌜x1⌝ + ⌜x2⌝ = ⌜x⌝ + σf(n0), ⌜x0⌝ = lim(⌜σf⌝ ), ⌜x0⌝ = σf(n1), ⌜x1⌝ = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), x + x0 ≠ x1, mon(⌜σf⌝ ), ⌜x1⌝ ⊑ ⌜x⌝ + σf(n2), ⌜x⌝ + ⌜x0⌝ ⊑ ⌜x1⌝ , ∀ n. ⌜x⌝ + σf(n) ⊑ ⌜x1⌝ , ∀ n. σf(n) ⊑ ⌜x0⌝ , ∀ m. ¬ # x0 < m → (∃ n2. ⌜x0 to m⌝ ⊑ σf(n2)), ∀ m. ¬ # x1 < m → (∃ n2. ⌜x1 to m⌝ ⊑ ⌜x⌝ + σf(n2)) ⊦

Rule argument

Additional information

Proof step 183 from proof lim-extractprefix (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 184:

  ⌜x1⌝ + ⌜x2⌝ = ⌜x⌝ + σf(n0), ⌜x0⌝ + σ1 = σf(n2), ⌜x1⌝ = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ⌜x0⌝ = σf(n1), ⌜x0⌝ = lim(⌜σf⌝ ), x + x0 ≠ x1, mon(⌜σf⌝ ), ⌜x⌝ + ⌜x0⌝ ⊑ ⌜x1⌝ , ⌜x1⌝ ⊑ ⌜x⌝ + σf(n2), σf(n2) ⊑ ⌜x0⌝ , ∀ m. ¬ # x1 < m → (∃ n2. ⌜x1 to m⌝ ⊑ ⌜x⌝ + σf(n2)), ∀ m. ¬ # x0 < m → (∃ n2. ⌜x0 to m⌝ ⊑ σf(n2)), ∀ n. σf(n) ⊑ ⌜x0⌝ , ∀ n. ⌜x⌝ + σf(n) ⊑ ⌜x1⌝ ⊦

Rule argument

Additional information

Used terms: n2

Proof step 184 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 185:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 186:

  ⌜x0⌝ + σ1 = σf(n2), ⌜x1⌝ + ⌜x2⌝ = ⌜x⌝ + σf(n0), ⌜x0⌝ = lim(⌜σf⌝ ), ⌜x0⌝ = σf(n1), ⌜x1⌝ = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), x + x0 ≠ x1, mon(⌜σf⌝ ), ⌜x1⌝ ⊑ ⌜x⌝ + σf(n2), ⌜x⌝ + ⌜x0⌝ ⊑ ⌜x1⌝ , ∀ n. ⌜x⌝ + σf(n) ⊑ ⌜x1⌝ , ∀ n. σf(n) ⊑ ⌜x0⌝ , ∀ m. ¬ # x0 < m → (∃ n2. ⌜x0 to m⌝ ⊑ σf(n2)), ∀ m. ¬ # x1 < m → (∃ n2. ⌜x1 to m⌝ ⊑ ⌜x⌝ + σf(n2)), ∃ σ1. σf(n2) + σ1 = ⌜x0⌝ ⊦

Rule argument

Additional information

Used terms: σf(n2), ⌜x0⌝

Proof step 185 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 186 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 187:

  σf(n2) + σ = ⌜x0⌝ , ⌜x1⌝ + ⌜x2⌝ = ⌜x⌝ + σf(n0), ⌜x0⌝ + σ1 = σf(n2), ⌜x1⌝ = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ⌜x0⌝ = σf(n1), ⌜x0⌝ = lim(⌜σf⌝ ), x + x0 ≠ x1, mon(⌜σf⌝ ), ⌜x⌝ + ⌜x0⌝ ⊑ ⌜x1⌝ , ⌜x1⌝ ⊑ ⌜x⌝ + σf(n2), ∀ m. ¬ # x1 < m → (∃ n2. ⌜x1 to m⌝ ⊑ ⌜x⌝ + σf(n2)), ∀ m. ¬ # x0 < m → (∃ n2. ⌜x0 to m⌝ ⊑ σf(n2)), ∀ n. σf(n) ⊑ ⌜x0⌝ , ∀ n. ⌜x⌝ + σf(n) ⊑ ⌜x1⌝ ⊦

Rule argument

Additional information

Proof step 187 from proof lim-extractprefix (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 188:

  ⌜x0⌝ + σ1 = σf(n2), ⌜x1⌝ + ⌜x2⌝ = ⌜x⌝ + σf(n0), σf(n2) + σ = ⌜x0⌝ , σ = ⌜x3⌝ , ⌜x0⌝ = lim(⌜σf⌝ ), ⌜x0⌝ = σf(n1), ⌜x1⌝ = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), x + x0 ≠ x1, mon(⌜σf⌝ ), ⌜x1⌝ ⊑ ⌜x⌝ + σf(n2), ⌜x⌝ + ⌜x0⌝ ⊑ ⌜x1⌝ , ∀ n. ⌜x⌝ + σf(n) ⊑ ⌜x1⌝ , ∀ n. σf(n) ⊑ ⌜x0⌝ , ∀ m. ¬ # x0 < m → (∃ n2. ⌜x0 to m⌝ ⊑ σf(n2)), ∀ m. ¬ # x1 < m → (∃ n2. ⌜x1 to m⌝ ⊑ ⌜x⌝ + σf(n2)) ⊦

Rule argument

Additional information

Used simplifier rules: stream('a) freely generated by ⌜ ⌝, ⌜ ⌝;; ⊦ ⌜x⌝ ≠ σ + ⌜nf0⌝ ; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream

Proof step 188 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 189:

  σf(n2) + ⌜x3⌝ = ⌜x0⌝ , ⌜x1⌝ + ⌜x2⌝ = ⌜x⌝ + σf(n0), ⌜x0⌝ + σ1 = σf(n2), ⌜x1⌝ = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ⌜x0⌝ = σf(n1), ⌜x0⌝ = lim(⌜σf⌝ ), x + x0 ≠ x1, mon(⌜σf⌝ ), ⌜x⌝ + ⌜x0⌝ ⊑ ⌜x1⌝ , ⌜x1⌝ ⊑ ⌜x⌝ + σf(n2), ∀ m. ¬ # x1 < m → (∃ n2. ⌜x1 to m⌝ ⊑ ⌜x⌝ + σf(n2)), ∀ m. ¬ # x0 < m → (∃ n2. ⌜x0 to m⌝ ⊑ σf(n2)), ∀ n. σf(n) ⊑ ⌜x0⌝ , ∀ n. ⌜x⌝ + σf(n) ⊑ ⌜x1⌝ ⊦

Rule argument

Additional information

Proof step 189 from proof lim-extractprefix (previous step) (back to tree)

User applied rule structural induction

And closed the goal!

Rule argument

Additional information

Used simplifier rules: stream('a) freely generated by ⌜ ⌝, ⌜ ⌝;; leq-trans: ⊦ σ ⊑ σ1 ∧ σ1 ⊑ σ2 → σ ⊑ σ2; (forward) from specification stream ⊦ ⌜nf⌝ + σ = ⌜nf⌝ ; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ ⌜x⌝ ≠ ⌜nf⌝ ; (s) ⊦ σ + σ0 = σ ↔ σ0 = [] ∨ # σ = ∞; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s) ⊦ ⌜n⌝ ≠ ∞; (s) ⊦ σ0 + σ1 = [] ↔ σ0 = [] ∧ σ1 = []; (ws) ⊦ σ + σ0 ⊑ σ ↔ σ0 = [] ∨ # σ = ∞; (s) ⊦ ⌜x⌝ = [] ↔ x = []; (s) ⊦ #(x + y) = # x + # y; (s) ⊦ ⌜x + y⌝ = ⌜x⌝ + ⌜y⌝ ; (s) ⊦ ⌜[]⌝ = []; (s) ⊦ σ + [] = σ; (s) ⊦ # [] = 0; (s) ⊦ n + 0 = n; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ x + [] = x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ ⌜x⌝ + σ1 ⊑ ⌜x⌝ + σ2 ↔ σ1 ⊑ σ2; (s) ⊦ σ ⊑ σ; (s) ⊦ σ1 ⊑ σ2 → (σ2 ⊑ σ1 ↔ σ1 = σ2); (s) ⊦ ⌜x⌝ + ⌜y⌝ = ⌜z⌝ ↔ x + y = z; (s)

Proof step 190 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 191:

  ⌜x1⌝ + ⌜nf⌝ = ⌜x⌝ + σf(n0), ⌜x1⌝ = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ⌜x0⌝ = σf(n1), ⌜x0⌝ = lim(⌜σf⌝ ), x + x0 ≠ x1, mon(⌜σf⌝ ), ∀ m. ¬ # x1 < m → (∃ n2. ⌜x1 to m⌝ ⊑ ⌜x⌝ + σf(n2)), ∀ m. ¬ # x0 < m → (∃ n2. ⌜x0 to m⌝ ⊑ σf(n2)), ∀ n. σf(n) ⊑ ⌜x0⌝ , ∀ n. ⌜x⌝ + σf(n) ⊑ ⌜x1⌝ ⊦

Rule argument

Additional information

Proof step 191 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 192:

  ⊦ σ0 = σ1 ↔ # σ0 = # σ1 ∧ (∀ n. ⌜n⌝ < # σ0 → σ0[n] = σ1[n])

• Proof step 193:

  #(⌜x1⌝ + ⌜nf⌝ ) = #(⌜x⌝ + σf(n0)), ⌜x0⌝ = lim(⌜σf⌝ ), ⌜x0⌝ = σf(n1), ⌜x1⌝ = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), x + x0 ≠ x1, mon(⌜σf⌝ ), ∀ n. ⌜x⌝ + σf(n) ⊑ ⌜x1⌝ , ∀ n. σf(n) ⊑ ⌜x0⌝ , ∀ m. ¬ # x0 < m → (∃ n2. ⌜x0 to m⌝ ⊑ σf(n2)), ∀ m. ¬ # x1 < m → (∃ n2. ⌜x1 to m⌝ ⊑ ⌜x⌝ + σf(n2)), ∀ n. ⌜n⌝ < #(⌜x1⌝ + ⌜nf⌝ ) → (⌜x1⌝ + ⌜nf⌝ )[n] = (⌜x⌝ + σf(n0))[n] ⊦

Rule argument

Additional information

Used terms: ⌜x1⌝ + ⌜nf⌝ , ⌜x⌝ + σf(n0)

Proof step 192 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 193 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 194:

  # σf(n0) = ∞, ⌜x1⌝ = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ⌜x0⌝ = σf(n1), ⌜x0⌝ = lim(⌜σf⌝ ), x + x0 ≠ x1, mon(⌜σf⌝ ), ∀ m. ¬ # x1 < m → (∃ n2. ⌜x1 to m⌝ ⊑ ⌜x⌝ + σf(n2)), ∀ m. ¬ # x0 < m → (∃ n2. ⌜x0 to m⌝ ⊑ σf(n2)), ∀ n. σf(n) ⊑ ⌜x0⌝ , ∀ n. ⌜x⌝ + σf(n) ⊑ ⌜x1⌝ , ∀ n. (⌜x1⌝ + ⌜nf⌝ )[n] = (⌜x⌝ + σf(n0))[n] ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ∞; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; 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) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s) ⊦ #(σ + σ1) = # σ + # σ1; (s) ⊦ ⌜m⌝ + n∞ = ∞ ↔ n∞ = ∞; (s)

Proof step 194 from proof lim-extractprefix (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 195:

  # σf(n0) = ∞, ⌜x0⌝ = lim(⌜σf⌝ ), ⌜x0⌝ = σf(n1), ⌜x1⌝ = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), x + x0 ≠ x1, mon(⌜σf⌝ ), σf(n0) ⊑ ⌜x0⌝ , ∀ n. (⌜x1⌝ + ⌜nf⌝ )[n] = (⌜x⌝ + σf(n0))[n], ∀ n. ⌜x⌝ + σf(n) ⊑ ⌜x1⌝ , ∀ m. ¬ # x0 < m → (∃ n2. ⌜x0 to m⌝ ⊑ σf(n2)), ∀ m. ¬ # x1 < m → (∃ n2. ⌜x1 to m⌝ ⊑ ⌜x⌝ + σf(n2)) ⊦

Rule argument

Additional information

Used terms: n0

Proof step 195 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 196:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 197:

  # σf(n0) = ∞, ⌜x1⌝ = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), ⌜x0⌝ = σf(n1), ⌜x0⌝ = lim(⌜σf⌝ ), x + x0 ≠ x1, mon(⌜σf⌝ ), ∀ m. ¬ # x1 < m → (∃ n2. ⌜x1 to m⌝ ⊑ ⌜x⌝ + σf(n2)), ∀ m. ¬ # x0 < m → (∃ n2. ⌜x0 to m⌝ ⊑ σf(n2)), ∀ n. ⌜x⌝ + σf(n) ⊑ ⌜x1⌝ , ∀ n. (⌜x1⌝ + ⌜nf⌝ )[n] = (⌜x⌝ + σf(n0))[n], ∃ σ1. σf(n0) + σ1 = ⌜x0⌝ ⊦

Rule argument

Additional information

Used terms: σf(n0), ⌜x0⌝

Proof step 196 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma

Proof step 197 from proof lim-extractprefix (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 198:

  # σf(n0) = ∞, σf(n0) + σ1 = ⌜x0⌝ , ⌜x0⌝ = lim(⌜σf⌝ ), ⌜x0⌝ = σf(n1), ⌜x1⌝ = lim(⌜λ n. ⌜x⌝ + σf(n)⌝ ), x + x0 ≠ x1, mon(⌜σf⌝ ), ∀ n. (⌜x1⌝ + ⌜nf⌝ )[n] = (⌜x⌝ + σf(n0))[n], ∀ n. ⌜x⌝ + σf(n) ⊑ ⌜x1⌝ , ∀ m. ¬ # x0 < m → (∃ n2. ⌜x0 to m⌝ ⊑ σf(n2)), ∀ m. ¬ # x1 < m → (∃ n2. ⌜x1 to m⌝ ⊑ ⌜x⌝ + σf(n2)) ⊦

Rule argument

Additional information

Proof step 198 from proof lim-extractprefix (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 199:

  ⊦ σ0 = σ1 ↔ # σ0 = # σ1 ∧ (∀ n. ⌜n⌝ < # σ0 → σ0[n] = σ1[n])

Rule argument

Additional information

Used simplifier rules: ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s) ⊦ #(σ + σ1) = # σ + # σ1; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ ⌜x⌝ [n] = x[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 ⊦ ⌜n⌝ ≠ ∞; (s) Used terms: σf(n0) + σ1, ⌜x0⌝

Proof step 199 from proof lim-extractprefix (previous step) (back to tree)

A spec-lemma