Proof step 1 from proof lifttr-cont (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  mon(trf), ¬ cont(lift(trf)) ⊦

Rule argument

Additional information

Proof step 2 from proof lifttr-cont (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 3:

  ⊦ cont(βf) ↔ mon(βf) ∧ pbd(βf)

• Proof step 4:

  mon(trf), ¬ mon(lift(trf)) ∨ ¬ pbd(lift(trf)) ⊦

Rule argument

Additional information

Used terms: lift(trf)

Proof step 3 from proof lifttr-cont (previous step) (back to tree)

Lemma conttr-def

Proof step 4 from proof lifttr-cont (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  mon(trf), ¬(# σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 → lift(trf)(σ1, α1) ⊑ lift(trf)(σ2, α2)) ∨ ¬ pbd(lift(trf)) ⊦

Rule argument

Additional information

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

Proof step 5 from proof lifttr-cont (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 6:

  # σ1 = # α1 + ⌜1⌝ , # σ2 = # α2 + ⌜1⌝ , mon(trf), σ1 ⊑ σ2, α1 ⊑ α2, ¬ lift(trf)(σ1, α1) ⊑ lift(trf)(σ2, α2) ⊦

• Proof step 44:

  mon(trf), ¬ pbd(lift(trf)) ⊦

Rule argument

Additional information

Proof step 6 from proof lifttr-cont (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 7:

  # σ2 = # α2 + ⌜1⌝ , # σ1 = # α1 + ⌜1⌝ , σ2 = ⌜sl⌝ , mon(trf), α1 ⊑ α2, σ1 ⊑ σ2, ¬ lift(trf)(σ1, α1) ⊑ lift(trf)(σ2, α2) ⊦

• Proof step 17:

  # σ2 = # α2 + ⌜1⌝ , # σ1 = # α1 + ⌜1⌝ , σ2 = ⌜sf⌝ , mon(trf), α1 ⊑ α2, σ1 ⊑ σ2, ¬ lift(trf)(σ1, α1) ⊑ lift(trf)(σ2, α2) ⊦

Rule argument

Additional information

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

Proof step 7 from proof lifttr-cont (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  # σ1 = # α1 + ⌜1⌝ , ⌜# sl⌝ = # α2 + ⌜1⌝ , mon(trf), σ1 ⊑ ⌜sl⌝ , α1 ⊑ α2, ¬ lift(trf)(σ1, α1) ⊑ lift(trf)(⌜sl⌝ , α2) ⊦

Rule argument

Additional information

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

Proof step 8 from proof lifttr-cont (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 9:

  # σ1 = # α1 + ⌜1⌝ , σ1 = ⌜sl0⌝ , ⌜# sl⌝ = # α2 + ⌜1⌝ , mon(trf), α1 ⊑ α2, σ1 ⊑ ⌜sl⌝ , ¬ lift(trf)(σ1, α1) ⊑ lift(trf)(⌜sl⌝ , α2) ⊦

Rule argument

Additional information

Used simplifier rules: Stream[A] freely generated by ⌝, ⌝; ⊦ # ⌜nf⌝ = ∞; (s) ⊦ ⌜m⌝ + 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 ⊦ ¬ ⌜nf⌝ ⊑ ⌜x⌝ ; (s)

Proof step 9 from proof lifttr-cont (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  ⌜# sl⌝ = # α2 + ⌜1⌝ , ⌜# sl0⌝ = # α1 + ⌜1⌝ , mon(trf), α1 ⊑ α2, sl0 ⊑ sl, ¬ lift(trf)(⌜sl0⌝ , α1) ⊑ lift(trf)(⌜sl⌝ , α2) ⊦

Rule argument

Additional information

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

Proof step 10 from proof lifttr-cont (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 11:

  α2 = ⌜al⌝ , ⌜# sl0⌝ = # α1 + ⌜1⌝ , ⌜# sl⌝ = # α2 + ⌜1⌝ , mon(trf), sl0 ⊑ sl, α1 ⊑ α2, ¬ lift(trf)(⌜sl0⌝ , α1) ⊑ lift(trf)(⌜sl⌝ , α2) ⊦

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 11 from proof lifttr-cont (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  # al + 1 = # sl, ⌜# sl0⌝ = # α1 + ⌜1⌝ , mon(trf), α1 ⊑ ⌜al⌝ , sl0 ⊑ sl, ¬ lift(trf)(⌜sl0⌝ , α1) ⊑ ⌜trf(sl, al)⌝ ⊦

Rule argument

Additional information

Used simplifier rules: lifttr-fin: ⊦ lift(trf)(⌜sl⌝ , ⌜al⌝ ) = ⌜trf(sl, al)⌝ ; 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 12 from proof lifttr-cont (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 13:

  # al + 1 = # sl, α1 = ⌜al0⌝ , ⌜# sl0⌝ = # α1 + ⌜1⌝ , mon(trf), sl0 ⊑ sl, α1 ⊑ ⌜al⌝ , ¬ lift(trf)(⌜sl0⌝ , α1) ⊑ ⌜trf(sl, al)⌝ ⊦

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 13 from proof lifttr-cont (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 14:

  # al + 1 = # sl, # al0 + 1 = # sl0, mon(trf), sl0 ⊑ sl, al0 ⊑ al, ¬ trf(sl0, al0) ⊑ trf(sl, al) ⊦

Rule argument

Additional information

Used simplifier rules: lifttr-fin: ⊦ lift(trf)(⌜sl⌝ , ⌜al⌝ ) = ⌜trf(sl, al)⌝ ; ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s) add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜n⌝ + ⌜m⌝ = ⌜n0⌝ ↔ n + m = n0; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s)

Proof step 14 from proof lifttr-cont (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 15:

  ⊦ mon(trf) ↔ (∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 → trf(sl1, al1) ⊑ trf(sl2, al2))

• Proof step 16:

  # al0 + 1 = # sl0, # al + 1 = # sl, al0 ⊑ al, sl0 ⊑ sl, ¬ trf(sl0, al0) ⊑ trf(sl, al), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 → trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

Rule argument

Additional information

Used terms: trf

Proof step 15 from proof lifttr-cont (previous step) (back to tree)

Lemma montrl-def

Proof step 16 from proof lifttr-cont (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: sl0, al0, sl, al

Proof step 17 from proof lifttr-cont (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  # σ1 = # α1 + ⌜1⌝ , # α2 = ∞, mon(trf), σ1 ⊑ ⌜sf⌝ , α1 ⊑ α2, ¬ lift(trf)(σ1, α1) ⊑ lift(trf)(⌜sf⌝ , α2) ⊦

Rule argument

Additional information

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

Proof step 18 from proof lifttr-cont (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 19:

  # α2 = ∞, # σ1 = # α1 + ⌜1⌝ , α2 = ⌜af⌝ , mon(trf), α1 ⊑ α2, σ1 ⊑ ⌜sf⌝ , ¬ lift(trf)(σ1, α1) ⊑ lift(trf)(⌜sf⌝ , α2) ⊦

Rule argument

Additional information

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

Proof step 19 from proof lifttr-cont (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

  # σ1 = # α1 + ⌜1⌝ , mon(trf), σ1 ⊑ ⌜sf⌝ , α1 ⊑ ⌜af⌝ , ¬ lift(trf)(σ1, α1) ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ⊦

Rule argument

Additional information

Used simplifier rules: lifttr-inf: ⊦ lift(trf)(⌜sf⌝ , ⌜af⌝ ) = lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ); ⊦ # ⌜nf⌝ = ∞; (s)

Proof step 20 from proof lifttr-cont (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 21:

  # σ1 = # α1 + ⌜1⌝ , α1 = ⌜al⌝ , mon(trf), α1 ⊑ ⌜af⌝ , σ1 ⊑ ⌜sf⌝ , ¬ lift(trf)(σ1, α1) ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ⊦

• Proof step 42:

  # σ1 = # α1 + ⌜1⌝ , α1 = ⌜af0⌝ , mon(trf), α1 ⊑ ⌜af⌝ , σ1 ⊑ ⌜sf⌝ , ¬ lift(trf)(σ1, α1) ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ⊦

Rule argument

Additional information

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

Proof step 21 from proof lifttr-cont (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 22:

  # σ1 = ⌜# al⌝ + ⌜1⌝ , mon(trf), σ1 ⊑ ⌜sf⌝ , ⌜al⌝ ⊑ ⌜af⌝ , ¬ lift(trf)(σ1, ⌜al⌝ ) ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ⊦

Rule argument

Additional information

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

Proof step 22 from proof lifttr-cont (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 23:

  # σ1 = ⌜# al⌝ + ⌜1⌝ , σ1 = ⌜sl⌝ , mon(trf), ⌜al⌝ ⊑ ⌜af⌝ , σ1 ⊑ ⌜sf⌝ , ¬ lift(trf)(σ1, ⌜al⌝ ) ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ⊦

Rule argument

Additional information

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

Proof step 23 from proof lifttr-cont (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

  # al + 1 = # sl, mon(trf), ⌜sl⌝ ⊑ ⌜sf⌝ , ⌜al⌝ ⊑ ⌜af⌝ , ¬ ⌜trf(sl, al)⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ⊦

Rule argument

Additional information

Used simplifier rules: lifttr-fin: ⊦ lift(trf)(⌜sl⌝ , ⌜al⌝ ) = ⌜trf(sl, al)⌝ ; 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 24 from proof lifttr-cont (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 25:

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

• Proof step 26:

  # al + 1 = # sl, mon(trf), ⌜al⌝ ⊑ ⌜af⌝ , ⌜sl⌝ ⊑ ⌜sf⌝ , ¬ ⌜trf(sl, al)⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), mon(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ∧ ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ≠ [] → (∀ n. ⌜n⌝ < # ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ → ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ (n) ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ )) ∧ (∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. ⌜n⌝ < # ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ∧ ⌜lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m)⌝ ⊑ ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ (n))) ⊦

Rule argument

Additional information

Used terms: ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝

Proof step 25 from proof lifttr-cont (previous step) (back to tree)

A spec-lemma

Proof step 26 from proof lifttr-cont (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 27:

  # al + 1 = # sl, mon(trf), ⌜sl⌝ ⊑ ⌜sf⌝ , ⌜al⌝ ⊑ ⌜af⌝ , ¬ ⌜trf(sl, al)⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), mon(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) → (∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ )) ∧ (∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n)))) ⊦

Rule argument

Additional information

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

Proof step 27 from proof lifttr-cont (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 28:

  # al + 1 = # sl, mon(trf), ⌜al⌝ ⊑ ⌜af⌝ , ⌜sl⌝ ⊑ ⌜sf⌝ , ¬ mon(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), ¬ ⌜trf(sl, al)⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ⊦

• Proof step 31:

  # al + 1 = # sl, mon(trf), ⌜al⌝ ⊑ ⌜af⌝ , ⌜sl⌝ ⊑ ⌜sf⌝ , ¬ ⌜trf(sl, al)⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), ∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))) ⊦

Rule argument

Additional information

Proof step 28 from proof lifttr-cont (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 29:

  ⊦ mon(trf) ↔ (∀ sf, af. mon(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ))

• Proof step 30:

  # al + 1 = # sl, ⌜sl⌝ ⊑ ⌜sf⌝ , ⌜al⌝ ⊑ ⌜af⌝ , ¬ mon(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), ¬ ⌜trf(sl, al)⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), ∀ sf, af. mon(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ⊦

Rule argument

Additional information

Used terms: trf

Proof step 29 from proof lifttr-cont (previous step) (back to tree)

Lemma montrl-equiv

Proof step 30 from proof lifttr-cont (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: sf, af

Proof step 31 from proof lifttr-cont (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 32:

  # al + 1 = # sl, mon(trf), ⌜sl⌝ ⊑ ⌜sf⌝ , ⌜al⌝ ⊑ ⌜af⌝ , ⌜trf(sf.take(# al + 1), af.take(# al))⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), ¬ ⌜trf(sl, al)⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), ∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))) ⊦

Rule argument

Additional information

Used terms: # al

Proof step 32 from proof lifttr-cont (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 33:

  # al + 1 = # sl, mon(trf), ⌜trf(sf.take(# sl), af.take(# al))⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), ⌜al⌝ ⊑ ⌜af⌝ , ⌜sl⌝ ⊑ ⌜sf⌝ , ¬ ⌜trf(sl, al)⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), ∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))) ⊦

Rule argument

Additional information

Proof step 33 from proof lifttr-cont (previous step) (back to tree)

User applied rule apply elim lemma

And obtained the following new goals

• Proof step 34:

  # sl0 = # sl, # al + 1 = # sl, sf = (⌜sl0⌝ + ⌜nf0⌝ ).f, mon(trf), ⌜sl⌝ ⊑ ⌜sf⌝ , ⌜al⌝ ⊑ ⌜af⌝ , ⌜trf(sl0, af.take(# al))⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), ¬ ⌜trf(sl, al)⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), ∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ x0 = nf.take(n) ∧ nf0 = nf.drop(n) ↔ nf = (⌜x0⌝ + ⌜nf0⌝ ).f ∧ # x0 = n; (elim) Used terms: sf, # sl, sl0, nf0

Proof step 34 from proof lifttr-cont (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 35:

  # sl0 = # sl, # al + 1 = # sl, mon(trf), ⌜trf(sl0, af.take(# al))⌝ ⊑ lim(⌜λ n. ⌜trf((sl0 + nf0).take(n + 1), af.take(n))⌝ ⌝ ), ⌜al⌝ ⊑ ⌜af⌝ , ⌜sl⌝ ⊑ ⌜sl0⌝ + ⌜nf0⌝ , ¬ ⌜trf(sl, al)⌝ ⊑ lim(⌜λ n. ⌜trf((sl0 + nf0).take(n + 1), af.take(n))⌝ ⌝ ), ∀ m. ¬ # lim(⌜λ n. ⌜trf((sl0 + nf0).take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf((sl0 + nf0).take(n + 1), af.take(n))⌝ ⌝ ).take(m) ⊑ trf((sl0 + nf0).take(n + 1), af.take(n))) ⊦

Rule argument

Additional information

Used simplifier rules: append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ (⌜x⌝ + ⌜nf⌝ ).f = x + nf; (s) ⊦ ⌜x + nf⌝ = ⌜x⌝ + ⌜nf⌝ ; (s)

Proof step 35 from proof lifttr-cont (previous step) (back to tree)

User applied rule apply elim lemma

And obtained the following new goals

• Proof step 36:

  # sl0 = # sl, # al0 = # al, # al + 1 = # sl, af = (⌜al0⌝ + ⌜af0⌝ ).f, mon(trf), ⌜sl⌝ ⊑ ⌜sl0⌝ + ⌜nf0⌝ , ⌜al⌝ ⊑ ⌜af⌝ , ⌜trf(sl0, al0)⌝ ⊑ lim(⌜λ n. ⌜trf((sl0 + nf0).take(n + 1), af.take(n))⌝ ⌝ ), ¬ ⌜trf(sl, al)⌝ ⊑ lim(⌜λ n. ⌜trf((sl0 + nf0).take(n + 1), af.take(n))⌝ ⌝ ), ∀ m. ¬ # lim(⌜λ n. ⌜trf((sl0 + nf0).take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf((sl0 + nf0).take(n + 1), af.take(n))⌝ ⌝ ).take(m) ⊑ trf((sl0 + nf0).take(n + 1), af.take(n))) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ x0 = nf.take(n) ∧ nf0 = nf.drop(n) ↔ nf = (⌜x0⌝ + ⌜nf0⌝ ).f ∧ # x0 = n; (elim) Used terms: af, # al, al0, af0

Proof step 36 from proof lifttr-cont (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 37:

  # al0 = # al, # sl0 = # sl, # al + 1 = # sl, mon(trf), ⌜sl⌝ ⊑ ⌜sl0⌝ + ⌜nf0⌝ , ⌜trf(sl0, al0)⌝ ⊑ lim(⌜λ n. ⌜trf((sl0 + nf0).take(n + 1), (al0 + af0).take(n))⌝ ⌝ ), ⌜al⌝ ⊑ ⌜al0⌝ + ⌜af0⌝ , ¬ ⌜trf(sl, al)⌝ ⊑ lim(⌜λ n. ⌜trf((sl0 + nf0).take(n + 1), (al0 + af0).take(n))⌝ ⌝ ), ∀ m. ¬ # lim(⌜λ n. ⌜trf((sl0 + nf0).take(n + 1), (al0 + af0).take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf((sl0 + nf0).take(n + 1), (al0 + af0).take(n))⌝ ⌝ ).take(m) ⊑ trf((sl0 + nf0).take(n + 1), (al0 + af0).take(n))) ⊦

Rule argument

Additional information

Used simplifier rules: append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ (⌜x⌝ + ⌜nf⌝ ).f = x + nf; (s) ⊦ ⌜x + nf⌝ = ⌜x⌝ + ⌜nf⌝ ; (s)

Proof step 37 from proof lifttr-cont (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 38:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 39:

  # sl0 = # sl, # al0 = # al, # al + 1 = # sl, mon(trf), ⌜al⌝ ⊑ ⌜al0⌝ + ⌜af0⌝ , ⌜trf(sl0, al0)⌝ ⊑ lim(⌜λ n. ⌜trf((sl0 + nf0).take(n + 1), (al0 + af0).take(n))⌝ ⌝ ), ¬ ⌜trf(sl, al)⌝ ⊑ lim(⌜λ n. ⌜trf((sl0 + nf0).take(n + 1), (al0 + af0).take(n))⌝ ⌝ ), ∀ m. ¬ # lim(⌜λ n. ⌜trf((sl0 + nf0).take(n + 1), (al0 + af0).take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf((sl0 + nf0).take(n + 1), (al0 + af0).take(n))⌝ ⌝ ).take(m) ⊑ trf((sl0 + nf0).take(n + 1), (al0 + af0).take(n))), ∃ σ1. ⌜sl⌝ + σ1 = ⌜sl0⌝ + ⌜nf0⌝ ⊦

Rule argument

Additional information

Used terms: ⌜sl⌝ , ⌜sl0⌝ + ⌜nf0⌝

Proof step 38 from proof lifttr-cont (previous step) (back to tree)

A spec-lemma

Proof step 39 from proof lifttr-cont (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 40:

  # al0 = # al, # al + 1 = # sl0, mon(trf), ⌜al⌝ ⊑ ⌜al0⌝ + ⌜af0⌝ , ⌜trf(sl0, al0)⌝ ⊑ lim(⌜λ n. ⌜trf((sl0 + nf0).take(n + 1), (al0 + af0).take(n))⌝ ⌝ ), ¬ ⌜trf(sl0, al)⌝ ⊑ lim(⌜λ n. ⌜trf((sl0 + nf0).take(n + 1), (al0 + af0).take(n))⌝ ⌝ ), ∀ m. ¬ # lim(⌜λ n. ⌜trf((sl0 + nf0).take(n + 1), (al0 + af0).take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf((sl0 + nf0).take(n + 1), (al0 + af0).take(n))⌝ ⌝ ).take(m) ⊑ trf((sl0 + nf0).take(n + 1), (al0 + af0).take(n))) ⊦

Rule argument

Additional information

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

Proof step 40 from proof lifttr-cont (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 41:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

Rule argument

Additional information

Used simplifier rules: ⊦ # x = # y → (⌜x⌝ + σ = ⌜y⌝ + σ0 ↔ x = y ∧ σ = σ0); (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream Used terms: ⌜al⌝ , ⌜al0⌝ + ⌜af0⌝

Proof step 41 from proof lifttr-cont (previous step) (back to tree)

A spec-lemma

Proof step 42 from proof lifttr-cont (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 43:

  # σ1 = ∞, mon(trf), σ1 ⊑ ⌜sf⌝ , ¬ lift(trf)(σ1, ⌜af⌝ ) ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ ⊑ ⌜nf0⌝ ↔ nf = nf0; (s) add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜m⌝ + ∞ = ∞; (s) ⊦ # ⌜nf⌝ = ∞; (s)

Proof step 43 from proof lifttr-cont (previous step) (back to tree)

User applied rule constructor cut

And closed the goal!

Rule argument

Additional information

Used simplifier rules: lifttr-inf: ⊦ lift(trf)(⌜sf⌝ , ⌜af⌝ ) = lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ); Stream[A] freely generated by ⌝, ⌝; ⊦ # ⌜nf⌝ = ∞; (s) ⊦ ⌜nf⌝ ⊑ ⌜nf0⌝ ↔ nf = nf0; (s) ⊦ σ ⊑ σ; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s) ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 44 from proof lifttr-cont (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 45:

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

• Proof step 46:

  mon(trf), ∃ sf, af, bl. ¬((∀ n. lift(trf)(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ lift(trf)(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝ ) ⊦

Rule argument

Additional information

Used terms: lift(trf)

Proof step 45 from proof lifttr-cont (previous step) (back to tree)

Lemma pbdtr-def

Proof step 46 from proof lifttr-cont (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 47:

  mon(trf) ⊦ (∀ n. trf(sf.take(n + 1), af.take(n)) ⊑ bl) ↔ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ⊑ ⌜bl⌝

Rule argument

Additional information

Used simplifier rules: lifttr-fin: ⊦ lift(trf)(⌜sl⌝ , ⌜al⌝ ) = ⌜trf(sl, al)⌝ ; lifttr-inf: ⊦ lift(trf)(⌜sf⌝ , ⌜af⌝ ) = lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ); ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s)

Proof step 47 from proof lifttr-cont (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 48:

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

• Proof step 49:

  mon(trf), mon(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ∧ ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ≠ [] → (∀ n. ⌜n⌝ < # ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ → ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ (n) ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ )) ∧ (∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. ⌜n⌝ < # ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ∧ ⌜lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m)⌝ ⊑ ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ (n))) ⊦ (∀ n. trf(sf.take(n + 1), af.take(n)) ⊑ bl) ↔ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ⊑ ⌜bl⌝

Rule argument

Additional information

Used terms: ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝

Proof step 48 from proof lifttr-cont (previous step) (back to tree)

A spec-lemma

Proof step 49 from proof lifttr-cont (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 50:

  mon(trf), mon(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) → (∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ )) ∧ (∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n)))) ⊦ (∀ n. trf(sf.take(n + 1), af.take(n)) ⊑ bl) ↔ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ⊑ ⌜bl⌝

Rule argument

Additional information

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

Proof step 50 from proof lifttr-cont (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 51:

  mon(trf), ¬ mon(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ⊦ (∀ n. trf(sf.take(n + 1), af.take(n)) ⊑ bl) ↔ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ⊑ ⌜bl⌝

• Proof step 54:

  mon(trf), ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), ∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))) ⊦ (∀ n. trf(sf.take(n + 1), af.take(n)) ⊑ bl) ↔ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ⊑ ⌜bl⌝

Rule argument

Additional information

Proof step 51 from proof lifttr-cont (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 52:

  ⊦ mon(trf) ↔ (∀ sf, af. mon(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ))

• Proof step 53:

  ¬ mon(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), ∀ sf, af. mon(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ⊦ (∀ n. trf(sf.take(n + 1), af.take(n)) ⊑ bl) ↔ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ⊑ ⌜bl⌝

Rule argument

Additional information

Used terms: trf

Proof step 52 from proof lifttr-cont (previous step) (back to tree)

Lemma montrl-equiv

Proof step 53 from proof lifttr-cont (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: sf, af

Proof step 54 from proof lifttr-cont (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 55:

  mon(trf), ¬ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ⊑ ⌜bl⌝ , ∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))), ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), ∀ n. trf(sf.take(n + 1), af.take(n)) ⊑ bl ⊦

• Proof step 72:

  mon(trf), lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ⊑ ⌜bl⌝ , ∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))), ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ⊦ ∀ n. trf(sf.take(n + 1), af.take(n)) ⊑ bl

Rule argument

Additional information

Proof step 55 from proof lifttr-cont (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 56:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bl0⌝ , mon(trf), ¬ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ⊑ ⌜bl⌝ , ∀ n. trf(sf.take(n + 1), af.take(n)) ⊑ bl, ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), ∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))) ⊦

• Proof step 62:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bf⌝ , mon(trf), ¬ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ⊑ ⌜bl⌝ , ∀ n. trf(sf.take(n + 1), af.take(n)) ⊑ bl, ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), ∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))) ⊦

Rule argument

Additional information

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

Proof step 56 from proof lifttr-cont (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 57:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bl0⌝ , mon(trf), ¬ bl0 ⊑ bl, ∀ n. trf(sf.take(n + 1), af.take(n)) ⊑ bl, ∀ n. trf(sf.take(n + 1), af.take(n)) ⊑ bl0, ∀ m. ¬ # bl0 < m → (∃ n. bl0.take(m) ⊑ trf(sf.take(n + 1), af.take(n))) ⊦

Rule argument

Additional information

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

Proof step 57 from proof lifttr-cont (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 58:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bl0⌝ , mon(trf), ¬ bl0 ⊑ bl, ¬ # bl0 < # bl0 → (∃ n. bl0.take(# bl0) ⊑ trf(sf.take(n + 1), af.take(n))), ∀ n. trf(sf.take(n + 1), af.take(n)) ⊑ bl0, ∀ n. trf(sf.take(n + 1), af.take(n)) ⊑ bl ⊦

Rule argument

Additional information

Used terms: # bl0

Proof step 58 from proof lifttr-cont (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 59:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bl0⌝ , mon(trf), bl0 ⊑ trf(sf.take(n + 1), af.take(n)), ¬ bl0 ⊑ bl, ∀ n. trf(sf.take(n + 1), af.take(n)) ⊑ bl, ∀ n. trf(sf.take(n + 1), af.take(n)) ⊑ bl0 ⊦

Rule argument

Additional information

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

Proof step 59 from proof lifttr-cont (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 60:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bl0⌝ , mon(trf), bl0 ⊑ trf(sf.take(n + 1), af.take(n)), trf(sf.take(n + 1), af.take(n)) ⊑ bl0, ¬ bl0 ⊑ bl, ∀ n. trf(sf.take(n + 1), af.take(n)) ⊑ bl0, ∀ n. trf(sf.take(n + 1), af.take(n)) ⊑ bl ⊦

Rule argument

Additional information

Used terms: n

Proof step 60 from proof lifttr-cont (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 61:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜trf(sf.take(n + 1), af.take(n))⌝ , mon(trf), ¬ trf(sf.take(n + 1), af.take(n)) ⊑ bl, ∀ n0. trf(sf.take(n0 + 1), af.take(n0)) ⊑ trf(sf.take(n + 1), af.take(n)), ∀ n. trf(sf.take(n + 1), af.take(n)) ⊑ bl ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x0 = # x1 → (x0 ⊑ x1 ↔ x0 = x1); (s) ⊦ x ⊑ y → (y ⊑ x ↔ x = y); (s)

Proof step 61 from proof lifttr-cont (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n

Proof step 62 from proof lifttr-cont (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 63:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bf⌝ , mon(trf), ∀ n. trf(sf.take(n + 1), af.take(n)) ⊑ bl, ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ ⌜bf⌝ , ∀ m. ∃ n. bf.take(m) ⊑ trf(sf.take(n + 1), af.take(n)) ⊦

Rule argument

Additional information

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

Proof step 63 from proof lifttr-cont (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 64:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bf⌝ , mon(trf), ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ ⌜bf⌝ , ∀ n. trf(sf.take(n + 1), af.take(n)) ⊑ bl, ∃ n. bf.take(# bl + 1) ⊑ trf(sf.take(n + 1), af.take(n)) ⊦

Rule argument

Additional information

Used terms: # bl + 1

Proof step 64 from proof lifttr-cont (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 65:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bf⌝ , mon(trf), bf.take(# bl + 1) ⊑ trf(sf.take(n + 1), af.take(n)), ∀ n. trf(sf.take(n + 1), af.take(n)) ⊑ bl, ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Proof step 65 from proof lifttr-cont (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 66:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bf⌝ , mon(trf), bf.take(# bl + 1) ⊑ trf(sf.take(n + 1), af.take(n)), trf(sf.take(n + 1), af.take(n)) ⊑ bl, ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Used terms: n

Proof step 66 from proof lifttr-cont (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 67:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bf⌝ , mon(trf), trf(sf.take(n + 1), af.take(n)) ⊑ bl, bf.take(# bl + 1) ⊑ trf(sf.take(n + 1), af.take(n)), bf.take(# bl + 1) ⊑ bl, ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Used simplifier rules: prefix-trans: ⊦ x ⊑ y ∧ y ⊑ z → x ⊑ z; (forward) from specification list-last

Proof step 67 from proof lifttr-cont (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 68:

  ⊦ x ⊑ y ↔ (∃ z. x + z = y)

• Proof step 69:

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

Rule argument

Additional information

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

Proof step 68 from proof lifttr-cont (previous step) (back to tree)

A spec-lemma

Proof step 69 from proof lifttr-cont (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 70:

  bf.take(# bl + 1) + bl0 = bl, lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bf⌝ , mon(trf), trf(sf.take(n + 1), af.take(n)) ⊑ bl, bf.take(# bl + 1) ⊑ trf(sf.take(n + 1), af.take(n)), ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Proof step 70 from proof lifttr-cont (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 71:

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

Rule argument

Additional information

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

Proof step 71 from proof lifttr-cont (previous step) (back to tree)

A spec-lemma

Proof step 72 from proof lifttr-cont (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 73:

  mon(trf), lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ⊑ ⌜bl⌝ , ¬ trf(sf.take(n + 1), af.take(n)) ⊑ bl, ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), ∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))) ⊦

Rule argument

Additional information

Proof step 73 from proof lifttr-cont (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

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