Proof step 1 from proof pcont-lifted (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  lifted(alp)(lf, f), pmon(alp)(lf), ¬ pcont(lift(alp))(f) ⊦

Rule argument

Additional information

Proof step 2 from proof pcont-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α) ⊦ (lift(alp)(α1) ∧ lift(alp)(α2) ∧ α1 ⊑ α2 → f(α1) ⊑ f(α2)) ∧ ppbd(lift(alp))(f)

Rule argument

Additional information

Used simplifier rules: monl-relative-def: ⊦ pmon(alp)(lf) ↔ (∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2)); lifted-def: ⊦ lifted(alp)(lf, f) ↔ (∀ α. lift(alp)(α) → lift(lf)(α) = f(α)); cont-relative-def: ⊦ pcont(αp)(f) ↔ pmon(αp)(f) ∧ ppbd(αp)(f); mon-relative-def: ⊦ pmon(αp)(f) ↔ (∀ α1, α2. αp(α1) ∧ αp(α2) ∧ α1 ⊑ α2 → f(α1) ⊑ f(α2));

Proof step 3 from proof pcont-lifted (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 4:

  lift(alp)(α1), lift(alp)(α2), α1 ⊑ α2, ¬ f(α1) ⊑ f(α2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2) ⊦

• Proof step 35:

  ¬ ppbd(lift(alp))(f), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2) ⊦

Rule argument

Additional information

Proof step 4 from proof pcont-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 5:

  lift(alp)(α2), lift(alp)(α1), α1 ⊑ α2, ¬ f(α1) ⊑ f(α2), lift(alp)(α2) → lift(lf)(α2) = f(α2), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α) ⊦

Rule argument

Additional information

Used terms: α2

Proof step 5 from proof pcont-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  lift(lf)(α2) = f(α2), lift(alp)(α1), lift(alp)(α2), α1 ⊑ α2, ¬ f(α1) ⊑ f(α2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2) ⊦

Rule argument

Additional information

Proof step 6 from proof pcont-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 7:

  lift(lf)(α2) = f(α2), lift(alp)(α2), lift(alp)(α1), α1 ⊑ α2, ¬ f(α1) ⊑ f(α2), lift(alp)(α1) → lift(lf)(α1) = f(α1), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α) ⊦

Rule argument

Additional information

Used terms: α1

Proof step 7 from proof pcont-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  lift(lf)(α2) = f(α2), lift(lf)(α1) = f(α1), lift(alp)(α1), lift(alp)(α2), α1 ⊑ α2, ¬ f(α1) ⊑ f(α2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2) ⊦

Rule argument

Additional information

Proof step 8 from proof pcont-lifted (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 9:

  lift(lf)(α1) = f(α1), lift(lf)(α2) = f(α2), α2 = ⌜al⌝ , lift(alp)(α2), lift(alp)(α1), α1 ⊑ α2, ¬ f(α1) ⊑ f(α2), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α) ⊦

• Proof step 13:

  lift(lf)(α1) = f(α1), lift(lf)(α2) = f(α2), α2 = ⌜af⌝ , lift(alp)(α2), lift(alp)(α1), α1 ⊑ α2, ¬ f(α1) ⊑ f(α2), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α) ⊦

Rule argument

Additional information

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

Proof step 9 from proof pcont-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  lift(lf)(α1) = f(α1), ⌜lf(al)⌝ = f(⌜al⌝ ), alp(al), lift(alp)(α1), α1 ⊑ ⌜al⌝ , ¬ f(α1) ⊑ ⌜lf(al)⌝ , ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2) ⊦

Rule argument

Additional information

Used simplifier rules: lift-fin: ⊦ lift(lf)(⌜al⌝ ) = ⌜lf(al)⌝ ; lift-pred-fin: ⊦ lift(alp)(⌜al⌝ ) ↔ alp(al);

Proof step 10 from proof pcont-lifted (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 11:

  lift(lf)(α1) = f(α1), α1 = ⌜al0⌝ , ⌜lf(al)⌝ = f(⌜al⌝ ), alp(al), lift(alp)(α1), α1 ⊑ ⌜al⌝ , ¬ f(α1) ⊑ ⌜lf(al)⌝ , ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α) ⊦

Rule argument

Additional information

Used simplifier rules: lift-inf: ⊦ lift(lf)(⌜af⌝ ) = lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ); Stream[A] freely generated by ⌝, ⌝; ⊦ ¬ ⌜nf⌝ ⊑ ⌜x⌝ ; (s)

Proof step 11 from proof pcont-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  ⌜lf(al0)⌝ = f(⌜al0⌝ ), ⌜lf(al)⌝ = f(⌜al⌝ ), alp(al0), alp(al), al0 ⊑ al, ¬ lf(al0) ⊑ lf(al), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α) ⊦

Rule argument

Additional information

Used simplifier rules: lift-fin: ⊦ lift(lf)(⌜al⌝ ) = ⌜lf(al)⌝ ; lift-pred-fin: ⊦ lift(alp)(⌜al⌝ ) ↔ alp(al); ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s)

Proof step 12 from proof pcont-lifted (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: al0, al

Proof step 13 from proof pcont-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 14:

  lift(lf)(α1) = f(α1), lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), lift(alp)(α1), α1 ⊑ ⌜af⌝ , ¬ f(α1) ⊑ f(⌜af⌝ ), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. alp(af.take(n)) ⊦

Rule argument

Additional information

Used simplifier rules: lift-inf: ⊦ lift(lf)(⌜af⌝ ) = lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ); lift-pred-inf: ⊦ lift(alp)(⌜af⌝ ) ↔ (∀ n. alp(af.take(n)));

Proof step 14 from proof pcont-lifted (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 15:

  lift(lf)(α1) = f(α1), lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), α1 = ⌜al⌝ , lift(alp)(α1), α1 ⊑ ⌜af⌝ , ¬ f(α1) ⊑ f(⌜af⌝ ), ∀ n. alp(af.take(n)), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α) ⊦

Rule argument

Additional information

Used simplifier rules: lift-inf: ⊦ lift(lf)(⌜af⌝ ) = lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ); lift-pred-inf: ⊦ lift(alp)(⌜af⌝ ) ↔ (∀ n. alp(af.take(n))); Stream[A] freely generated by ⌝, ⌝; ⊦ ⌜nf⌝ ⊑ ⌜nf0⌝ ↔ nf = nf0; (s) ⊦ σ ⊑ σ; (s)

Proof step 15 from proof pcont-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), ⌜lf(al)⌝ = f(⌜al⌝ ), alp(al), ⌜al⌝ ⊑ ⌜af⌝ , ¬ ⌜lf(al)⌝ ⊑ f(⌜af⌝ ), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. alp(af.take(n)) ⊦

Rule argument

Additional information

Used simplifier rules: lift-fin: ⊦ lift(lf)(⌜al⌝ ) = ⌜lf(al)⌝ ; lift-pred-fin: ⊦ lift(alp)(⌜al⌝ ) ↔ alp(al);

Proof step 16 from proof pcont-lifted (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 17:

  ⊦ 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 18:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), ⌜lf(al)⌝ = f(⌜al⌝ ), alp(al), ⌜al⌝ ⊑ ⌜af⌝ , ¬ ⌜lf(al)⌝ ⊑ f(⌜af⌝ ), mon(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) ∧ ⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ≠ [] → (∀ n. ⌜n⌝ < # ⌜λ n. ⌜lf(af.take(n))⌝ ⌝ → ⌜λ n. ⌜lf(af.take(n))⌝ ⌝ (n) ⊑ lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ )) ∧ (∀ m. ¬ # lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. ⌜n⌝ < # ⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ∧ ⌜lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ).take(m)⌝ ⊑ ⌜λ n. ⌜lf(af.take(n))⌝ ⌝ (n))), ∀ n. alp(af.take(n)), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α) ⊦

Rule argument

Additional information

Used terms: ⌜λ n. ⌜lf(af.take(n))⌝ ⌝

Proof step 17 from proof pcont-lifted (previous step) (back to tree)

A spec-lemma

Proof step 18 from proof pcont-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 19:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), ⌜lf(al)⌝ = f(⌜al⌝ ), alp(al), ⌜al⌝ ⊑ ⌜af⌝ , ¬ ⌜lf(al)⌝ ⊑ f(⌜af⌝ ), mon(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) → (∀ n. ⌜lf(af.take(n))⌝ ⊑ f(⌜af⌝ )) ∧ (∀ m. ¬ # f(⌜af⌝ ) < ⌜m⌝ → (∃ n. f(⌜af⌝ ).take(m) ⊑ lf(af.take(n)))), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. alp(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 19 from proof pcont-lifted (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 20:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), ⌜lf(al)⌝ = f(⌜al⌝ ), alp(al), ⌜al⌝ ⊑ ⌜af⌝ , ¬ mon(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ), ¬ ⌜lf(al)⌝ ⊑ f(⌜af⌝ ), ∀ n. alp(af.take(n)), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α) ⊦

• Proof step 30:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), ⌜lf(al)⌝ = f(⌜al⌝ ), alp(al), ⌜al⌝ ⊑ ⌜af⌝ , ¬ ⌜lf(al)⌝ ⊑ f(⌜af⌝ ), ∀ n. alp(af.take(n)), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ n. ⌜lf(af.take(n))⌝ ⊑ f(⌜af⌝ ), ∀ m. ¬ # f(⌜af⌝ ) < ⌜m⌝ → (∃ n. f(⌜af⌝ ).take(m) ⊑ lf(af.take(n))) ⊦

Rule argument

Additional information

Proof step 20 from proof pcont-lifted (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 21:

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

• Proof step 22:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), ⌜lf(al)⌝ = f(⌜al⌝ ), alp(al), ⌜al⌝ ⊑ ⌜af⌝ , ¬ ⌜lf(al)⌝ ⊑ f(⌜af⌝ ), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. alp(af.take(n)), ∃ n. ¬(⌜n⌝ + ⌜1⌝ < # ⌜λ n. ⌜lf(af.take(n))⌝ ⌝ → ⌜λ n. ⌜lf(af.take(n))⌝ ⌝ (n) ⊑ ⌜λ n. ⌜lf(af.take(n))⌝ ⌝ (n + 1)) ⊦

Rule argument

Additional information

Used terms: ⌜λ n. ⌜lf(af.take(n))⌝ ⌝

Proof step 21 from proof pcont-lifted (previous step) (back to tree)

A spec-lemma

Proof step 22 from proof pcont-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 23:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), ⌜lf(al)⌝ = f(⌜al⌝ ), alp(al), ⌜al⌝ ⊑ ⌜af⌝ , ¬ ⌜lf(al)⌝ ⊑ f(⌜af⌝ ), ¬ lf(af.take(n)) ⊑ lf(af.take(n + 1)), ∀ n. alp(af.take(n)), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α) ⊦

Rule argument

Additional information

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

Proof step 23 from proof pcont-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 24:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), ⌜lf(al)⌝ = f(⌜al⌝ ), alp(al), ⌜al⌝ ⊑ ⌜af⌝ , ¬ ⌜lf(al)⌝ ⊑ f(⌜af⌝ ), ¬ lf(af.take(n)) ⊑ lf(af.take(n + 1)), alp(af.take(n)) ∧ alp(af.take(n + 1)) ∧ af.take(n) ⊑ af.take(n + 1) → lf(af.take(n)) ⊑ lf(af.take(n + 1)), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ n. alp(af.take(n)) ⊦

Rule argument

Additional information

Used terms: af.take(n), af.take(n + 1)

Proof step 24 from proof pcont-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 25:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), ⌜lf(al)⌝ = f(⌜al⌝ ), alp(al), ⌜al⌝ ⊑ ⌜af⌝ , ¬ ⌜lf(al)⌝ ⊑ f(⌜af⌝ ), ¬ lf(af.take(n)) ⊑ lf(af.take(n + 1)), ∀ n. alp(af.take(n)), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α) ⊦ alp(af.take(n)) ∧ alp(af.take(n + 1)) ∧ af.take(n) ⊑ af.take(n + 1)

Rule argument

Additional information

Proof step 25 from proof pcont-lifted (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 26:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), ⌜lf(al)⌝ = f(⌜al⌝ ), alp(al), ⌜al⌝ ⊑ ⌜af⌝ , ¬ alp(af.take(n)), ¬ ⌜lf(al)⌝ ⊑ f(⌜af⌝ ), ¬ lf(af.take(n)) ⊑ lf(af.take(n + 1)), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ n. alp(af.take(n)) ⊦

• Proof step 27:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), ⌜lf(al)⌝ = f(⌜al⌝ ), alp(al), ⌜al⌝ ⊑ ⌜af⌝ , ¬ alp(af.take(n + 1)), ¬ ⌜lf(al)⌝ ⊑ f(⌜af⌝ ), ¬ lf(af.take(n)) ⊑ lf(af.take(n + 1)), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ n. alp(af.take(n)) ⊦

• Proof step 28:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), ⌜lf(al)⌝ = f(⌜al⌝ ), alp(al), ⌜al⌝ ⊑ ⌜af⌝ , ¬ af.take(n) ⊑ af.take(n + 1), ¬ ⌜lf(al)⌝ ⊑ f(⌜af⌝ ), ¬ lf(af.take(n)) ⊑ lf(af.take(n + 1)), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ n. alp(af.take(n)) ⊦

Rule argument

Additional information

Proof step 26 from proof pcont-lifted (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 27 from proof pcont-lifted (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n + 1

Proof step 28 from proof pcont-lifted (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 29:

  ⊦ nf.take(m + 1) = nf.take(m) + nf(m) '

Rule argument

Additional information

Used simplifier rules: ⊦ # x0 = # x1 → (x0 ⊑ x1 + x2 ↔ x0 = x1); (s) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: af, n

Proof step 29 from proof pcont-lifted (previous step) (back to tree)

A spec-lemma

Proof step 30 from proof pcont-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 31:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), ⌜lf(al)⌝ = f(⌜al⌝ ), alp(al), ⌜al⌝ ⊑ ⌜af⌝ , ⌜lf(af.take(# al))⌝ ⊑ f(⌜af⌝ ), ¬ ⌜lf(al)⌝ ⊑ f(⌜af⌝ ), ∀ m. ¬ # f(⌜af⌝ ) < ⌜m⌝ → (∃ n. f(⌜af⌝ ).take(m) ⊑ lf(af.take(n))), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. alp(af.take(n)) ⊦

Rule argument

Additional information

Used terms: # al

Proof step 31 from proof pcont-lifted (previous step) (back to tree)

User applied rule apply elim lemma

And obtained the following new goals

• Proof step 32:

  # al0 = # al, af = (⌜al0⌝ + ⌜nf0⌝ ).f, lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), ⌜lf(al)⌝ = f(⌜al⌝ ), alp(al), ⌜lf(al0)⌝ ⊑ f(⌜af⌝ ), ⌜al⌝ ⊑ ⌜af⌝ , ¬ ⌜lf(al)⌝ ⊑ f(⌜af⌝ ), ∀ n. alp(af.take(n)), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ m. ¬ # f(⌜af⌝ ) < ⌜m⌝ → (∃ n. f(⌜af⌝ ).take(m) ⊑ lf(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, nf0

Proof step 32 from proof pcont-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 33:

  # al0 = # al, lim(⌜λ n. ⌜lf((al0 + nf0).take(n))⌝ ⌝ ) = f(⌜al0⌝ + ⌜nf0⌝ ), ⌜lf(al)⌝ = f(⌜al⌝ ), alp(al), ⌜al⌝ ⊑ ⌜al0⌝ + ⌜nf0⌝ , ⌜lf(al0)⌝ ⊑ f(⌜al0⌝ + ⌜nf0⌝ ), ¬ ⌜lf(al)⌝ ⊑ f(⌜al0⌝ + ⌜nf0⌝ ), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ m. ¬ # f(⌜al0⌝ + ⌜nf0⌝ ) < ⌜m⌝ → (∃ n. f(⌜al0⌝ + ⌜nf0⌝ ).take(m) ⊑ lf((al0 + nf0).take(n))), ∀ n. alp((al0 + nf0).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 33 from proof pcont-lifted (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 34:

  ⊦ 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⌝ + ⌜nf0⌝

Proof step 34 from proof pcont-lifted (previous step) (back to tree)

A spec-lemma

Proof step 35 from proof pcont-lifted (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 36:

  ⊦ ppbd(αp)(f) ↔ (∀ af, bl. αp(⌜af⌝ ) → ((∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ f(⌜af⌝ ) ⊑ ⌜bl⌝ ))

• Proof step 37:

  ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∃ af, bl. ¬(lift(alp)(⌜af⌝ ) → ((∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ f(⌜af⌝ ) ⊑ ⌜bl⌝ )) ⊦

Rule argument

Additional information

Used terms: lift(alp), f

Proof step 36 from proof pcont-lifted (previous step) (back to tree)

Lemma pbd-relative-def

Proof step 37 from proof pcont-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 38:

  ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. alp(af.take(n)) ⊦ (∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ f(⌜af⌝ ) ⊑ ⌜bl⌝

Rule argument

Additional information

Used simplifier rules: lift-pred-inf: ⊦ lift(alp)(⌜af⌝ ) ↔ (∀ n. alp(af.take(n)));

Proof step 38 from proof pcont-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 39:

  lift(alp)(⌜af⌝ ) → lift(lf)(⌜af⌝ ) = f(⌜af⌝ ), ∀ n. alp(af.take(n)), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α) ⊦ (∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ f(⌜af⌝ ) ⊑ ⌜bl⌝

Rule argument

Additional information

Used terms: ⌜af⌝

Proof step 39 from proof pcont-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 40:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. alp(af.take(n)) ⊦ (∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ f(⌜af⌝ ) ⊑ ⌜bl⌝

Rule argument

Additional information

Used simplifier rules: lift-pred-inf: ⊦ lift(alp)(⌜af⌝ ) ↔ (∀ n. alp(af.take(n))); lift-inf: ⊦ lift(lf)(⌜af⌝ ) = lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ );

Proof step 40 from proof pcont-lifted (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 41:

  ⊦ 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 42:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), mon(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) ∧ ⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ≠ [] → (∀ n. ⌜n⌝ < # ⌜λ n. ⌜lf(af.take(n))⌝ ⌝ → ⌜λ n. ⌜lf(af.take(n))⌝ ⌝ (n) ⊑ lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ )) ∧ (∀ m. ¬ # lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. ⌜n⌝ < # ⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ∧ ⌜lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ).take(m)⌝ ⊑ ⌜λ n. ⌜lf(af.take(n))⌝ ⌝ (n))), ∀ n. alp(af.take(n)), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α) ⊦ (∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ f(⌜af⌝ ) ⊑ ⌜bl⌝

Rule argument

Additional information

Used terms: ⌜λ n. ⌜lf(af.take(n))⌝ ⌝

Proof step 41 from proof pcont-lifted (previous step) (back to tree)

A spec-lemma

Proof step 42 from proof pcont-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 43:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), mon(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) → (∀ n. ⌜lf(af.take(n))⌝ ⊑ f(⌜af⌝ )) ∧ (∀ m. ¬ # f(⌜af⌝ ) < ⌜m⌝ → (∃ n. f(⌜af⌝ ).take(m) ⊑ lf(af.take(n)))), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. alp(af.take(n)) ⊦ (∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ f(⌜af⌝ ) ⊑ ⌜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 43 from proof pcont-lifted (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 44:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), ¬ mon(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ), ∀ n. alp(af.take(n)), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α) ⊦ (∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ f(⌜af⌝ ) ⊑ ⌜bl⌝

• Proof step 54:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), ∀ n. alp(af.take(n)), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ n. ⌜lf(af.take(n))⌝ ⊑ f(⌜af⌝ ), ∀ m. ¬ # f(⌜af⌝ ) < ⌜m⌝ → (∃ n. f(⌜af⌝ ).take(m) ⊑ lf(af.take(n))) ⊦ (∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ f(⌜af⌝ ) ⊑ ⌜bl⌝

Rule argument

Additional information

Proof step 44 from proof pcont-lifted (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 45:

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

• Proof step 46:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. alp(af.take(n)), ∃ n. ¬(⌜n⌝ + ⌜1⌝ < # ⌜λ n. ⌜lf(af.take(n))⌝ ⌝ → ⌜λ n. ⌜lf(af.take(n))⌝ ⌝ (n) ⊑ ⌜λ n. ⌜lf(af.take(n))⌝ ⌝ (n + 1)) ⊦ (∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ f(⌜af⌝ ) ⊑ ⌜bl⌝

Rule argument

Additional information

Used terms: ⌜λ n. ⌜lf(af.take(n))⌝ ⌝

Proof step 45 from proof pcont-lifted (previous step) (back to tree)

A spec-lemma

Proof step 46 from proof pcont-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 47:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), ¬ lf(af.take(n)) ⊑ lf(af.take(n + 1)), ∀ n. alp(af.take(n)), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α) ⊦ (∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ f(⌜af⌝ ) ⊑ ⌜bl⌝

Rule argument

Additional information

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

Proof step 47 from proof pcont-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 48:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), ¬ lf(af.take(n)) ⊑ lf(af.take(n + 1)), alp(af.take(n)) ∧ alp(af.take(n + 1)) ∧ af.take(n) ⊑ af.take(n + 1) → lf(af.take(n)) ⊑ lf(af.take(n + 1)), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ n. alp(af.take(n)) ⊦ (∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ f(⌜af⌝ ) ⊑ ⌜bl⌝

Rule argument

Additional information

Used terms: af.take(n), af.take(n + 1)

Proof step 48 from proof pcont-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 49:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), ¬ lf(af.take(n)) ⊑ lf(af.take(n + 1)), ∀ n. alp(af.take(n)), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α) ⊦ (∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ f(⌜af⌝ ) ⊑ ⌜bl⌝ , alp(af.take(n)) ∧ alp(af.take(n + 1)) ∧ af.take(n) ⊑ af.take(n + 1)

Rule argument

Additional information

Proof step 49 from proof pcont-lifted (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 50:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), ¬ alp(af.take(n)), ¬ lf(af.take(n)) ⊑ lf(af.take(n + 1)), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ n. alp(af.take(n)) ⊦ (∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ f(⌜af⌝ ) ⊑ ⌜bl⌝

• Proof step 51:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), ¬ alp(af.take(n + 1)), ¬ lf(af.take(n)) ⊑ lf(af.take(n + 1)), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ n. alp(af.take(n)) ⊦ (∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ f(⌜af⌝ ) ⊑ ⌜bl⌝

• Proof step 52:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), ¬ af.take(n) ⊑ af.take(n + 1), ¬ lf(af.take(n)) ⊑ lf(af.take(n + 1)), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ n. alp(af.take(n)) ⊦ (∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ f(⌜af⌝ ) ⊑ ⌜bl⌝

Rule argument

Additional information

Proof step 50 from proof pcont-lifted (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 51 from proof pcont-lifted (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n + 1

Proof step 52 from proof pcont-lifted (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 53:

  ⊦ nf.take(m + 1) = nf.take(m) + nf(m) '

Rule argument

Additional information

Used simplifier rules: ⊦ # x0 = # x1 → (x0 ⊑ x1 + x2 ↔ x0 = x1); (s) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: af, n

Proof step 53 from proof pcont-lifted (previous step) (back to tree)

A spec-lemma

Proof step 54 from proof pcont-lifted (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 55:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), ¬ f(⌜af⌝ ) ⊑ ⌜bl⌝ , ∀ m. ¬ # f(⌜af⌝ ) < ⌜m⌝ → (∃ n. f(⌜af⌝ ).take(m) ⊑ lf(af.take(n))), ∀ n. ⌜lf(af.take(n))⌝ ⊑ f(⌜af⌝ ), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. alp(af.take(n)), ∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ⊦

• Proof step 78:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), f(⌜af⌝ ) ⊑ ⌜bl⌝ , ∀ m. ¬ # f(⌜af⌝ ) < ⌜m⌝ → (∃ n. f(⌜af⌝ ).take(m) ⊑ lf(af.take(n))), ∀ n. ⌜lf(af.take(n))⌝ ⊑ f(⌜af⌝ ), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. alp(af.take(n)) ⊦ ∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝

Rule argument

Additional information

Proof step 55 from proof pcont-lifted (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 56:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = ⌜bl0⌝ , ¬ f(⌜af⌝ ) ⊑ ⌜bl⌝ , ∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∀ n. alp(af.take(n)), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ n. ⌜lf(af.take(n))⌝ ⊑ f(⌜af⌝ ), ∀ m. ¬ # f(⌜af⌝ ) < ⌜m⌝ → (∃ n. f(⌜af⌝ ).take(m) ⊑ lf(af.take(n))) ⊦

• Proof step 64:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = ⌜bf⌝ , ¬ f(⌜af⌝ ) ⊑ ⌜bl⌝ , ∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∀ n. alp(af.take(n)), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ n. ⌜lf(af.take(n))⌝ ⊑ f(⌜af⌝ ), ∀ m. ¬ # f(⌜af⌝ ) < ⌜m⌝ → (∃ n. f(⌜af⌝ ).take(m) ⊑ lf(af.take(n))) ⊦

Rule argument

Additional information

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

Proof step 56 from proof pcont-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 57:

  f(⌜af⌝ ) = ⌜bl0⌝ , lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = ⌜bl0⌝ , ¬ bl0 ⊑ bl, ∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∀ n. alp(af.take(n)), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ n. lf(af.take(n)) ⊑ bl0, ∀ m. ¬ # bl0 < m → (∃ n. bl0.take(m) ⊑ lf(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 pcont-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 58:

  f(⌜af⌝ ) = ⌜bl0⌝ , lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = ⌜bl0⌝ , ¬ bl0 ⊑ bl, ¬ # bl0 < # bl0 → (∃ n. bl0.take(# bl0) ⊑ lf(af.take(n))), ∀ n. lf(af.take(n)) ⊑ bl0, ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. alp(af.take(n)), ∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used terms: # bl0

Proof step 58 from proof pcont-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 59:

  f(⌜af⌝ ) = ⌜bl0⌝ , lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = ⌜bl0⌝ , bl0 ⊑ lf(af.take(n)), ¬ bl0 ⊑ bl, ∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∀ n. alp(af.take(n)), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ n. lf(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 pcont-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 60:

  f(⌜af⌝ ) = ⌜bl0⌝ , lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = ⌜bl0⌝ , bl0 ⊑ lf(af.take(n)), ¬ bl0 ⊑ bl, lift(alp)(⌜af.take(n)⌝ ) → lift(lf)(⌜af.take(n)⌝ ) = f(⌜af.take(n)⌝ ), ∀ n. lf(af.take(n)) ⊑ bl0, ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. alp(af.take(n)), ∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used terms: ⌜af.take(n)⌝

Proof step 60 from proof pcont-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 61:

  f(⌜af⌝ ) = ⌜bl0⌝ , lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = ⌜bl0⌝ , bl0 ⊑ lf(af.take(n)), ¬ bl0 ⊑ bl, alp(af.take(n)) → ⌜lf(af.take(n))⌝ = f(⌜af.take(n)⌝ ), ∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∀ n. alp(af.take(n)), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. lf(af.take(n)) ⊑ bl0 ⊦

Rule argument

Additional information

Used simplifier rules: lift-pred-fin: ⊦ lift(alp)(⌜al⌝ ) ↔ alp(al); lift-fin: ⊦ lift(lf)(⌜al⌝ ) = ⌜lf(al)⌝ ;

Proof step 61 from proof pcont-lifted (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 62:

  f(⌜af⌝ ) = ⌜bl0⌝ , lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = ⌜bl0⌝ , bl0 ⊑ lf(af.take(n)), ¬ alp(af.take(n)), ¬ bl0 ⊑ bl, ∀ n. lf(af.take(n)) ⊑ bl0, ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. alp(af.take(n)), ∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ⊦

• Proof step 63:

  f(⌜af⌝ ) = ⌜bl0⌝ , lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = ⌜bl0⌝ , ⌜lf(af.take(n))⌝ = f(⌜af.take(n)⌝ ), bl0 ⊑ lf(af.take(n)), ¬ bl0 ⊑ bl, ∀ n. lf(af.take(n)) ⊑ bl0, ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. alp(af.take(n)), ∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Proof step 62 from proof pcont-lifted (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 63 from proof pcont-lifted (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s) prefix-trans: ⊦ x ⊑ y ∧ y ⊑ z → x ⊑ z; (forward) from specification list-last Used terms: n

Proof step 64 from proof pcont-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 65:

  f(⌜af⌝ ) = ⌜bf⌝ , lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = ⌜bf⌝ , ∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∀ n. alp(af.take(n)), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ n. ⌜lf(af.take(n))⌝ ⊑ ⌜bf⌝ , ∀ m. ∃ n. bf.take(m) ⊑ lf(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 65 from proof pcont-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 66:

  f(⌜af⌝ ) = ⌜bf⌝ , lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = ⌜bf⌝ , ∀ n. ⌜lf(af.take(n))⌝ ⊑ ⌜bf⌝ , ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. alp(af.take(n)), ∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∃ n. bf.take(# bl + 1) ⊑ lf(af.take(n)) ⊦

Rule argument

Additional information

Used terms: # bl + 1

Proof step 66 from proof pcont-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 67:

  f(⌜af⌝ ) = ⌜bf⌝ , lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = ⌜bf⌝ , bf.take(# bl + 1) ⊑ lf(af.take(n)), ∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∀ n. alp(af.take(n)), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ n. ⌜lf(af.take(n))⌝ ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Proof step 67 from proof pcont-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 68:

  f(⌜af⌝ ) = ⌜bf⌝ , lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = ⌜bf⌝ , bf.take(# bl + 1) ⊑ lf(af.take(n)), lift(alp)(⌜af.take(n)⌝ ) → lift(lf)(⌜af.take(n)⌝ ) = f(⌜af.take(n)⌝ ), ∀ n. ⌜lf(af.take(n))⌝ ⊑ ⌜bf⌝ , ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. alp(af.take(n)), ∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used terms: ⌜af.take(n)⌝

Proof step 68 from proof pcont-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 69:

  f(⌜af⌝ ) = ⌜bf⌝ , lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = ⌜bf⌝ , bf.take(# bl + 1) ⊑ lf(af.take(n)), alp(af.take(n)) → ⌜lf(af.take(n))⌝ = f(⌜af.take(n)⌝ ), ∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∀ n. alp(af.take(n)), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. ⌜lf(af.take(n))⌝ ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Used simplifier rules: lift-pred-fin: ⊦ lift(alp)(⌜al⌝ ) ↔ alp(al); lift-fin: ⊦ lift(lf)(⌜al⌝ ) = ⌜lf(al)⌝ ;

Proof step 69 from proof pcont-lifted (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 70:

  f(⌜af⌝ ) = ⌜bf⌝ , lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = ⌜bf⌝ , bf.take(# bl + 1) ⊑ lf(af.take(n)), ¬ alp(af.take(n)), ∀ n. ⌜lf(af.take(n))⌝ ⊑ ⌜bf⌝ , ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. alp(af.take(n)), ∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ⊦

• Proof step 71:

  f(⌜af⌝ ) = ⌜bf⌝ , lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = ⌜bf⌝ , ⌜lf(af.take(n))⌝ = f(⌜af.take(n)⌝ ), bf.take(# bl + 1) ⊑ lf(af.take(n)), ∀ n. ⌜lf(af.take(n))⌝ ⊑ ⌜bf⌝ , ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. alp(af.take(n)), ∀ n. f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Proof step 70 from proof pcont-lifted (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 71 from proof pcont-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 72:

  f(⌜af⌝ ) = ⌜bf⌝ , lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = ⌜bf⌝ , ⌜lf(af.take(n))⌝ = f(⌜af.take(n)⌝ ), bf.take(# bl + 1) ⊑ lf(af.take(n)), f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∀ n. alp(af.take(n)), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. ⌜lf(af.take(n))⌝ ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Used terms: n

Proof step 72 from proof pcont-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 73:

  f(⌜af⌝ ) = ⌜bf⌝ , lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = ⌜bf⌝ , ⌜lf(af.take(n))⌝ = f(⌜af.take(n)⌝ ), bf.take(# bl + 1) ⊑ lf(af.take(n)), lf(af.take(n)) ⊑ bl, bf.take(# bl + 1) ⊑ bl, ∀ n. alp(af.take(n)), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. ⌜lf(af.take(n))⌝ ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

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

Proof step 73 from proof pcont-lifted (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 74:

  ⊦ x ⊑ y ↔ (∃ z. x + z = y)

• Proof step 75:

  f(⌜af⌝ ) = ⌜bf⌝ , lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = ⌜bf⌝ , ⌜lf(af.take(n))⌝ = f(⌜af.take(n)⌝ ), lf(af.take(n)) ⊑ bl, bf.take(# bl + 1) ⊑ lf(af.take(n)), ∀ n. ⌜lf(af.take(n))⌝ ⊑ ⌜bf⌝ , ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. alp(af.take(n)), ∃ z. bf.take(# bl + 1) + z = bl ⊦

Rule argument

Additional information

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

Proof step 74 from proof pcont-lifted (previous step) (back to tree)

A spec-lemma

Proof step 75 from proof pcont-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 76:

  bf.take(# bl + 1) + bl0 = bl, f(⌜af⌝ ) = ⌜bf⌝ , lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = ⌜bf⌝ , ⌜lf(af.take(n))⌝ = f(⌜af.take(n)⌝ ), bf.take(# bl + 1) ⊑ lf(af.take(n)), lf(af.take(n)) ⊑ bl, ∀ n. alp(af.take(n)), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. ⌜lf(af.take(n))⌝ ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Proof step 76 from proof pcont-lifted (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 77:

  ⊦ 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 77 from proof pcont-lifted (previous step) (back to tree)

A spec-lemma

Proof step 78 from proof pcont-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 79:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), f(⌜af⌝ ) ⊑ ⌜bl⌝ , ¬ f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∀ n. alp(af.take(n)), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ n. ⌜lf(af.take(n))⌝ ⊑ f(⌜af⌝ ), ∀ m. ¬ # f(⌜af⌝ ) < ⌜m⌝ → (∃ n. f(⌜af⌝ ).take(m) ⊑ lf(af.take(n))) ⊦

Rule argument

Additional information

Proof step 79 from proof pcont-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 80:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), f(⌜af⌝ ) ⊑ ⌜bl⌝ , ⌜lf(af.take(n))⌝ ⊑ f(⌜af⌝ ), ¬ f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∀ m. ¬ # f(⌜af⌝ ) < ⌜m⌝ → (∃ n. f(⌜af⌝ ).take(m) ⊑ lf(af.take(n))), ∀ n. ⌜lf(af.take(n))⌝ ⊑ f(⌜af⌝ ), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. alp(af.take(n)) ⊦

Rule argument

Additional information

Used terms: n

Proof step 80 from proof pcont-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 81:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), f(⌜af⌝ ) ⊑ ⌜bl⌝ , ⌜lf(af.take(n))⌝ ⊑ f(⌜af⌝ ), lf(af.take(n)) ⊑ bl, ¬ f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∀ m. ¬ # f(⌜af⌝ ) < ⌜m⌝ → (∃ n. f(⌜af⌝ ).take(m) ⊑ lf(af.take(n))), ∀ n. ⌜lf(af.take(n))⌝ ⊑ f(⌜af⌝ ), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. alp(af.take(n)) ⊦

Rule argument

Additional information

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

Proof step 81 from proof pcont-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 82:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), lf(af.take(n)) ⊑ bl, ⌜lf(af.take(n))⌝ ⊑ f(⌜af⌝ ), f(⌜af⌝ ) ⊑ ⌜bl⌝ , ¬ f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , lift(alp)(⌜af.take(n)⌝ ) → lift(lf)(⌜af.take(n)⌝ ) = f(⌜af.take(n)⌝ ), ∀ n. alp(af.take(n)), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ n. ⌜lf(af.take(n))⌝ ⊑ f(⌜af⌝ ), ∀ m. ¬ # f(⌜af⌝ ) < ⌜m⌝ → (∃ n. f(⌜af⌝ ).take(m) ⊑ lf(af.take(n))) ⊦

Rule argument

Additional information

Used terms: ⌜af.take(n)⌝

Proof step 82 from proof pcont-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 83:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), f(⌜af⌝ ) ⊑ ⌜bl⌝ , ⌜lf(af.take(n))⌝ ⊑ f(⌜af⌝ ), lf(af.take(n)) ⊑ bl, ¬ f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , alp(af.take(n)) → ⌜lf(af.take(n))⌝ = f(⌜af.take(n)⌝ ), ∀ m. ¬ # f(⌜af⌝ ) < ⌜m⌝ → (∃ n. f(⌜af⌝ ).take(m) ⊑ lf(af.take(n))), ∀ n. ⌜lf(af.take(n))⌝ ⊑ f(⌜af⌝ ), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ n. alp(af.take(n)) ⊦

Rule argument

Additional information

Used simplifier rules: lift-pred-fin: ⊦ lift(alp)(⌜al⌝ ) ↔ alp(al); lift-fin: ⊦ lift(lf)(⌜al⌝ ) = ⌜lf(al)⌝ ;

Proof step 83 from proof pcont-lifted (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 84:

  lim(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) = f(⌜af⌝ ), lf(af.take(n)) ⊑ bl, ⌜lf(af.take(n))⌝ ⊑ f(⌜af⌝ ), f(⌜af⌝ ) ⊑ ⌜bl⌝ , ¬ alp(af.take(n)), ¬ f(⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∀ n. alp(af.take(n)), ∀ al1, al2. alp(al1) ∧ alp(al2) ∧ al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∀ α. lift(alp)(α) → lift(lf)(α) = f(α), ∀ n. ⌜lf(af.take(n))⌝ ⊑ f(⌜af⌝ ), ∀ m. ¬ # f(⌜af⌝ ) < ⌜m⌝ → (∃ n. f(⌜af⌝ ).take(m) ⊑ lf(af.take(n))) ⊦

Rule argument

Additional information

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

Proof step 84 from proof pcont-lifted (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n