Proof step 1 from proof montrl-equiv (conclusion) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 2:

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

• Proof step 3:

  ⊦ (∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 → trf(sl1, al1) ⊑ trf(sl2, al2)) ↔ (∀ sf, af. mon(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ))

Rule argument

Additional information

Used terms: trf

Proof step 2 from proof montrl-equiv (previous step) (back to tree)

Lemma montrl-def

Proof step 3 from proof montrl-equiv (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 4:

  ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 → trf(sl1, al1) ⊑ trf(sl2, al2) ⊦ ∀ sf, af. mon(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ )

• Proof step 15:

  ∀ sf, af. mon(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ⊦ ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 → trf(sl1, al1) ⊑ trf(sl2, al2)

Rule argument

Additional information

Proof step 4 from proof montrl-equiv (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  ¬ mon(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 → trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

Rule argument

Additional information

Proof step 5 from proof montrl-equiv (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 6:

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

• Proof step 7:

  ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 → trf(sl1, al1) ⊑ trf(sl2, al2), ∃ n. ¬ ( ⌜n⌝ + ⌜1⌝ < # ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ → ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ (n) ⊑ ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ (n + 1)) ⊦

Rule argument

Additional information

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

Proof step 6 from proof montrl-equiv (previous step) (back to tree)

A spec-lemma

Proof step 7 from proof montrl-equiv (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  ¬ trf(sf.take(n + 1), af.take(n)) ⊑ trf(sf.take(2 + n), af.take(n + 1)), ∀ 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 simplifier rules: ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ ⌜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 8 from proof montrl-equiv (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 9:

  ¬ trf(sf.take(n + 1), af.take(n)) ⊑ trf(sf.take(2 + n), af.take(n + 1)), # sf.take(n + 1) = # af.take(n) + 1 ∧ # sf.take(2 + n) = # af.take(n + 1) + 1 ∧ af.take(n) ⊑ af.take(n + 1) ∧ sf.take(n + 1) ⊑ sf.take(2 + n) → trf(sf.take(n + 1), af.take(n)) ⊑ trf(sf.take(2 + n), af.take(n + 1)) ⊦

Rule argument

Additional information

Used terms: sf.take(n + 1), af.take(n), sf.take(2 + n), af.take(n + 1)

Proof step 9 from proof montrl-equiv (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  ¬ trf(sf.take(n + 1), af.take(n)) ⊑ trf(sf.take(2 + n), af.take(n + 1)) ⊦ af.take(n) ⊑ af.take(n + 1) ∧ sf.take(n + 1) ⊑ sf.take(2 + 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 + n = m + n0 ↔ n = n0; (s) ⊦ # nf.take(n) = n; (s)

Proof step 10 from proof montrl-equiv (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 11:

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

• Proof step 12:

  ¬ trf(sf.take(n + 1), af.take(n)) ⊑ trf(sf.take(2 + n), af.take(n) + af(n) ') ⊦ af.take(n) ⊑ af.take(n) + af(n) ' ∧ sf.take(n + 1) ⊑ sf.take(2 + n)

Rule argument

Additional information

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

Proof step 11 from proof montrl-equiv (previous step) (back to tree)

A spec-lemma

Proof step 12 from proof montrl-equiv (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  ¬ trf(sf.take(n + 1), af.take(n)) ⊑ trf(sf.take(2 + n), af.take(n) + af(n) '), ¬ sf.take(n + 1) ⊑ sf.take(2 + n) ⊦

Rule argument

Additional information

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

Proof step 13 from proof montrl-equiv (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 14:

  ⊦ nf.take(m + 2) = nf.take(m + 1) + nf(m + 1) '

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 a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: sf, n

Proof step 14 from proof montrl-equiv (previous step) (back to tree)

A spec-lemma

Proof step 15 from proof montrl-equiv (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  # sl1 = # al1 + 1, # sl2 = # al2 + 1, al1 ⊑ al2, sl1 ⊑ sl2, ¬ trf(sl1, al1) ⊑ trf(sl2, al2), ∀ sf, af. mon(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ⊦

Rule argument

Additional information

Proof step 16 from proof montrl-equiv (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 17:

  # sl2 = # al2 + 1, # sl1 = # al1 + 1, mon(⌜λ n. ⌜trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(n + 1), (λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ ), sl1 ⊑ sl2, al1 ⊑ al2, ¬ trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

Rule argument

Additional information

Used terms: λ n. (n < # sl2 ⊃ sl2(n);s), λ n. (n < # al2 ⊃ al2(n);a)

Proof step 17 from proof montrl-equiv (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 18:

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

• Proof step 19:

  # sl1 = # al1 + 1, # sl2 = # al2 + 1, mon(⌜λ n. ⌜trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(n + 1), (λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ ), al1 ⊑ al2, sl1 ⊑ sl2, ⌜λ n. ⌜trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(n + 1), (λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ (# al1) ⊑ ⌜λ n. ⌜trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(n + 1), (λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ (# al2), ¬ trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # ⌜nf⌝ = ∞; (s) ⊦ ⌜m⌝ < ∞; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ # y < # x → ¬ x ⊑ y; (s) Used terms: ⌜λ n. ⌜trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(n + 1), (λ n. (n < # al2 ⊃ al2(n);a)).take(n)) ⌝ ⌝ , # al1, # al2

Proof step 18 from proof montrl-equiv (previous step) (back to tree)

A spec-lemma

Proof step 19 from proof montrl-equiv (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

  # sl2 = # al2 + 1, # sl1 = # al1 + 1, mon(⌜λ n. ⌜trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(n + 1), (λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ ), sl1 ⊑ sl2, al1 ⊑ al2, trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl1), (λ n. (n < # al2 ⊃ al2(n);a)).take(# al1)) ⊑ trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl2), (λ n. (n < # al2 ⊃ al2(n);a)).take(# al2)), ¬ trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

Rule argument

Additional information

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

Proof step 20 from proof montrl-equiv (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 21:

  ⊦ x ⊑ y ↔ (∃ z. x + z = y)

• Proof step 22:

  # sl1 = # al1 + 1, # sl2 = # al2 + 1, mon(⌜λ n. ⌜trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(n + 1), (λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ ), al1 ⊑ al2, sl1 ⊑ sl2, ¬ trf(sl1, al1) ⊑ trf(sl2, al2), ∃ z. trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl1), (λ n. (n < # al2 ⊃ al2(n);a)).take(# al1)) + z = trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl2), (λ n. (n < # al2 ⊃ al2(n);a)).take(# al2)) ⊦

Rule argument

Additional information

Used terms: trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl1), (λ n. (n < # al2 ⊃ al2(n);a)).take(# al1)), trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl2), (λ n. (n < # al2 ⊃ al2(n);a)).take(# al2))

Proof step 21 from proof montrl-equiv (previous step) (back to tree)

A spec-lemma

Proof step 22 from proof montrl-equiv (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 23:

  # sl2 = # al2 + 1, # sl1 = # al1 + 1, trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl1), (λ n. (n < # al2 ⊃ al2(n);a)).take(# al1)) + bl = trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl2), (λ n. (n < # al2 ⊃ al2(n);a)).take(# al2)), mon(⌜λ n. ⌜trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(n + 1), (λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ ), sl1 ⊑ sl2, al1 ⊑ al2, ¬ trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

Rule argument

Additional information

Proof step 23 from proof montrl-equiv (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 24:

  # sl1 = # al1 + 1, # sl2 = # al2 + 1, trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl1), (λ n. (n < # al2 ⊃ al2(n);a)).take(# al1)) + bl = trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl2), (λ n. (n < # al2 ⊃ al2(n);a)).take(# al2)), (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl2) = sl2, mon(⌜λ n. ⌜trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(n + 1), (λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ ), al1 ⊑ al2, sl1 ⊑ sl2, ¬ trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

• Proof step 48:

  # sl1 = # al1 + 1, # sl2 = # al2 + 1, trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl1), (λ n. (n < # al2 ⊃ al2(n);a)).take(# al1)) + bl = trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl2), (λ n. (n < # al2 ⊃ al2(n);a)).take(# al2)), (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl2) ≠ sl2, mon(⌜λ n. ⌜trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(n + 1), (λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ ), al1 ⊑ al2, sl1 ⊑ sl2, ¬ trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

Rule argument

Additional information

Used formulas: (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl2) = sl2

Proof step 24 from proof montrl-equiv (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 25:

  # sl2 = # al2 + 1, # sl1 = # al1 + 1, trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl1), (λ n. (n < # al2 ⊃ al2(n);a)).take(# al1)) + bl = trf(sl2, (λ n. (n < # al2 ⊃ al2(n);a)).take(# al2)), (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl2) = sl2, mon(⌜λ n. ⌜trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(n + 1), (λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ ), sl1 ⊑ sl2, al1 ⊑ al2, ¬ trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

Rule argument

Additional information

Proof step 25 from proof montrl-equiv (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 26:

  # sl1 = # al1 + 1, # sl2 = # al2 + 1, trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl1), (λ n. (n < # al2 ⊃ al2(n);a)).take(# al1)) + bl = trf(sl2, (λ n. (n < # al2 ⊃ al2(n);a)).take(# al2)), (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl2) = sl2, (λ n. (n < # al2 ⊃ al2(n);a)).take(# al2) = al2, mon(⌜λ n. ⌜trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(n + 1), (λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ ), al1 ⊑ al2, sl1 ⊑ sl2, ¬ trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

• Proof step 46:

  # sl1 = # al1 + 1, # sl2 = # al2 + 1, trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl1), (λ n. (n < # al2 ⊃ al2(n);a)).take(# al1)) + bl = trf(sl2, (λ n. (n < # al2 ⊃ al2(n);a)).take(# al2)), (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl2) = sl2, (λ n. (n < # al2 ⊃ al2(n);a)).take(# al2) ≠ al2, mon(⌜λ n. ⌜trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(n + 1), (λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ ), al1 ⊑ al2, sl1 ⊑ sl2, ¬ trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

Rule argument

Additional information

Used formulas: (λ n. (n < # al2 ⊃ al2(n);a)).take(# al2) = al2

Proof step 26 from proof montrl-equiv (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 27:

  # sl2 = # al2 + 1, # sl1 = # al1 + 1, trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl1), (λ n. (n < # al2 ⊃ al2(n);a)).take(# al1)) + bl = trf(sl2, al2), (λ n. (n < # al2 ⊃ al2(n);a)).take(# al2) = al2, (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl2) = sl2, mon(⌜λ n. ⌜trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(n + 1), (λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ ), sl1 ⊑ sl2, al1 ⊑ al2, ¬ trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

Rule argument

Additional information

Proof step 27 from proof montrl-equiv (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 28:

  # sl1 = # al1 + 1, # sl2 = # al2 + 1, trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl1), (λ n. (n < # al2 ⊃ al2(n);a)).take(# al1)) + bl = trf(sl2, al2), (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl2) = sl2, (λ n. (n < # al2 ⊃ al2(n);a)).take(# al2) = al2, (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl1) = sl1, mon(⌜λ n. ⌜trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(n + 1), (λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ ), al1 ⊑ al2, sl1 ⊑ sl2, ¬ trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

• Proof step 39:

  # sl1 = # al1 + 1, # sl2 = # al2 + 1, trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl1), (λ n. (n < # al2 ⊃ al2(n);a)).take(# al1)) + bl = trf(sl2, al2), (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl2) = sl2, (λ n. (n < # al2 ⊃ al2(n);a)).take(# al2) = al2, (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl1) ≠ sl1, mon(⌜λ n. ⌜trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(n + 1), (λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ ), al1 ⊑ al2, sl1 ⊑ sl2, ¬ trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

Rule argument

Additional information

Used formulas: (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl1) = sl1

Proof step 28 from proof montrl-equiv (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 29:

  # sl2 = # al2 + 1, # sl1 = # al1 + 1, trf(sl1, (λ n. (n < # al2 ⊃ al2(n);a)).take(# al1)) + bl = trf(sl2, al2), (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl1) = sl1, (λ n. (n < # al2 ⊃ al2(n);a)).take(# al2) = al2, (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl2) = sl2, mon(⌜λ n. ⌜trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(n + 1), (λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ ), sl1 ⊑ sl2, al1 ⊑ al2, ¬ trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

Rule argument

Additional information

Proof step 29 from proof montrl-equiv (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 30:

  # sl1 = # al1 + 1, # sl2 = # al2 + 1, trf(sl1, (λ n. (n < # al2 ⊃ al2(n);a)).take(# al1)) + bl = trf(sl2, al2), (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl2) = sl2, (λ n. (n < # al2 ⊃ al2(n);a)).take(# al2) = al2, (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl1) = sl1, (λ n. (n < # al2 ⊃ al2(n);a)).take(# al1) = al1, mon(⌜λ n. ⌜trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(n + 1), (λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ ), al1 ⊑ al2, sl1 ⊑ sl2, ¬ trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

• Proof step 34:

  # sl1 = # al1 + 1, # sl2 = # al2 + 1, trf(sl1, (λ n. (n < # al2 ⊃ al2(n);a)).take(# al1)) + bl = trf(sl2, al2), (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl2) = sl2, (λ n. (n < # al2 ⊃ al2(n);a)).take(# al2) = al2, (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl1) = sl1, (λ n. (n < # al2 ⊃ al2(n);a)).take(# al1) ≠ al1, mon(⌜λ n. ⌜trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(n + 1), (λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ ), al1 ⊑ al2, sl1 ⊑ sl2, ¬ trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

Rule argument

Additional information

Used formulas: (λ n. (n < # al2 ⊃ al2(n);a)).take(# al1) = al1

Proof step 30 from proof montrl-equiv (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 31:

  # sl2 = # al2 + 1, # sl1 = # al1 + 1, trf(sl1, al1) + bl = trf(sl2, al2), (λ n. (n < # al2 ⊃ al2(n);a)).take(# al1) = al1, (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl1) = sl1, (λ n. (n < # al2 ⊃ al2(n);a)).take(# al2) = al2, (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl2) = sl2, mon(⌜λ n. ⌜trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(n + 1), (λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ ), sl1 ⊑ sl2, al1 ⊑ al2, ¬ trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

Rule argument

Additional information

Proof step 31 from proof montrl-equiv (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 32:

  ⊦ x ⊑ y ↔ (∃ z. x + z = y)

• Proof step 33:

  # sl1 = # al1 + 1, # sl2 = # al2 + 1, trf(sl1, al1) + bl = trf(sl2, al2), (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl2) = sl2, (λ n. (n < # al2 ⊃ al2(n);a)).take(# al2) = al2, (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl1) = sl1, (λ n. (n < # al2 ⊃ al2(n);a)).take(# al1) = al1, mon(⌜λ n. ⌜trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(n + 1), (λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ ), al1 ⊑ al2, sl1 ⊑ sl2, ∀ z. trf(sl1, al1) + z ≠ trf(sl2, al2) ⊦

Rule argument

Additional information

Used terms: trf(sl1, al1), trf(sl2, al2)

Proof step 32 from proof montrl-equiv (previous step) (back to tree)

A spec-lemma

Proof step 33 from proof montrl-equiv (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: bl

Proof step 34 from proof montrl-equiv (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 35:

  ⊦ x ⊑ y ↔ (∃ z. x + z = y)

• Proof step 36:

  # sl2 = # al2 + 1, # sl1 = # al1 + 1, trf(sl1, (λ n. (n < # al2 ⊃ al2(n);a)).take(# al1)) + bl = trf(sl2, al2), (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl1) = sl1, (λ n. (n < # al2 ⊃ al2(n);a)).take(# al2) = al2, (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl2) = sl2, (λ n. (n < # al2 ⊃ al2(n);a)).take(# al1) ≠ al1, mon(⌜λ n. ⌜trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(n + 1), (λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ ), sl1 ⊑ sl2, ¬ trf(sl1, al1) ⊑ trf(sl2, al2), ∃ z. al1 + z = al2 ⊦

Rule argument

Additional information

Used terms: al1, al2

Proof step 35 from proof montrl-equiv (previous step) (back to tree)

A spec-lemma

Proof step 36 from proof montrl-equiv (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 37:

  # sl1 = # al1 + 1, # sl2 = (# al1 + # al) + 1, trf(sl1, (λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1)) + bl = trf(sl2, al1 + al), (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl2) = sl2, (λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1 + # al) = al1 + al, (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl1) = sl1, (λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1) ≠ al1, mon(⌜λ n. ⌜trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(n + 1), (λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(n))⌝ ⌝ ), sl1 ⊑ sl2, ¬ trf(sl1, al1) ⊑ trf(sl2, al1 + al) ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ #(x + y) = # x + # y; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic

Proof step 37 from proof montrl-equiv (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 38:

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

Rule argument

Additional information

Used simplifier rules: ⊦ # nf.take(n) = n; (s) ⊦ m < n → (n ≤ m ↔ false); (s) m < n ⊦ nf.take(n)(m) = nf(m); (s) ⊦ n < # x → (x + y)(n) = x(n); (s) a: ⊦ (x + y) + z = x + y + z; from specification list f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ # x = 0 ↔ x = []; (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat ⊦ ¬ m + n < m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: (λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1), al1

Proof step 38 from proof montrl-equiv (previous step) (back to tree)

A spec-lemma

Proof step 39 from proof montrl-equiv (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 40:

  ⊦ x ⊑ y ↔ (∃ z. x + z = y)

• Proof step 41:

  # sl2 = # al2 + 1, # sl1 = # al1 + 1, trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl1), (λ n. (n < # al2 ⊃ al2(n);a)).take(# al1)) + bl = trf(sl2, al2), (λ n. (n < # al2 ⊃ al2(n);a)).take(# al2) = al2, (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl2) = sl2, (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl1) ≠ sl1, mon(⌜λ n. ⌜trf((λ n. (n < # sl2 ⊃ sl2(n);s)).take(n + 1), (λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ ), al1 ⊑ al2, ¬ trf(sl1, al1) ⊑ trf(sl2, al2), ∃ z. sl1 + z = sl2 ⊦

Rule argument

Additional information

Used terms: sl1, sl2

Proof step 40 from proof montrl-equiv (previous step) (back to tree)

A spec-lemma

Proof step 41 from proof montrl-equiv (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 42:

  # sl1 = # al1 + 1, # sl1 + # sl = # al2 + 1, trf((λ n. (¬ # al2 < n ⊃ (sl1 + sl)(n);s)).take(# sl1), (λ n. (n < # al2 ⊃ al2(n);a)).take(# al1)) + bl = trf(sl1 + sl, al2), (λ n. (n < # al2 ⊃ al2(n);a)).take(# al2) = al2, (λ n. (¬ # al2 < n ⊃ (sl1 + sl)(n);s)).take(# al2 + 1) = sl1 + sl, (λ n. (¬ # al2 < n ⊃ (sl1 + sl)(n);s)).take(# sl1) ≠ sl1, mon(⌜λ n. ⌜trf((λ n. (¬ # al2 < n ⊃ (sl1 + sl)(n);s)).take(n + 1), (λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ ), al1 ⊑ al2, ¬ trf(sl1, al1) ⊑ trf(sl1 + sl, al2) ⊦

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 < n + 1 ↔ ¬ n < m; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ #(x + y) = # x + # y; (s)

Proof step 42 from proof montrl-equiv (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 43:

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

• Proof step 44:

  # sl1 = # al1 + 1, trf((λ n. (¬ # al2 < n ⊃ (sl1 + sl)(n);s)).take(# sl1), (λ n. (n < # al2 ⊃ al2(n);a)).take(# al1)) + bl = trf(sl1 + sl, al2), # sl1 + # sl = # al2 + 1, (λ n. (¬ # al2 < n ⊃ (sl1 + sl)(n);s)).take(# al2 + 1) = sl1 + sl, (λ n. (n < # al2 ⊃ al2(n);a)).take(# al2) = al2, mon(⌜λ n. ⌜trf((λ n. (¬ # al2 < n ⊃ (sl1 + sl)(n);s)).take(n + 1), (λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ ), al1 ⊑ al2, ¬ trf(sl1, al1) ⊑ trf(sl1 + sl, al2), # sl1 ≠ # (λ n. (¬ # al2 < n ⊃ (sl1 + sl)(n);s)).take(# sl1) ∨ (∃ n. ¬(n < # sl1 → sl1(n) = (λ n. (¬ # al2 < n ⊃ (sl1 + sl)(n);s)).take(# sl1)(n))) ⊦

Rule argument

Additional information

Used terms: (λ n. (¬ # al2 < n ⊃ (sl1 + sl)(n);s)).take(# sl1), sl1

Proof step 43 from proof montrl-equiv (previous step) (back to tree)

A spec-lemma

Proof step 44 from proof montrl-equiv (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 45:

  # sl1 = # al1 + 1, trf((λ n. (¬ # al2 < n ⊃ (sl1 + sl)(n);s)).take(# sl1), (λ n. (n < # al2 ⊃ al2(n);a)).take(# al1)) + bl = trf(sl1 + sl, al2), # sl1 + # sl = # al2 + 1, (λ n. (¬ # al2 < n ⊃ (sl1 + sl)(n);s)).take(# al2 + 1) = sl1 + sl, (λ n. (n < # al2 ⊃ al2(n);a)).take(# al2) = al2, sl1(n) ≠ (¬ # al2 < n ⊃ sl1(n);s), sl1 ≠ [], n < # sl1, mon(⌜λ n. ⌜trf((λ n. (¬ # al2 < n ⊃ (sl1 + sl)(n);s)).take(n + 1), (λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ ), n ≤ # al1, al1 ⊑ al2, ¬ trf(sl1, al1) ⊑ trf(sl1 + sl, al2) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ # x = 0 ↔ x = []; (s) lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ n < # x → (x + y)(n) = x(n); (s) m < n ⊦ nf.take(n)(m) = nf(m); (s) ⊦ m < n → (n ≤ m ↔ false); (s) ⊦ # nf.take(n) = n; (s)

Proof step 45 from proof montrl-equiv (previous step) (back to tree)

User applied rule case distinction

And closed the goal!

Rule argument

Additional information

Used simplifier rules: eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic lf: ⊦ m < n ∧ n < n0 → m.+1 < n0; (forward) from specification nat-basic fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat trans-Nat: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic ⊦ m < n → n ≠ 0; (s) ⊦ m.+1 = m + 1; (s) ⊦ m = m0 + n0 ∧ n < n0 → m0 + n < m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ # y < # x → ¬ x ⊑ y; (s)

Proof step 46 from proof montrl-equiv (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 47:

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

Rule argument

Additional information

Used simplifier rules: ⊦ # nf.take(n) = n; (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m < n → (n ≤ m ↔ false); (s) m < n ⊦ nf.take(n)(m) = nf(m); (s) Used terms: (λ n. (n < # al2 ⊃ al2(n);a)).take(# al2), al2

Proof step 47 from proof montrl-equiv (previous step) (back to tree)

A spec-lemma

Proof step 48 from proof montrl-equiv (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 49:

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

Rule argument

Additional information

Used simplifier rules: ⊦ # nf.take(n) = n; (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m < n → (n ≤ m ↔ false); (s) m < n ⊦ nf.take(n)(m) = nf(m); (s) Used terms: (λ n. (n < # sl2 ⊃ sl2(n);s)).take(# sl2), sl2

Proof step 49 from proof montrl-equiv (previous step) (back to tree)

A spec-lemma