Proof step 1 from proof inv-inverts-leq (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ¬(inv(nσ)[nσ[n]] < n ↔ false), ⌜n⌝ < # nσ, ⌜nσ[n]⌝ < leninv(nσ), stutf(nσ) ⊦

Rule argument

Additional information

Proof step 2 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  inv(nσ)[nσ[n]] < n, ⌜n⌝ < # nσ, ⌜nσ[n]⌝ < leninv(nσ), stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 3 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 4:

  ⊦ stutf(nσ) → inv(nσ) = (leninv(nσ) = ∞ ⊃ ⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n⌝ ;⌜(λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to leninv(nσ).n⌝ )

• Proof step 5:

  ⌜nσ[n]⌝ < leninv(nσ), ⌜n⌝ < # nσ, (leninv(nσ) = ∞ ⊃ ⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n⌝ ;⌜(λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to leninv(nσ).n⌝ ) [nσ[n] ] < n, stutf(nσ) ⊦

Rule argument

Additional information

Used terms: nσ

Proof step 4 from proof inv-inverts-leq (previous step) (back to tree)

Lemma inv-def

Proof step 5 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  (leninv(nσ) = ∞ ⊃ ⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n⌝ ;⌜(λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to leninv(nσ).n⌝ ) [nσ[n] ] < n, ⌜n⌝ < # nσ, ⌜nσ[n]⌝ < leninv(nσ), stutf(nσ) ⊦

Rule argument

Additional information

Proof step 6 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 7:

  leninv(nσ) = ∞, ⌜nσ[n]⌝ < leninv(nσ), ⌜n⌝ < # nσ, ⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n⌝ [nσ[n]] < n, stutf(nσ) ⊦

• Proof step 29:

  leninv(nσ) ≠ ∞, ⌜nσ[n]⌝ < leninv(nσ), ⌜n⌝ < # nσ, ⌜(λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to leninv(nσ).n⌝ [nσ[n]] < n, stutf(nσ) ⊦

Rule argument

Additional information

Proof step 7 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  # nσ = ∞, ⌜nσ[n]⌝ < leninv(nσ), max∞(λ n0. nσ[n0] = nσ[n]).n < n, stutf(nσ), ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦

Rule argument

Additional information

Used simplifier rules: leninv-inf-strong: ⊦ stutf(nσ) → ( leninv(nσ) = ∞ ↔ # nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0]) ∧ (∀ m. ∃ m0. m < nσ[m0])); ⊦ ⌜m⌝ < ∞; (s) ⊦ ⌜nf⌝ [n] = nf(n); (s)

Proof step 8 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 9:

  ⊦ (max∞(np) = ∞ ⊃ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n));(∃ n. np(max∞(np).n)) ∧ (∀ m. np(m) → ¬ max∞(np) < ⌜m⌝ ))

• Proof step 10:

  # nσ = ∞, max∞(λ n0. nσ[n0] = nσ[n]).n < n, ⌜nσ[n]⌝ < leninv(nσ), stutf(nσ), (max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]) = ∞ ⊃ (∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = nσ[n]) ∨ (∀ n0. ¬(⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n])) ; (∃ n0. ⌜max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]).n⌝ < # nσ ∧ nσ[max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]).n] = nσ[n]) ∧ (∀ m. ⌜m⌝ < # nσ ∧ nσ[m] = nσ[n] → ¬ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]) < ⌜m⌝ ) ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0] ⊦

Rule argument

Additional information

Used terms: λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]

Proof step 9 from proof inv-inverts-leq (previous step) (back to tree)

Lemma maxinf-def

Proof step 10 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  # nσ = ∞, max∞(λ n0. nσ[n0] = nσ[n]).n < n, ⌜nσ[n]⌝ < leninv(nσ), stutf(nσ), (max∞(λ n0. nσ[n0] = nσ[n]) = ∞ ⊃ (∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[n]) ∨ (∀ n0. nσ[n0] ≠ nσ[n]) ;nσ[max∞(λ n0. nσ[n0] = nσ[n]).n] = nσ[n] ∧ (∀ m. nσ[m] = nσ[n] → ¬ max∞(λ n0. nσ[n0] = nσ[n]) < ⌜m⌝ ) ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0] ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ∞; (s) ⊦ n∞ < ∞ ↔ n∞ ≠ ∞; (s) n∞ ≠ ∞ ⊦ ⌜n∞.n⌝ = n∞; (s)

Proof step 11 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 12:

  # nσ = ∞, max∞(λ n0. nσ[n0] = nσ[n]) = ∞, ⌜nσ[n]⌝ < leninv(nσ), max∞(λ n0. nσ[n0] = nσ[n]).n < n, stutf(nσ), ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], (∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[n]) ∨ (∀ n0. nσ[n0] ≠ nσ[n]) ⊦

• Proof step 25:

  # nσ = ∞, nσ[max∞(λ n0. nσ[n0] = nσ[n]).n] = nσ[n], max∞(λ n0. nσ[n0] = nσ[n]) ≠ ∞, ⌜nσ[n]⌝ < leninv(nσ), max∞(λ n0. nσ[n0] = nσ[n]).n < n, stutf(nσ), ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. nσ[m] = nσ[n] → ¬ max∞(λ n0. nσ[n0] = nσ[n]) < ⌜m⌝ ⊦

Rule argument

Additional information

Proof step 12 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  # nσ = ∞, max∞(λ n0. nσ[n0] = nσ[n]) = ∞, ∞.n < n, ⌜nσ[n]⌝ < leninv(nσ), stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0], (∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[n]) ∨ (∀ n0. nσ[n0] ≠ nσ[n]) ⊦

Rule argument

Additional information

Proof step 13 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 14:

  ⊦ max∞(np) = ∞ ↔ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n))

• Proof step 15:

  # nσ = ∞, ⌜nσ[n]⌝ < leninv(nσ), ∞.n < n, stutf(nσ), ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], (∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[n]) ∨ (∀ n0. nσ[n0] ≠ nσ[n]) ⊦

Rule argument

Additional information

Used terms: λ n0. nσ[n0] = nσ[n]

Proof step 14 from proof inv-inverts-leq (previous step) (back to tree)

Lemma maxinf-inf

Proof step 15 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  # nσ = ∞, ∞.n < n, ⌜nσ[n]⌝ < leninv(nσ), stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0], (∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[n]) ∨ (∀ n0. nσ[n0] ≠ nσ[n]) ⊦

Rule argument

Additional information

Proof step 16 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 17:

  # nσ = ∞, ⌜nσ[n]⌝ < leninv(nσ), ∞.n < n, stutf(nσ), ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[n] ⊦

• Proof step 24:

  # nσ = ∞, ⌜nσ[n]⌝ < leninv(nσ), ∞.n < n, stutf(nσ), ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ n0. nσ[n0] ≠ nσ[n] ⊦

Rule argument

Additional information

Proof step 17 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  # nσ = ∞, ∞.n < n, ⌜nσ[n]⌝ < leninv(nσ), stutf(nσ), ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[n], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0] ⊦

Rule argument

Additional information

Proof step 18 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 19:

  # nσ = ∞, ⌜nσ[n]⌝ < leninv(nσ), ∞.n < n, stutf(nσ), ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[n], ∃ m0. n < m0 ∧ nσ[n] < nσ[m0] ⊦

Rule argument

Additional information

Used terms: n

Proof step 19 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

  # nσ = ∞, ∞.n < m0, ⌜nσ[n]⌝ < leninv(nσ), ∞.n < n, nσ[n] < nσ[m0], n < m0, ∞.n + 1 < m0, stutf(nσ), ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[n] ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s) ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic trans-nat: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic

Proof step 20 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 21:

  # nσ = ∞, ∞.n + 1 < m0, n < m0, nσ[n] < nσ[m0], ∞.n < n, ⌜nσ[n]⌝ < leninv(nσ), ∞.n < m0, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0], ∃ n1. m0 < n1 ∧ nσ[n1] = nσ[n] ⊦

Rule argument

Additional information

Used terms: m0

Proof step 21 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 22:

  # nσ = ∞, nσ[n1] = nσ[n], ∞.n < n1, ∞.n + 1 < m0, n < m0, ∞.n < n, ⌜nσ[n]⌝ < leninv(nσ), ∞.n < m0, m0 < n1, 2 + ∞.n < n1, n + 1 < n1, n < n1, nσ[n] < nσ[m0], ∞.n + 1 < n1, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0] ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic trans-nat: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic

Proof step 22 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 23:

  ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n]

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); ⊦ m < n → m ≤ n; (s) ⊦ ⌜m⌝ < ∞; (s) ⊦ m < n → (n ≤ m ↔ false); (s) Used terms: nσ, m0, n1

Proof step 23 from proof inv-inverts-leq (previous step) (back to tree)

Lemma monle-leq

Proof step 24 from proof inv-inverts-leq (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 25 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 26:

  # nσ = ∞, nσ[max∞(λ n0. nσ[n0] = nσ[n]).n] = nσ[n], max∞(λ n0. nσ[n0] = nσ[n]) ≠ ∞, max∞(λ n0. nσ[n0] = nσ[n]).n < n, ⌜nσ[n]⌝ < leninv(nσ), stutf(nσ), ∀ m. nσ[m] = nσ[n] → ¬ max∞(λ n0. nσ[n0] = nσ[n]) < ⌜m⌝ , ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0] ⊦

Rule argument

Additional information

Proof step 26 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 27:

  # nσ = ∞, nσ[n1] = nσ[n], max∞(λ n0. nσ[n0] = nσ[n]) = ⌜n1⌝ , max∞(λ n0. nσ[n0] = nσ[n]) ≠ ∞, ⌜nσ[n]⌝ < leninv(nσ), n1 < n, stutf(nσ), ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. nσ[m] = nσ[n] → ¬ max∞(λ n0. nσ[n0] = nσ[n]) < ⌜m⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: max∞(λ n0. nσ[n0] = nσ[n]), n1

Proof step 27 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 28:

  # nσ = ∞, nσ[n1] = nσ[n], max∞(λ n0. nσ[n0] = nσ[n]) = ⌜n1⌝ , ⌜nσ[n]⌝ < leninv(nσ), n1 < n, stutf(nσ), ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. nσ[m] = nσ[n] → ¬ n1 < m ⊦

Rule argument

Additional information

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

Proof step 28 from proof inv-inverts-leq (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 29 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 30:

  leninv(nσ) ≠ ∞, ⌜n⌝ < # nσ, ⌜nσ[n]⌝ < leninv(nσ), ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to leninv(nσ).n)[nσ[n]] < n, stutf(nσ) ⊦

Rule argument

Additional information

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

Proof step 30 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 31:

  leninv(nσ) = ⌜n0⌝ , leninv(nσ) ≠ ∞, ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to n0)[nσ[n]] < n, ⌜nσ[n]⌝ < leninv(nσ), ⌜n⌝ < # nσ, stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: leninv(nσ), n0

Proof step 31 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 32:

  leninv(nσ) = ⌜n0⌝ , nσ[n] < n0, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]).n < n, ⌜n⌝ < # nσ, stutf(nσ) ⊦

Rule argument

Additional information

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

Proof step 32 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 33:

  leninv(nσ) = ⌜n0⌝ , n0 = n1 +1, ⌜n⌝ < # nσ, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]).n < n, nσ[n] < n0, stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: nat freely generated by 0, +1;; ⊦ ¬ n < 0; (s)

Proof step 33 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 34:

  leninv(nσ) = ⌜n1⌝ + ⌜1⌝ , max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]).n < n, ⌜n⌝ < # nσ, stutf(nσ), nσ[n] ≤ n1 ⊦

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) ⊦ m +1 = m + 1; (s) ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)

Proof step 34 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 35:

  ⊦ stutf(nσ) → (leninv(nσ) = ⌜n⌝ + ⌜1⌝ ↔ (# nσ = ∞ ⊃ ∃ m. nσ[m] = n ∧ (∀ m0. m < m0 → nσ[m0] = n + 1);nσ[(# nσ).n - 1] = n))

• Proof step 36:

  ⌜n⌝ < # nσ, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]).n < n, stutf(nσ), nσ[n] ≤ n1, (# nσ = ∞ ⊃ ∃ m. nσ[m] = n1 ∧ (∀ m0. m < m0 → nσ[m0] = n1 + 1);nσ[(# nσ).n - 1] = n1) ⊦

Rule argument

Additional information

Used terms: n1, nσ

Proof step 35 from proof inv-inverts-leq (previous step) (back to tree)

Lemma leninv-succ

Proof step 36 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 37:

  max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]).n < n, ⌜n⌝ < # nσ, stutf(nσ), nσ[n] ≤ n1, (# nσ = ∞ ⊃ nσ[m] = n1 ∧ (∀ m0. m < m0 → nσ[m0] = nσ[m] + 1);nσ[(# nσ).n - 1] = n1) ⊦

Rule argument

Additional information

Proof step 37 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 38:

  # nσ = ∞, nσ[m] = n1, ⌜n⌝ < # nσ, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]).n < n, stutf(nσ), nσ[n] ≤ n1, ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦

• Proof step 58:

  nσ[(# nσ).n - 1] = n1, # nσ ≠ ∞, ⌜n⌝ < # nσ, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]).n < n, stutf(nσ), nσ[n] ≤ n1 ⊦

Rule argument

Additional information

Proof step 38 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 39:

  # nσ = ∞, max∞(λ n0. nσ[n0] = nσ[n]).n < n, stutf(nσ), nσ[n] ≤ nσ[m], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ∞; (s)

Proof step 39 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule apply elim lemma

And obtained the following new goals

• Proof step 40:

  # nσ = ∞, max∞(λ n0. nσ[n0] = nσ[n]) = ∞, max∞(λ n0. nσ[n0] = nσ[n]).n < n, stutf(nσ), nσ[n] ≤ nσ[m], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦

• Proof step 53:

  # nσ = ∞, max∞(λ n0. nσ[n0] = nσ[n]) = ⌜n1⌝ , max∞(λ n0. nσ[n0] = nσ[n]) ≠ ∞, n1 < n, stutf(nσ), nσ[n] ≤ nσ[m], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: max∞(λ n0. nσ[n0] = nσ[n]), n1

Proof step 40 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 41:

  # nσ = ∞, max∞(λ n0. nσ[n0] = nσ[n]) = ∞, ∞.n < n, stutf(nσ), nσ[n] ≤ nσ[m], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦

Rule argument

Additional information

Proof step 41 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 42:

  ⊦ max∞(np) = ∞ ↔ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n))

• Proof step 43:

  # nσ = ∞, ∞.n < n, stutf(nσ), nσ[n] ≤ nσ[m], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1, (∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[n]) ∨ (∀ n0. nσ[n0] ≠ nσ[n]) ⊦

Rule argument

Additional information

Used terms: λ n0. nσ[n0] = nσ[n]

Proof step 42 from proof inv-inverts-leq (previous step) (back to tree)

Lemma maxinf-inf

Proof step 43 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 44:

  # nσ = ∞, ∞.n < n, stutf(nσ), nσ[n] ≤ nσ[m], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1, ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[n] ⊦

• Proof step 52:

  # nσ = ∞, ∞.n < n, stutf(nσ), nσ[n] ≤ nσ[m], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1, ∀ n0. nσ[n0] ≠ nσ[n] ⊦

Rule argument

Additional information

Proof step 44 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 45:

  # nσ = ∞, ∞.n < n, stutf(nσ), nσ[n] ≤ nσ[m], ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[n], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1, ∃ n1. m + 1 < n1 ∧ nσ[n1] = nσ[n] ⊦

Rule argument

Additional information

Used terms: m + 1

Proof step 45 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 46:

  # nσ = ∞, nσ[n1] = nσ[n], ∞.n < n, m + 1 < n1, stutf(nσ), nσ[n] ≤ nσ[m], ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[n], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 46 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 47:

  # nσ = ∞, nσ[n1] = nσ[n], m + 1 < n1, ∞.n < n, stutf(nσ), nσ[n] ≤ nσ[m], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1, ∃ n1. m + 1 < n1 ∧ nσ[n1] = nσ[n] ⊦

Rule argument

Additional information

Used terms: m + 1

Proof step 47 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 48:

  # nσ = ∞, nσ[n1] = nσ[n], nσ[n0] = nσ[n], m + 1 < n1, ∞.n < n, m + 1 < n0, stutf(nσ), nσ[n] ≤ nσ[m], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 48 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 49:

  ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n]

• Proof step 50:

  # nσ = ∞, nσ[n0] = nσ[n], nσ[n1] = nσ[n], m + 1 < n0, ∞.n < n, m + 1 < n1, stutf(nσ), nσ[n] ≤ nσ[m], mon≤(nσ) ∧ m + 1 ≤ n1 ∧ ⌜n1⌝ < # nσ → nσ[m + 1] ≤ nσ[n1], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦

Rule argument

Additional information

Used terms: nσ, m + 1, n1

Proof step 49 from proof inv-inverts-leq (previous step) (back to tree)

Lemma monle-leq

Proof step 50 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 51:

  # nσ = ∞, nσ[n0] = nσ[n], nσ[n1] = nσ[n], m + 1 < n0, ∞.n < n, m + 1 < n1, m < n1, stutf(nσ), nσ[n] ≤ nσ[m], nσ[m + 1] ≤ nσ[n], nσ[m + 1] ≤ nσ[m], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ ¬ m < n → ¬ m + m0 < n; (s) ⊦ ⌜m⌝ < ∞; (s) ⊦ m + 1 ≤ n ↔ m < n; (s)

Proof step 51 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ ¬ n < n; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ m +1 = m + 1; (s) ⊦ m = n + 1 → ¬ m ≤ n; (s) Used terms: m +1

Proof step 52 from proof inv-inverts-leq (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 53 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 54:

  # nσ = ∞, max∞(λ n0. nσ[n0] = nσ[n]) = ⌜n1⌝ , n1 < n, stutf(nσ), nσ[n] ≤ nσ[m], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 54 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 55:

  ⊦ max∞(np) = ⌜n⌝ ↔ np(n) ∧ (∀ m. np(m) → ¬ n < m)

• Proof step 56:

  # nσ = ∞, nσ[n1] = nσ[n], n1 < n, stutf(nσ), nσ[n] ≤ nσ[m], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1, ∀ m. nσ[m] = nσ[n] → ¬ n1 < m ⊦

Rule argument

Additional information

Used terms: n1, λ n0. nσ[n0] = nσ[n]

Proof step 55 from proof inv-inverts-leq (previous step) (back to tree)

Lemma maxinf-fin

Proof step 56 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 57:

  # nσ = ∞, nσ[n1] = nσ[n], n1 < n, stutf(nσ), nσ[n] ≤ nσ[m], ∀ m. nσ[m] = nσ[n] → ¬ n1 < m, ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦

Rule argument

Additional information

Proof step 57 from proof inv-inverts-leq (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 58 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 59:

  # nσ ≠ ∞, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]).n < n, ⌜n⌝ < # nσ, stutf(nσ), nσ[n] ≤ nσ[(# nσ).n - 1] ⊦

Rule argument

Additional information

Proof step 59 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 60:

  # nσ = ⌜n1⌝ , # nσ ≠ ∞, ⌜n⌝ < # nσ, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]).n < n, stutf(nσ), nσ[n] ≤ nσ[n1 - 1] ⊦

Rule argument

Additional information

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

Proof step 60 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 61:

  # nσ = ⌜n1⌝ , max∞(λ n0. n0 < n1 ∧ nσ[n0] = nσ[n]).n < n1, max∞(λ n0. n0 < n1 ∧ nσ[n0] = nσ[n]).n < n, n < n1, max∞(λ n0. n0 < n1 ∧ nσ[n0] = nσ[n]).n + 1 < n1, stutf(nσ), nσ[n] ≤ nσ[n1 - 1] ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s) ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic trans-nat: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 61 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 62:

  # nσ = ⌜n1⌝ , n1 = m + 1, n1 ≠ 0, max∞(λ n0. n0 < n1 ∧ nσ[n0] = nσ[n]).n + 1 < n1, n < n1, max∞(λ n0. n0 < n1 ∧ nσ[n0] = nσ[n]).n < n, max∞(λ n0. n0 < n1 ∧ nσ[n0] = nσ[n]).n < n1, stutf(nσ), nσ[n] ≤ nσ[m] ⊦

Rule argument

Additional information

Used simplifier rules: stutf-nonempty: ⊦ stutf(nσ) → (nσ = [] ↔ false); ⊦ # σ = ⌜0⌝ ↔ σ = []; (s) n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n1, m

Proof step 62 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 63:

  # nσ = ⌜m⌝ + ⌜1⌝ , max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = nσ[n]).n < m, max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = nσ[n]).n < n, stutf(nσ), n ≤ m, nσ[n] ≤ nσ[m] ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m < n → ¬ n < m; (s) 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) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ m + 1 ≠ 0; (s) ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)

Proof step 63 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule apply elim lemma

And obtained the following new goals

• Proof step 64:

  # nσ = ⌜m⌝ + ⌜1⌝ , max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = nσ[n]) = ∞, max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = nσ[n]).n < n, max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = nσ[n]).n < m, stutf(nσ), nσ[n] ≤ nσ[m], n ≤ m ⊦

• Proof step 71:

  # nσ = ⌜m⌝ + ⌜1⌝ , max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = nσ[n]) = ⌜n1⌝ , max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = nσ[n]) ≠ ∞, n1 < n, n1 < m, stutf(nσ), nσ[n] ≤ nσ[m], n ≤ m ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = nσ[n]), n1

Proof step 64 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 65:

  # nσ = ⌜m⌝ + ⌜1⌝ , max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = nσ[n]) = ∞, ∞.n < n, ∞.n < m, stutf(nσ), n ≤ m, nσ[n] ≤ nσ[m] ⊦

Rule argument

Additional information

Used simplifier rules: fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat

Proof step 65 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 66:

  ⊦ max∞(np) = ∞ ↔ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n))

• Proof step 67:

  # nσ = ⌜m⌝ + ⌜1⌝ , ∞.n < m, ∞.n < n, stutf(nσ), nσ[n] ≤ nσ[m], n ≤ m, (∀ n0. ∃ n1. n0 < n1 ∧ ¬ m < n1 ∧ nσ[n1] = nσ[n]) ∨ (∀ n0. ¬(¬ m < n0 ∧ nσ[n0] = nσ[n])) ⊦

Rule argument

Additional information

Used terms: λ n0. ¬ m < n0 ∧ nσ[n0] = nσ[n]

Proof step 66 from proof inv-inverts-leq (previous step) (back to tree)

Lemma maxinf-inf

Proof step 67 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 68:

  # nσ = ⌜m⌝ + ⌜1⌝ , ∞.n < n, ∞.n < m, stutf(nσ), n ≤ m, nσ[n] ≤ nσ[m], (∀ n0. ∃ n1. n0 < n1 ∧ ¬ m < n1 ∧ nσ[n1] = nσ[n]) ∨ (∀ n0. m < n0 ∨ nσ[n0] ≠ nσ[n]) ⊦

Rule argument

Additional information

Proof step 68 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 69:

  # nσ = ⌜m⌝ + ⌜1⌝ , ∞.n < m, ∞.n < n, stutf(nσ), nσ[n] ≤ nσ[m], n ≤ m, ∀ n0. ∃ n1. n0 < n1 ∧ ¬ m < n1 ∧ nσ[n1] = nσ[n] ⊦

• Proof step 70:

  # nσ = ⌜m⌝ + ⌜1⌝ , ∞.n < m, ∞.n < n, stutf(nσ), nσ[n] ≤ nσ[m], n ≤ m, ∀ n0. m < n0 ∨ nσ[n0] ≠ nσ[n] ⊦

Rule argument

Additional information

Proof step 69 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ m < n → ¬ m + m0 < n; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: m + 1

Proof step 70 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ n → (n < m ↔ false); (s) Used terms: n

Proof step 71 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 72:

  # nσ = ⌜m⌝ + ⌜1⌝ , max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = nσ[n]) = ⌜n1⌝ , n1 < m, n1 < n, stutf(nσ), nσ[n] ≤ nσ[m], n ≤ m ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜n⌝ ≠ ∞; (s) fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat

Proof step 72 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 73:

  ⊦ max∞(np) = ⌜n⌝ ↔ np(n) ∧ (∀ m. np(m) → ¬ n < m)

• Proof step 74:

  # nσ = ⌜m⌝ + ⌜1⌝ , nσ[n1] = nσ[n], n1 < n, n1 < m, stutf(nσ), n ≤ m, nσ[n] ≤ nσ[m], ¬ m < n1, ∀ m0. ¬ m < m0 ∧ nσ[m0] = nσ[n] → ¬ n1 < m0 ⊦

Rule argument

Additional information

Used terms: n1, λ n0. ¬ m < n0 ∧ nσ[n0] = nσ[n]

Proof step 73 from proof inv-inverts-leq (previous step) (back to tree)

Lemma maxinf-fin

Proof step 74 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 75:

  # nσ = ⌜m⌝ + ⌜1⌝ , nσ[n1] = nσ[n], n1 < m, n1 < n, stutf(nσ), nσ[n] ≤ nσ[m], n ≤ m, ∀ m0. ¬ m < m0 ∧ nσ[m0] = nσ[n] → ¬ n1 < m0 ⊦

Rule argument

Additional information

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

Proof step 75 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ n → (n < m ↔ false); (s) Used terms: n