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

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  # nσ = ⌜n⌝ + ⌜1⌝ , inv(nσ)[nσ[n]] ≠ n, stutf(nσ) ⊦

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  # nσ = ⌜n⌝ + ⌜1⌝ , inv(nσ)[nσ[n]] ≠ n, stutf(nσ) ⊦

Rule argument

Additional information

Proof step 3 from proof inv-inverts-last (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⌝ + ⌜1⌝ , (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-last (previous step) (back to tree)

Lemma inv-def

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

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

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ to n = ⌜nf to n⌝ ; (s) 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) ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); (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

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

User applied rule case distinction

And obtained the following new goals

• Proof step 7:

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

• Proof step 23:

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

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

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

Rule argument

Additional information

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

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

User applied rule cut formula

And obtained the following new goals

• Proof step 9:

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

• Proof step 16:

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

Rule argument

Additional information

Used formulas: max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]) = ∞

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

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

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) ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); (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

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 11:

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

• Proof step 12:

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

Rule argument

Additional information

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

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

Lemma maxinf-inf

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

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

Rule argument

Additional information

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

User applied rule case distinction

And obtained the following new goals

• Proof step 14:

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

• Proof step 15:

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

Rule argument

Additional information

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

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 17:

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

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) ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); (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

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

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 18:

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

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 19:

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

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜n⌝ ≠ ∞; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 20:

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

• Proof step 21:

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

Rule argument

Additional information

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

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

Lemma maxinf-fin

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 22:

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

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 ↔ ¬ n < m); (ss)

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

User applied rule all left

And closed the goal!

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

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

Rule argument

Additional information

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

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

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 25:

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

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 26:

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

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜n⌝ ≠ ∞; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 27:

  m < n ⊦ (nf to n)[m] = nf(m)

• Proof step 28:

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

• Proof step 33:

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

Rule argument

Additional information

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

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 29:

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

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)

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

User applied rule constructor cut

And obtained the following new goals

• Proof step 30:

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

Rule argument

Additional information

Used simplifier rules: leninv-invlast: ⊦ # nσ = ⌜m⌝ + ⌜1⌝ ∧ leninv(nσ) = ⌜n⌝ + ⌜1⌝ ∧ stutf(nσ) → nσ[m] = n; nat freely generated by 0, +1;; ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ ¬ n < n; (s) m < n ⊦ (nf to n)[m] = nf(m); (s) ⊦ m +1 = m + 1; (s) ⊦ m + m0 ≤ m ↔ m0 = 0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 31:

  # nσ = ⌜n⌝ + ⌜1⌝ , leninv(nσ) = ⌜0⌝ , [][nσ[n]] ≠ n, stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ 0 ≤ m; (s) ⊦ nf to 0 = []; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 32:

  ⊦ stutf(nσ) → (leninv(nσ) = ⌜0⌝ ↔ nσ = ⌜λ n. 0⌝ )

Rule argument

Additional information

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

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

Lemma leninv-zero

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

User applied rule cut formula

And obtained the following new goals

• Proof step 34:

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

• Proof step 41:

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

Rule argument

Additional information

Used formulas: max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]) = ∞

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 35:

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

Rule argument

Additional information

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 36:

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

• Proof step 37:

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

Rule argument

Additional information

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

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

Lemma maxinf-inf

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 38:

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

Rule argument

Additional information

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

User applied rule case distinction

And obtained the following new goals

• Proof step 39:

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

• Proof step 40:

  # nσ = ⌜n⌝ + ⌜1⌝ , leninv(nσ) = ⌜n1⌝ , ∞.n ≠ n, nσ[n] < n1, stutf(nσ), ∀ n0. n < n0 ∨ nσ[n0] ≠ nσ[n] ⊦

Rule argument

Additional information

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

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 42:

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

Rule argument

Additional information

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

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

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 43:

  # nσ = ⌜n⌝ + ⌜1⌝ , leninv(nσ) = ⌜n1⌝ , max∞(λ n0. ¬ n < n0 ∧ nσ[n0] = nσ[n]) = ⌜n2⌝ , max∞(λ n0. ¬ n < n0 ∧ nσ[n0] = nσ[n]) ≠ ∞, n2 ≠ n, nσ[n] < n1, stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: max∞(λ n0. ¬ n < n0 ∧ nσ[n0] = nσ[n]), n2

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 44:

  # nσ = ⌜n⌝ + ⌜1⌝ , leninv(nσ) = ⌜n1⌝ , max∞(λ n0. ¬ n < n0 ∧ nσ[n0] = nσ[n]) = ⌜n2⌝ , n2 ≠ n, nσ[n] < n1, stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜n⌝ ≠ ∞; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 45:

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

• Proof step 46:

  # nσ = ⌜n⌝ + ⌜1⌝ , nσ[n2] = nσ[n], leninv(nσ) = ⌜n1⌝ , n2 ≠ n, nσ[n] < n1, stutf(nσ), ¬ n < n2, ∀ m. ¬ n < m ∧ nσ[m] = nσ[n] → ¬ n2 < m ⊦

Rule argument

Additional information

Used terms: n2, λ n0. ¬ n < n0 ∧ nσ[n0] = nσ[n]

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

Lemma maxinf-fin

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 47:

  # nσ = ⌜n⌝ + ⌜1⌝ , nσ[n2] = nσ[n], leninv(nσ) = ⌜n1⌝ , n2 ≠ n, n2 < n, nσ[n] < n1, stutf(nσ), ∀ m. ¬ n < m ∧ nσ[m] = nσ[n] → ¬ n2 < m ⊦

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 ↔ ¬ n < m); (ss)

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

User applied rule all left

And closed the goal!

Rule argument

Additional information

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