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

System applied rule normalize

And obtained the following new goals

• Proof step 2:

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

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

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

Rule argument

Additional information

Proof step 3 from proof inv-inverts (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] = 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, ⌜n⌝ + ⌜1⌝ < # nσ, stutf(nσ) ⊦

Rule argument

Additional information

Used terms: nσ

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

Lemma inv-def

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

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

Rule argument

Additional information

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

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

User applied rule cut formula

And obtained the following new goals

• Proof step 7:

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

• Proof step 39:

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

Rule argument

Additional information

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

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

User applied rule case distinction

And obtained the following new goals

• Proof step 8:

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

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])); ⊦ ⌜nf⌝ [n] = nf(n); (s) ⊦ ⌜m⌝ < ∞; (s) ⊦ ⌜n⌝ .n = n; (s)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

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

Rule argument

Additional information

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

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

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 10:

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

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

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

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜n⌝ ≠ ∞; (s)

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

User applied rule constructor cut

And obtained the following new goals

• Proof step 12:

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

• Proof step 15:

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

Rule argument

Additional information

Used simplifier rules: nat freely generated by 0, +1;;

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

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

Rule argument

Additional information

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 14:

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

Rule argument

Additional information

Used simplifier rules: ⊦ # ⌜nf⌝ = ∞; (s) ⊦ ⌜m⌝ < ∞; (s) ⊦ ⌜nf⌝ [n] = nf(n); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: nσ

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

Lemma leninv-zero

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

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

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⌝ = ⌜m⌝ + ⌜n⌝ ; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 17:

  ⊦ 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 18:

  nσ[n + 1] = nσ[n] + 1, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]) = ⌜n⌝ , ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to n2 + 1)[nσ[n]] ≠ n, ⌜n⌝ + ⌜1⌝ < # nσ, stutf(nσ), (# nσ = ∞ ⊃ ∃ m. nσ[m] = n2 ∧ (∀ m0. m < m0 → nσ[m0] = n2 + 1);nσ[(# nσ).n - 1] = n2) ⊦

Rule argument

Additional information

Used terms: n2, nσ

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

Lemma leninv-succ

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 19:

  nσ[n + 1] = nσ[n] + 1, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]) = ⌜n⌝ , ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to n2 + 1)[nσ[n]] ≠ n, ⌜n⌝ + ⌜1⌝ < # nσ, stutf(nσ), (# nσ = ∞ ⊃ nσ[n0] = n2 ∧ (∀ m0. n0 < m0 → nσ[m0] = nσ[n0] + 1);nσ[(# nσ).n - 1] = n2) ⊦

Rule argument

Additional information

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

User applied rule case distinction

And obtained the following new goals

• Proof step 20:

  # nσ = ∞, nσ[n + 1] = nσ[n] + 1, nσ[n0] = n2, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]) = ⌜n⌝ , ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to n2 + 1)[nσ[n]] ≠ n, ⌜n⌝ + ⌜1⌝ < # nσ, stutf(nσ), ∀ m0. n0 < m0 → nσ[m0] = nσ[n0] + 1 ⊦

• Proof step 32:

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

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 21:

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

Rule argument

Additional information

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

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

User applied rule all left

And obtained the following new goals

• Proof step 22:

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

Rule argument

Additional information

Used terms: n

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

User applied rule cut formula

And obtained the following new goals

• Proof step 23:

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

• Proof step 25:

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

Rule argument

Additional information

Used formulas: n0 < n

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

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

Rule argument

Additional information

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

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

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ m < n → ¬ n < m; (s) eqs: ⊦ m = n + n0 ∧ m = n + n1 → n0 = n1; (forward) from specification nat-basic ⊦ m = m + n ↔ n = 0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: n + 1

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 26:

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

Rule argument

Additional information

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

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

User applied rule all left

And obtained the following new goals

• Proof step 27:

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

Rule argument

Additional information

Used terms: n + 1

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 28:

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

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) ⊦ m < n + 1 ↔ ¬ n < m; (s)

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

User applied rule cut formula

And obtained the following new goals

• Proof step 29:

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

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m ≤ n → (n ≤ m ↔ m = n); (s) ⊦ 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 < n ⊦ (nf to n)[m] = nf(m); (s) ⊦ ⌜n⌝ .n = n; (s) Used formulas: n < n0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 30:

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

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)

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

User applied rule apply lemma

And obtained the following new goals

• Proof step 31:

  ⊦ 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 + 1 ↔ ¬ n < m; (s) ⊦ m ≤ n → (n < m ↔ false); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic m < n ⊦ (nf to n)[m] = nf(m); (s) ⊦ ⌜n⌝ .n = n; (s) Used terms: nσ, n, n0

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

Lemma monle-leq

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 33:

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

Rule argument

Additional information

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

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 34:

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

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 35:

  # nσ = ⌜n1⌝ , nσ[n + 1] = nσ[n] + 1, max∞(λ n0. n0 < n1 ∧ nσ[n0] = nσ[n]) = ⌜n⌝ , ((λ m. max∞(λ n0. n0 < n1 ∧ nσ[n0] = m).n) to nσ[n1 - 1] + 1)[nσ[n]] ≠ n, n + 1 < 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 add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; (s) ⊦ ⌜n⌝ ≠ ∞; (s) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)

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

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 36:

  # nσ = ⌜n1⌝ , nσ[n + 1] = nσ[n] + 1, max∞(λ n0. n0 < n1 ∧ nσ[n0] = nσ[n]) = ⌜n⌝ , n1 = n2 + 1, ((λ m. max∞(λ n0. n0 < n1 ∧ nσ[n0] = m).n) to nσ[n2] + 1)[nσ[n]] ≠ n, n1 ≠ 0, n + 1 < n1, stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: stutf-nonempty: ⊦ stutf(nσ) → (nσ = [] ↔ false); ⊦ ¬ n < 0; (s) ⊦ # σ = ⌜0⌝ ↔ σ = []; (s) n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n1, m0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 37:

  # nσ = ⌜n2⌝ + ⌜1⌝ , nσ[n + 1] = nσ[n] + 1, max∞(λ n0. ¬ n2 < n0 ∧ nσ[n0] = nσ[n]) = ⌜n⌝ , ((λ m. max∞(λ n0. ¬ n2 < n0 ∧ nσ[n0] = m).n) to nσ[n2] + 1)[nσ[n]] ≠ n, n < n2, 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 < m + n0 ↔ n < n0; (s) ⊦ m + 1 ≠ 0; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)

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

User applied rule apply lemma

And obtained the following new goals

• Proof step 38:

  ⊦ 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) add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); (s) ⊦ ¬ n < n; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ m ≤ n → (n < m ↔ false); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic m < n ⊦ (nf to n)[m] = nf(m); (s) ⊦ ⌜n⌝ .n = n; (s) Used terms: nσ, n, n2

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

Lemma monle-leq

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

User applied rule cut formula

And obtained the following new goals

• Proof step 40:

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

• Proof step 50:

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

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 41:

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

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 41 from proof inv-inverts (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 + 1] = 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, ⌜n⌝ + ⌜1⌝ < # nσ, stutf(nσ), (∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = nσ[n]) ∨ (∀ n0. ¬(⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n])) ⊦

Rule argument

Additional information

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

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

Lemma maxinf-inf

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 44:

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

Rule argument

Additional information

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

User applied rule case distinction

And obtained the following new goals

• Proof step 45:

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

• Proof step 49:

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

Rule argument

Additional information

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

User applied rule all left

And obtained the following new goals

• Proof step 46:

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

Rule argument

Additional information

Used terms: n + 1

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 47:

  nσ[n + 1] = nσ[n] + 1, nσ[n1] = nσ[n], (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⌝ + ⌜1⌝ < # nσ, ⌜n1⌝ < # nσ, n + 1 < 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

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

User applied rule apply lemma

And obtained the following new goals

• Proof step 48:

  ⊦ 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 + 1 ≤ n ↔ m < n; (s) ⊦ ¬ n < n; (s) ⊦ ¬ 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: nσ, n + 1, n1

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

Lemma monle-leq

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: lsle-trans: ⊦ n∞ < n∞1 ∧ ¬ n∞2 < n∞1 → n∞ < n∞2; (forward) from specification nat-inf ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; (s) ⊦ ¬ m + n < m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic 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: n

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 51:

  ⊦ n∞ = ∞ ↔ ¬(∃ n. n∞ = ⌜n⌝ )

• Proof step 52:

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

Rule argument

Additional information

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

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 53:

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

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜n⌝ = ⌜n0⌝ ↔ n = n0; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 54:

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

• Proof step 55:

  nσ[n0] = nσ[n], nσ[n + 1] = 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, n0 ≠ n, ⌜n⌝ + ⌜1⌝ < # nσ, ⌜n0⌝ < # nσ, stutf(nσ), ∀ m. ⌜m⌝ < # nσ ∧ nσ[m] = nσ[n] → ¬ n0 < m ⊦

Rule argument

Additional information

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

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

Lemma maxinf-fin

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 56:

  nσ[n + 1] = nσ[n] + 1, nσ[n0] = nσ[n], (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, n0 ≠ n, ⌜n0⌝ < # nσ, ⌜n⌝ + ⌜1⌝ < # nσ, stutf(nσ), ∀ m. ⌜m⌝ < # nσ ∧ nσ[m] = nσ[n] → ¬ n0 < m ⊦

Rule argument

Additional information

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

User applied rule all left

And obtained the following new goals

• Proof step 57:

  nσ[n0] = nσ[n], nσ[n + 1] = 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, n0 ≠ n, ⌜n⌝ + ⌜1⌝ < # nσ, ⌜n0⌝ < # nσ, stutf(nσ), ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n] → ¬ n0 < n0, ∀ m. ⌜m⌝ < # nσ ∧ nσ[m] = nσ[n] → ¬ n0 < m ⊦

Rule argument

Additional information

Used terms: n0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 58:

  nσ[n + 1] = nσ[n] + 1, nσ[n0] = nσ[n], (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, n0 ≠ n, ⌜n0⌝ < # nσ, ⌜n⌝ + ⌜1⌝ < # nσ, stutf(nσ), ∀ m. ⌜m⌝ < # nσ ∧ nσ[m] = nσ[n] → ¬ n0 < m ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s)

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

User applied rule apply lemma

And obtained the following new goals

• Proof step 59:

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

• Proof step 60:

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

Rule argument

Additional information

Used terms: nσ, n + 1, n0

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

Lemma monle-leq

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 61:

  nσ[n + 1] = nσ[n] + 1, nσ[n0] = nσ[n], (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, n0 ≠ n, n0 < n, ⌜n0⌝ < # nσ, ⌜n⌝ + ⌜1⌝ < # nσ, stutf(nσ), ∀ m. ⌜m⌝ < # nσ ∧ nσ[m] = nσ[n] → ¬ n0 < m ⊦

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m ≠ n → (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 ⊦ ¬ n < n; (s) ⊦ m + 1 ≤ n ↔ m < n; (s)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: lsle-trans: ⊦ n∞ < n∞1 ∧ ¬ n∞2 < n∞1 → n∞ < n∞2; (forward) from specification nat-inf ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; (s) ⊦ ¬ m + n < m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic 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: n