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

System applied rule normalize

And obtained the following new goals

• Proof step 2:

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

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

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

Rule argument

Additional information

Proof step 3 from proof inv-get-inverse-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).take(leninv(nσ).n)⌝ )

• Proof step 5:

  ⌜n⌝ + ⌜1⌝ = leninv(nσ), 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).take(leninv(nσ).n)⌝ ) (n) + m + 1) ≠ 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).take(leninv(nσ).n)⌝ ) (n) ⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, stutf(nσ) ⊦

Rule argument

Additional information

Used terms: nσ

Proof step 4 from proof inv-get-inverse-last (previous step) (back to tree)

Lemma inv-def

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

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

Rule argument

Additional information

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

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

User applied rule flip equation

And obtained the following new goals

• Proof step 7:

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

Rule argument

Additional information

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

User applied rule cut formula

And obtained the following new goals

• Proof step 8:

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

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∞ = ⌜m⌝ + ⌜n⌝ → (n∞ = ∞ ↔ false); (s) Used formulas: leninv(nσ) = ∞

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

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

Rule argument

Additional information

Used simplifier rules: ⊦ n∞ = ⌜m⌝ + ⌜n⌝ → (n∞ = ∞ ↔ false); (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

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

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 10:

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

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∞ = ⌜m⌝ + ⌜n⌝ → (n∞ = ∞ ↔ false); (s) ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: leninv(nσ), n0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  leninv(nσ) = ⌜n⌝ + ⌜1⌝ , nσ(max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n + m + 1) ≠ n + 1, ⌜max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, 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⌝ = ⌜m⌝ + ⌜n⌝ ; (s) add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf m < n ⊦ nf.take(n)(m) = nf(m); (s) ⊦ ¬ n < n; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf ⊦ ⌜n⌝ + ⌜m⌝ = ⌜n0⌝ ↔ n + m = n0; (s) ⊦ n∞ = ⌜m⌝ + ⌜n⌝ → (n∞ = ∞ ↔ false); (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 12:

  ⊦ 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 13:

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

Rule argument

Additional information

Used terms: n, nσ

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

Lemma leninv-succ

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 14:

  nσ(max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n + m + 1) ≠ n + 1, ⌜max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, stutf(nσ), (# nσ = ∞ ⊃ nσ(m0) = n ∧ (∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1);nσ((# nσ).n - 1) = n) ⊦

Rule argument

Additional information

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

User applied rule cut formula

And obtained the following new goals

• Proof step 15:

  max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) = ∞, nσ(max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n + m + 1) ≠ n + 1, ⌜max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, stutf(nσ), (# nσ = ∞ ⊃ nσ(m0) = n ∧ (∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1);nσ((# nσ).n - 1) = n) ⊦

• Proof step 40:

  max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) ≠ ∞, nσ(max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n + m + 1) ≠ n + 1, ⌜max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, stutf(nσ), (# nσ = ∞ ⊃ nσ(m0) = n ∧ (∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1);nσ((# nσ).n - 1) = n) ⊦

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) = ∞, nσ(∞.n + m + 1) ≠ n + 1, stutf(nσ), ⌜∞.n⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, (# nσ = ∞ ⊃ nσ(m0) = n ∧ (∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1);nσ((# nσ).n - 1) = n) ⊦

Rule argument

Additional information

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

User applied rule apply lemma

And obtained the following new goals

• Proof step 17:

  ⊦ stutf(nσ) → (max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m) = ∞ ↔ # nσ = ∞ ∧ (∃ n. ∀ n0. ¬ n0 < n → nσ(n0) = m) ∨ (∀ n. ⌜n⌝ < # nσ → nσ(n) < m))

• Proof step 18:

  max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) = ∞, nσ(∞.n + m + 1) ≠ n + 1, stutf(nσ), ⌜∞.n⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, (# nσ = ∞ ⊃ nσ(m0) = n ∧ (∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1);nσ((# nσ).n - 1) = n), stutf(nσ) → ( max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) = ∞ ↔ # nσ = ∞ ∧ (∃ n0. ∀ n1. ¬ n1 < n0 → nσ(n1) = n) ∨ (∀ n0. ⌜n0⌝ < # nσ → nσ(n0) < n)) ⊦

Rule argument

Additional information

Used terms: nσ, n

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

Lemma stutf-maxinf-is-fin

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 19:

  max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) = ∞, nσ(∞.n + m + 1) ≠ n + 1, stutf(nσ), ⌜∞.n⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, # nσ = ∞ ∧ (∀ n1. ¬ n1 < n0 → nσ(n1) = n) ∨ (∀ n0. ⌜n0⌝ < # nσ → nσ(n0) < n), (# nσ = ∞ ⊃ nσ(m0) = n ∧ (∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1);nσ((# nσ).n - 1) = n) ⊦

Rule argument

Additional information

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

User applied rule case distinction

And obtained the following new goals

• Proof step 20:

  # nσ = ∞, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) = ∞, nσ(∞.n + m + 1) ≠ n + 1, stutf(nσ), ⌜∞.n⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, (# nσ = ∞ ⊃ nσ(m0) = n ∧ (∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1);nσ((# nσ).n - 1) = n), ∀ n1. ¬ n1 < n0 → nσ(n1) = n ⊦

• Proof step 31:

  max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) = ∞, nσ(∞.n + m + 1) ≠ n + 1, stutf(nσ), ⌜∞.n⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, (# nσ = ∞ ⊃ nσ(m0) = n ∧ (∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1);nσ((# nσ).n - 1) = n), ∀ n0. ⌜n0⌝ < # nσ → nσ(n0) < n ⊦

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 21:

  # nσ = ∞, max∞(λ n. nσ(n) = nσ(m0)) = ∞, nσ(∞.n + m + 1) ≠ nσ(m0) + 1, stutf(nσ), ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1, ∀ n1. ¬ n1 < n0 → nσ(n1) = nσ(m0) ⊦

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-get-inverse-last (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 22:

  # nσ = ∞, max∞(λ n. nσ(n) = nσ(m0)) = ∞, nσ(∞.n + m + 1) ≠ nσ(m0) + 1, stutf(nσ), m0 < n0 → nσ(n0) = nσ(m0) + 1, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1, ∀ n1. ¬ n1 < n0 → nσ(n1) = nσ(m0) ⊦

Rule argument

Additional information

Used terms: n0

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

User applied rule cut formula

And obtained the following new goals

• Proof step 23:

  # nσ = ∞, max∞(λ n. nσ(n) = nσ(m0)) = ∞, nσ(∞.n + m + 1) ≠ nσ(m0) + 1, m0 < n0, stutf(nσ), m0 < n0 → nσ(n0) = nσ(m0) + 1, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1, ∀ n1. ¬ n1 < n0 → nσ(n1) = nσ(m0) ⊦

• Proof step 27:

  # nσ = ∞, max∞(λ n. nσ(n) = nσ(m0)) = ∞, nσ(∞.n + m + 1) ≠ nσ(m0) + 1, stutf(nσ), ¬ m0 < n0, m0 < n0 → nσ(n0) = nσ(m0) + 1, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1, ∀ n1. ¬ n1 < n0 → nσ(n1) = nσ(m0) ⊦

Rule argument

Additional information

Used formulas: m0 < n0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

  # nσ = ∞, max∞(λ n. nσ(n) = nσ(m0)) = ∞, nσ(n0) = nσ(m0) + 1, nσ(∞.n + m + 1) ≠ nσ(n0), m0 < n0, stutf(nσ), ∀ m1. m0 < m1 → nσ(m1) = nσ(n0), ∀ n1. ¬ n1 < n0 → nσ(n1) = nσ(m0) ⊦

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-get-inverse-last (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 25:

  # nσ = ∞, max∞(λ n. nσ(n) = nσ(m0)) = ∞, nσ(n0) = nσ(m0) + 1, nσ(∞.n + m + 1) ≠ nσ(n0), m0 < n0, stutf(nσ), m0 < n0 → nσ(n0) = nσ(n0), ∀ m1. m0 < m1 → nσ(m1) = nσ(n0), ∀ n1. ¬ n1 < n0 → nσ(n1) = nσ(m0) ⊦

Rule argument

Additional information

Used terms: n0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 26:

  # nσ = ∞, max∞(λ n. nσ(n) = nσ(m0)) = ∞, nσ(n0) = nσ(m0) + 1, nσ(∞.n + m + 1) ≠ nσ(n0), m0 < n0, stutf(nσ), ∀ m1. m0 < m1 → nσ(m1) = nσ(n0), ∀ n1. ¬ n1 < n0 → nσ(n1) = nσ(m0) ⊦

Rule argument

Additional information

Proof step 26 from proof inv-get-inverse-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) ⊦ 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: n0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 28:

  # nσ = ∞, max∞(λ n. nσ(n) = nσ(m0)) = ∞, nσ(∞.n + m + 1) ≠ nσ(m0) + 1, n0 ≤ m0, stutf(nσ), ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1, ∀ n1. ¬ n1 < n0 → nσ(n1) = nσ(m0) ⊦

Rule argument

Additional information

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

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

User applied rule all left

And obtained the following new goals

• Proof step 29:

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

Rule argument

Additional information

Used terms: m0 + 1

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 30:

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

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 ⊦ ¬ n < n; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s)

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

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m = m + n ↔ n = 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat ⊦ m < n → n ≠ 0; (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 Used terms: m0 + 1

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

User applied rule case distinction

And obtained the following new goals

• Proof step 32:

  # nσ = ∞, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) = ∞, nσ(m0) = n, nσ(∞.n + m + 1) ≠ n + 1, stutf(nσ), ⌜∞.n⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1, ∀ n0. ⌜n0⌝ < # nσ → nσ(n0) < n ⊦

• Proof step 34:

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

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 33:

  # nσ = ∞, max∞(λ n0. nσ(n0) = nσ(m0)) = ∞, nσ(∞.n + m + 1) ≠ nσ(m0) + 1, stutf(nσ), ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1, ∀ n0. nσ(n0) < nσ(m0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ∞; (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)

Proof step 33 from proof inv-get-inverse-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: m0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 35:

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

Rule argument

Additional information

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

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 36:

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

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 37:

  # nσ = ⌜n⌝ , max∞(λ n0. n0 < n ∧ nσ(n0) = nσ(n - 1)) = ∞, nσ(∞.n + m + 1) ≠ nσ(n - 1) + 1, stutf(nσ), ∞.n + m + 1 < n, ∀ n0. n0 < n → nσ(n0) < nσ(n - 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) 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 ⊦ ⌜n1⌝ + ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n1 + n + m < n0; (s) ⊦ ⌜n⌝ ≠ ∞; (s)

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

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 38:

  # nσ = ⌜n⌝ , max∞(λ n0. n0 < n ∧ nσ(n0) = nσ(m0)) = ∞, n = m0 + 1, n ≠ 0, nσ(∞.n + m + 1) ≠ nσ(m0) + 1, stutf(nσ), ∞.n + m + 1 < n, ∀ n0. n0 < n → nσ(n0) < nσ(m0) ⊦

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: n, m0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 39:

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

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) ⊦ m + n < m + n0 ↔ n < n0; (s) ⊦ m + 1 ≠ 0; (s) ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)

Proof step 39 from proof inv-get-inverse-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: m0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 41:

  max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) ≠ ∞, nσ(max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n + m + 1) ≠ n + 1, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) + ⌜m⌝ + ⌜1⌝ < # nσ, stutf(nσ), (# nσ = ∞ ⊃ nσ(m0) = n ∧ (∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1);nσ((# nσ).n - 1) = n) ⊦

Rule argument

Additional information

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

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

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 42:

  max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) = ⌜n1⌝ , max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) ≠ ∞, nσ(n1 + m + 1) ≠ n + 1, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) + ⌜m⌝ + ⌜1⌝ < # nσ, stutf(nσ), (# nσ = ∞ ⊃ nσ(m0) = n ∧ (∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1);nσ((# nσ).n - 1) = n) ⊦

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 43:

  max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) = ⌜n1⌝ , nσ(n1 + m + 1) ≠ n + 1, ⌜n1⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, stutf(nσ), (# nσ = ∞ ⊃ nσ(m0) = n ∧ (∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1);nσ((# nσ).n - 1) = n) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜n⌝ ≠ ∞; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 44:

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

• Proof step 45:

  nσ(n1) = n, nσ(n1 + m + 1) ≠ n + 1, ⌜n1⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, ⌜n1⌝ < # nσ, stutf(nσ), (# nσ = ∞ ⊃ nσ(m0) = n ∧ (∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1);nσ((# nσ).n - 1) = n), ∀ m. ⌜m⌝ < # nσ ∧ nσ(m) = n → ¬ n1 < m ⊦

Rule argument

Additional information

Used terms: n1, λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n

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

Lemma maxinf-fin

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 46:

  nσ(n1 + m + 1) ≠ nσ(n1) + 1, ⌜n1⌝ < # nσ, ⌜n1⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, stutf(nσ), (# nσ = ∞ ⊃ nσ(m0) = nσ(n1) ∧ (∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1);nσ((# nσ).n - 1) = nσ(n1)), ∀ m. ⌜m⌝ < # nσ ∧ nσ(m) = nσ(n1) → ¬ n1 < m ⊦

Rule argument

Additional information

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

User applied rule case distinction

And obtained the following new goals

• Proof step 47:

  # nσ = ∞, nσ(m0) = nσ(n1), nσ(n1 + m + 1) ≠ nσ(n1) + 1, ⌜n1⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, ⌜n1⌝ < # nσ, stutf(nσ), ∀ m. ⌜m⌝ < # nσ ∧ nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

• Proof step 58:

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

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 48:

  # nσ = ∞, nσ(m0) = nσ(n1), nσ(n1 + m + 1) ≠ nσ(n1) + 1, stutf(nσ), ∀ m1. m0 < m1 → nσ(m1) = nσ(n1) + 1, ∀ m1. nσ(m1) = nσ(n1) → ¬ n1 < m1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ∞; (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)

Proof step 48 from proof inv-get-inverse-last (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 49:

  # nσ = ∞, nσ(m0) = nσ(n1), nσ(n1 + m + 1) ≠ nσ(n1) + 1, stutf(nσ), m0 < n1 → nσ(n1) = nσ(n1) + 1, ∀ m1. nσ(m1) = nσ(n1) → ¬ n1 < m1, ∀ m1. m0 < m1 → nσ(m1) = nσ(n1) + 1 ⊦

Rule argument

Additional information

Used terms: n1

Proof step 49 from proof inv-get-inverse-last (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 50:

  # nσ = ∞, nσ(m0) = nσ(n1), nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 ≤ m0, stutf(nσ), ∀ m1. m0 < m1 → nσ(m1) = nσ(n1) + 1, ∀ m1. nσ(m1) = nσ(n1) → ¬ n1 < m1 ⊦

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 = m + n ↔ n = 0; (s)

Proof step 50 from proof inv-get-inverse-last (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 51:

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

• Proof step 52:

  # nσ = ∞, nσ(m0) = nσ(n1), nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 ≤ m0, stutf(nσ), ∀ m1. nσ(m1) = nσ(n1) → ¬ n1 < m1, ∀ m1. m0 < m1 → nσ(m1) = nσ(n1) + 1, mon≤(nσ) ∧ n1 ≤ m0 ∧ ⌜m0⌝ < # nσ → nσ(n1) ≤ nσ(m0) ⊦

Rule argument

Additional information

Used terms: nσ, n1, m0

Proof step 51 from proof inv-get-inverse-last (previous step) (back to tree)

Lemma monle-leq

Proof step 52 from proof inv-get-inverse-last (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 53:

  # nσ = ∞, nσ(m0) = nσ(n1), nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 ≤ m0, stutf(nσ), ∀ m1. nσ(m1) = nσ(n1) → ¬ n1 < m1, ∀ m1. m0 < m1 → nσ(m1) = nσ(n1) + 1 ⊦

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); ⊦ m ≤ m; (s)

Proof step 53 from proof inv-get-inverse-last (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 54:

  # nσ = ∞, nσ(m0) = nσ(n1), nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 ≤ m0, stutf(nσ), m0 < m0 + 1 → nσ(m0 + 1) = nσ(n1) + 1, ∀ m1. m0 < m1 → nσ(m1) = nσ(n1) + 1, ∀ m1. nσ(m1) = nσ(n1) → ¬ n1 < m1 ⊦

Rule argument

Additional information

Used terms: m0 + 1

Proof step 54 from proof inv-get-inverse-last (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 55:

  # nσ = ∞, nσ(m0) = nσ(n1), nσ(m0 + 1) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 ≤ m0, stutf(nσ), ∀ m1. nσ(m1) = nσ(n1) → ¬ n1 < m1, ∀ m1. m0 < m1 → nσ(m1) = nσ(n1) + 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 ⊦ ¬ n < n; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s)

Proof step 55 from proof inv-get-inverse-last (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 56:

  # nσ = ∞, nσ(m0 + 1) = nσ(n1) + 1, nσ(m0) = nσ(n1), nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 ≤ m0, stutf(nσ), ∀ m1. m0 < m1 → nσ(m1) = nσ(n1) + 1, ∀ m1. nσ(m1) = nσ(n1) → ¬ n1 < m1, nσ(m0) = nσ(n1) → ¬ n1 < m0 ⊦

Rule argument

Additional information

Used terms: m0

Proof step 56 from proof inv-get-inverse-last (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 57:

  # nσ = ∞, nσ(m0 + 1) = nσ(m0) + 1, nσ(m0 + m + 1) ≠ nσ(m0) + 1, stutf(nσ), ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1, ∀ m1. nσ(m1) = nσ(m0) → ¬ m0 < m1 ⊦

Rule argument

Additional information

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

Proof step 57 from proof inv-get-inverse-last (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) ⊦ ¬ 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: m0 + m + 1

Proof step 58 from proof inv-get-inverse-last (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 59:

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

Rule argument

Additional information

Proof step 59 from proof inv-get-inverse-last (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 60:

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

Rule argument

Additional information

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

Proof step 60 from proof inv-get-inverse-last (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 61:

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

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) 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 ⊦ ⌜n1⌝ + ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n1 + n + m < n0; (s) ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 61 from proof inv-get-inverse-last (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 62:

  # nσ = ⌜n⌝ , n = m1 + 1, nσ(m1) = nσ(n1), n ≠ 0, nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 + m + 1 < n, n1 < n, stutf(nσ), ∀ m0. m0 < n ∧ nσ(m0) = nσ(n1) → ¬ n1 < m0 ⊦

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: n, m1

Proof step 62 from proof inv-get-inverse-last (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 63:

  # nσ = ⌜m1⌝ + ⌜1⌝ , nσ(m1) = nσ(n1), nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 ≤ m1, n1 + m < m1, stutf(nσ), ∀ m0. ¬ m1 < m0 ∧ nσ(m0) = nσ(n1) → ¬ n1 < m0 ⊦

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) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m + n < m + n0 ↔ n < n0; (s) ⊦ m + 1 ≠ 0; (s) ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)

Proof step 63 from proof inv-get-inverse-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) ⊦ ¬ 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: m1