Proof step 1 from proof leninv-succ (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

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

Rule argument

Additional information

Proof step 2 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

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

Rule argument

Additional information

Proof step 3 from proof leninv-succ (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 4:

  ⊦ leninv(nσ) = max∞(λ n. n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n - 1 ∧ ¬(# nσ = ∞ ∧ (∀ m0. m < m0 → nσ(m0) = n - 1))))

• Proof step 5:

  stutf(nσ) ⊦ max∞(λ n. n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n - 1 ∧ ¬(# nσ = ∞ ∧ (∀ m0. m < m0 → nσ(m0) = n - 1)))) = ⌜n⌝ + ⌜1⌝ ↔ (# nσ = ∞ ⊃ ∃ m. nσ(m) = n ∧ (∀ m0. m < m0 → nσ(m0) = nσ(m) + 1);nσ((# nσ).n - 1) = n)

Rule argument

Additional information

Used terms: nσ

Proof step 4 from proof leninv-succ (previous step) (back to tree)

Lemma leninv-def

Proof step 5 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  stutf(nσ) ⊦ max∞(λ n. n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ(m0) = n - 1)))) = ⌜n⌝ + ⌜1⌝ ↔ (# nσ = ∞ ⊃ ∃ m. nσ(m) = n ∧ (∀ m0. m < m0 → nσ(m0) = nσ(m) + 1);nσ((# nσ).n - 1) = n)

Rule argument

Additional information

Proof step 6 from proof leninv-succ (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 7:

  ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝

• Proof step 8:

  stutf(nσ) ⊦ max∞(λ n. n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ(m0) = n - 1)))) = ⌜n + 1⌝ ↔ (# nσ = ∞ ⊃ ∃ m. nσ(m) = n ∧ (∀ m0. m < m0 → nσ(m0) = nσ(m) + 1);nσ((# nσ).n - 1) = n)

Rule argument

Additional information

Used terms: 1, n

Proof step 7 from proof leninv-succ (previous step) (back to tree)

A spec-lemma

Proof step 8 from proof leninv-succ (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 9:

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

• Proof step 10:

  stutf(nσ) ⊦ (n + 1 ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n + 1 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ(m0) = n + 1 - 1)))) ∧ (∀ m. (m ≠ 0 → (∃ m0. ⌜m0⌝ < # nσ ∧ nσ(m0) = m - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m1. m0 < m1 → nσ(m1) = m - 1)))) → ¬ n + 1 < m) ↔ (# nσ = ∞ ⊃ ∃ m. nσ(m) = n ∧ (∀ m0. m < m0 → nσ(m0) = nσ(m) + 1);nσ((# nσ).n - 1) = n)

Rule argument

Additional information

Used terms: n + 1, λ n. n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ(m0) = n - 1)))

Proof step 9 from proof leninv-succ (previous step) (back to tree)

Lemma maxinf-fin

Proof step 10 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

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

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ n + 0 = n; (s) ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s) ⊦ m + 1 ≠ 0; (s)

Proof step 11 from proof leninv-succ (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 12:

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

• Proof step 90:

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

Rule argument

Additional information

Used formulas: # nσ = ∞

Proof step 12 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  # nσ = ∞, stutf(nσ) ⊦ (∃ m. nσ(m) = n ∧ ¬(∀ m0. m < m0 → nσ(m0) = nσ(m))) ∧ (∀ n0. (∃ n1. nσ(n1) = n0 - 1 ∧ ¬(∀ m1. n1 < m1 → nσ(m1) = n0 - 1)) → ¬ n + 1 < n0) ↔ (∃ m. nσ(m) = n ∧ (∀ m0. m < m0 → nσ(m0) = nσ(m) + 1))

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ∞; (s)

Proof step 13 from proof leninv-succ (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 14:

  # nσ = ∞, stutf(nσ), ∃ m. nσ(m) = n ∧ ¬(∀ m0. m < m0 → nσ(m0) = nσ(m)), ∀ n0. (∃ n1. nσ(n1) = n0 - 1 ∧ ¬(∀ m1. n1 < m1 → nσ(m1) = n0 - 1)) → ¬ n + 1 < n0 ⊦ ∃ m. nσ(m) = n ∧ (∀ m0. m < m0 → nσ(m0) = nσ(m) + 1)

• Proof step 74:

  # nσ = ∞, stutf(nσ), ∃ m. nσ(m) = n ∧ (∀ m0. m < m0 → nσ(m0) = nσ(m) + 1) ⊦ (∃ m. nσ(m) = n ∧ ¬(∀ m0. m < m0 → nσ(m0) = nσ(m))) ∧ (∀ n0. (∃ n1. nσ(n1) = n0 - 1 ∧ ¬(∀ m1. n1 < m1 → nσ(m1) = n0 - 1)) → ¬ n + 1 < n0)

Rule argument

Additional information

Proof step 14 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  # nσ = ∞, nσ(n1) ≠ nσ(n0), n0 < n1, stutf(nσ), ∀ n1. (∃ n. nσ(n) = n1 - 1 ∧ ¬(∀ m1. n < m1 → nσ(m1) = n1 - 1)) → ¬ nσ(n0) + 1 < n1 ⊦ ∃ m. nσ(m) = nσ(n0) ∧ (∀ 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 15 from proof leninv-succ (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 16:

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

• Proof step 17:

  # nσ = ∞, nσ(n1) ≠ nσ(n0), n0 < n1, stutf(nσ), mon≤(nσ) ∧ n0 ≤ n1 ∧ ⌜n1⌝ < # nσ → nσ(n0) ≤ nσ(n1), ∀ n1. (∃ n. nσ(n) = n1 - 1 ∧ ¬(∀ m1. n < m1 → nσ(m1) = n1 - 1)) → ¬ nσ(n0) + 1 < n1 ⊦ ∃ m. nσ(m) = nσ(n0) ∧ (∀ m0. m < m0 → nσ(m0) = nσ(m) + 1)

Rule argument

Additional information

Used terms: nσ, n0, n1

Proof step 16 from proof leninv-succ (previous step) (back to tree)

Lemma monle-leq

Proof step 17 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

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

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⌝ < ∞; (s) ⊦ m ≠ n → (m ≤ n ↔ m < n); (s) ⊦ m < n → m ≤ n; (s)

Proof step 18 from proof leninv-succ (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 19:

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

Rule argument

Additional information

Used terms: nσ(n1)

Proof step 19 from proof leninv-succ (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 20:

  # nσ = ∞, nσ(n1) = n2 + 1, nσ(n1) ≠ 0, nσ(n1) ≠ nσ(n0), n0 < n1, nσ(n0) < nσ(n1), stutf(nσ), (∃ n. nσ(n) = n2 ∧ ¬(∀ m1. n < m1 → nσ(m1) = n2)) → ¬ nσ(n0) + 1 < nσ(n1), ∀ n1. (∃ n. nσ(n) = n1 - 1 ∧ ¬(∀ m1. n < m1 → nσ(m1) = n1 - 1)) → ¬ nσ(n0) + 1 < n1 ⊦ ∃ m. nσ(m) = nσ(n0) ∧ (∀ m0. m < m0 → nσ(m0) = nσ(m) + 1)

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: nσ(n1), n2

Proof step 20 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 21:

  # nσ = ∞, nσ(n1) = n2 + 1, n2 + 1 ≠ nσ(n0), n0 < n1, nσ(n0) ≤ n2, stutf(nσ), (∃ n. nσ(n) = n2 ∧ ¬(∀ m1. n < m1 → nσ(m1) = nσ(n))) → ¬ nσ(n0) < n2, ∀ n1. (∃ n. nσ(n) = n1 - 1 ∧ ¬(∀ m1. n < m1 → nσ(m1) = n1 - 1)) → ¬ nσ(n0) + 1 < n1 ⊦ ∃ m. nσ(m) = nσ(n0) ∧ (∀ m0. m < m0 → nσ(m0) = nσ(m) + 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 ↔ ¬ n ≤ m; (ss) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ m + 1 ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m < n → n ≠ 0; (s)

Proof step 21 from proof leninv-succ (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 22:

  # nσ = ∞, nσ(n1) = n2 + 1, n2 + 1 ≠ nσ(n0), n0 < n1, nσ(n0) ≤ n2, stutf(nσ), ∀ n1. (∃ n. nσ(n) = n1 - 1 ∧ ¬(∀ m1. n < m1 → nσ(m1) = n1 - 1)) → ¬ nσ(n0) + 1 < n1 ⊦ ∃ m. nσ(m) = nσ(n0) ∧ (∀ m0. m < m0 → nσ(m0) = nσ(m) + 1), ∃ n. nσ(n) = n2 ∧ ¬(∀ m1. n < m1 → nσ(m1) = nσ(n))

• Proof step 29:

  # nσ = ∞, nσ(n1) = n2 + 1, n2 + 1 ≠ nσ(n0), n0 < n1, nσ(n0) ≤ n2, stutf(nσ), ¬ nσ(n0) < n2, ∀ n1. (∃ n. nσ(n) = n1 - 1 ∧ ¬(∀ m1. n < m1 → nσ(m1) = n1 - 1)) → ¬ nσ(n0) + 1 < n1 ⊦ ∃ m. nσ(m) = nσ(n0) ∧ (∀ m0. m < m0 → nσ(m0) = nσ(m) + 1)

Rule argument

Additional information

Proof step 22 from proof leninv-succ (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 23:

  ⊦ stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ(m) → (∃ m0. m0 ≤ m ∧ nσ(m0) = n)

• Proof step 24:

  # nσ = ∞, nσ(n1) = n2 + 1, n2 + 1 ≠ nσ(n0), n0 < n1, nσ(n0) ≤ n2, stutf(nσ), ∀ n1. (∃ n. nσ(n) = n1 - 1 ∧ ¬(∀ m1. n < m1 → nσ(m1) = n1 - 1)) → ¬ nσ(n0) + 1 < n1, stutf(nσ) ∧ ⌜n1⌝ < # nσ ∧ n2 ≤ nσ(n1) → (∃ m0. m0 ≤ n1 ∧ nσ(m0) = n2) ⊦ ∃ m. nσ(m) = nσ(n0) ∧ (∀ m0. m < m0 → nσ(m0) = nσ(m) + 1), ∃ n. nσ(n) = n2 ∧ ¬(∀ m1. n < m1 → nσ(m1) = nσ(n))

Rule argument

Additional information

Used terms: nσ, n1, n2

Proof step 23 from proof leninv-succ (previous step) (back to tree)

Lemma stutf-alltaken

Proof step 24 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 25:

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

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 ≤ m + n; (s) ⊦ ⌜m⌝ < ∞; (s)

Proof step 25 from proof leninv-succ (previous step) (back to tree)

User applied rule exists right

And obtained the following new goals

• Proof step 26:

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

Rule argument

Additional information

Used terms: n

Proof step 26 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 27:

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

Rule argument

Additional information

Proof step 27 from proof leninv-succ (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 28:

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

Rule argument

Additional information

Used terms: n1

Proof step 28 from proof leninv-succ (previous step) (back to tree)

User applied rule case distinction

And closed the goal!

Rule argument

Additional information

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

Proof step 29 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 30:

  # nσ = ∞, nσ(n1) = nσ(n0) + 1, n0 < n1, stutf(nσ), ∀ n1. (∃ n. nσ(n) = n1 - 1 ∧ ¬(∀ m1. n < m1 → nσ(m1) = n1 - 1)) → ¬ nσ(n0) + 1 < n1 ⊦ ∃ m. nσ(m) = nσ(n0) ∧ (∀ m0. m < m0 → nσ(m0) = nσ(m) + 1)

Rule argument

Additional information

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

Proof step 30 from proof leninv-succ (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 31:

  # nσ = ∞, nσ(n1) = nσ(n0) + 1, n0 < n1, stutf(nσ), ∃ m1. n0 ≤ m1 ∧ m1 < n1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1, ∀ n1. (∃ n. nσ(n) = n1 - 1 ∧ ¬(∀ m1. n < m1 → nσ(m1) = n1 - 1)) → ¬ nσ(n0) + 1 < n1 ⊦ ∃ m. nσ(m) = nσ(n0) ∧ (∀ m0. m < m0 → nσ(m0) = nσ(m) + 1)

• Proof step 50:

  # nσ = ∞, nσ(n1) = nσ(n0) + 1, n0 < n1, stutf(nσ), ∀ n1. (∃ n. nσ(n) = n1 - 1 ∧ ¬(∀ m1. n < m1 → nσ(m1) = n1 - 1)) → ¬ nσ(n0) + 1 < n1 ⊦ ∃ m. nσ(m) = nσ(n0) ∧ (∀ m0. m < m0 → nσ(m0) = nσ(m) + 1), ∃ m1. n0 ≤ m1 ∧ m1 < n1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1

Rule argument

Additional information

Used formulas: ∃ m1. n0 ≤ m1 ∧ m1 < n1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1

Proof step 31 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 32:

  # nσ = ∞, nσ(n1) = nσ(n) + 1, nσ(n + 1) = nσ(n1), nσ(n0) = nσ(n), n < n1, n0 < n1, n0 ≤ n, stutf(nσ), ∀ n1. (∃ n. nσ(n) = n1 - 1 ∧ ¬(∀ m1. n < m1 → nσ(m1) = n1 - 1)) → ¬ nσ(n) + 1 < n1 ⊦ ∃ m. nσ(m) = nσ(n) ∧ (∀ m0. m < m0 → nσ(m0) = nσ(m) + 1)

Rule argument

Additional information

Used simplifier rules: fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat

Proof step 32 from proof leninv-succ (previous step) (back to tree)

User applied rule exists right

And obtained the following new goals

• Proof step 33:

  # nσ = ∞, nσ(n0) = nσ(n), nσ(n + 1) = nσ(n1), nσ(n1) = nσ(n) + 1, n < n1, n0 ≤ n, n0 < n1, stutf(nσ), ∀ n1. (∃ n. nσ(n) = n1 - 1 ∧ ¬(∀ m1. n < m1 → nσ(m1) = n1 - 1)) → ¬ nσ(n) + 1 < n1 ⊦ nσ(n) = nσ(n) ∧ (∀ m0. n < m0 → nσ(m0) = nσ(n) + 1)

Rule argument

Additional information

Used terms: n

Proof step 33 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 34:

  # nσ = ∞, nσ(n0) = nσ(n), nσ(n + 1) = nσ(n1), nσ(n1) = nσ(n) + 1, nσ(n2) ≠ nσ(n1), n < n2, n < n1, n0 < n2, n0 ≤ n, n0 < n1, stutf(nσ), ∀ n1. (∃ n. nσ(n) = n1 - 1 ∧ ¬(∀ m1. n < m1 → nσ(m1) = n1 - 1)) → ¬ nσ(n) + 1 < n1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 34 from proof leninv-succ (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 35:

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

• Proof step 36:

  # nσ = ∞, nσ(n1) = nσ(n) + 1, nσ(n + 1) = nσ(n1), nσ(n0) = nσ(n), nσ(n2) ≠ nσ(n1), n < n1, n < n2, n0 < n1, n0 ≤ n, n0 < n2, stutf(nσ), mon≤(nσ) ∧ n + 1 ≤ n2 ∧ ⌜n2⌝ < # nσ → nσ(n + 1) ≤ nσ(n2), ∀ n1. (∃ n. nσ(n) = n1 - 1 ∧ ¬(∀ m1. n < m1 → nσ(m1) = n1 - 1)) → ¬ nσ(n) + 1 < n1 ⊦

Rule argument

Additional information

Used terms: nσ, n + 1, n2

Proof step 35 from proof leninv-succ (previous step) (back to tree)

Lemma monle-leq

Proof step 36 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 37:

  # nσ = ∞, nσ(n0) = nσ(n), nσ(n + 1) = nσ(n1), nσ(n1) = nσ(n) + 1, nσ(n2) ≠ nσ(n1), n < n2, n < n1, n0 < n2, n0 ≤ n, n0 < n1, nσ(n1) < nσ(n2), stutf(nσ), ∀ n1. (∃ n. nσ(n) = n1 - 1 ∧ ¬(∀ m1. n < m1 → nσ(m1) = n1 - 1)) → ¬ nσ(n) + 1 < n1 ⊦

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); ⊦ m < n → n ≠ 0; (s) lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m ≠ n → (m ≤ n ↔ m < n); (s) ⊦ ⌜m⌝ < ∞; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + 1 ≤ n ↔ m < n; (s)

Proof step 37 from proof leninv-succ (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 38:

  # nσ = ∞, nσ(n1) = nσ(n) + 1, nσ(n + 1) = nσ(n1), nσ(n0) = nσ(n), nσ(n2) ≠ nσ(n1), n < n1, n < n2, n0 < n1, n0 ≤ n, n0 < n2, nσ(n1) < nσ(n2), stutf(nσ), (∃ n. nσ(n) = nσ(n2) - 1 ∧ ¬(∀ m1. n < m1 → nσ(m1) = nσ(n2) - 1)) → ¬ nσ(n) + 1 < nσ(n2) ⊦

Rule argument

Additional information

Used terms: nσ(n2)

Proof step 38 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 39:

  # nσ = ∞, nσ(n1) = nσ(n) + 1, nσ(n + 1) = nσ(n1), nσ(n0) = nσ(n), nσ(n2) ≠ nσ(n1), n < n1, n < n2, n0 < n1, n0 ≤ n, n0 < n2, nσ(n1) < nσ(n2), stutf(nσ) ⊦ ∃ n. nσ(n) = nσ(n2) - 1 ∧ ¬(∀ m1. n < m1 → nσ(m1) = nσ(n2) - 1)

Rule argument

Additional information

Proof step 39 from proof leninv-succ (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 40:

  # nσ = ∞, nσ(n0) = nσ(n), nσ(n + 1) = nσ(n1), nσ(n1) = nσ(n) + 1, nσ(n2) = n3 + 1, nσ(n2) ≠ 0, nσ(n2) ≠ nσ(n1), n < n2, n < n1, n0 < n2, n0 ≤ n, n0 < n1, nσ(n1) < nσ(n2), stutf(nσ) ⊦ ∃ n. nσ(n) = n3 ∧ ¬(∀ m1. n < m1 → nσ(m1) = n3)

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: nσ(n2), n3

Proof step 40 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 41:

  # nσ = ∞, nσ(n2) = n3 + 1, nσ(n1) = nσ(n) + 1, nσ(n + 1) = nσ(n1), nσ(n0) = nσ(n), n3 + 1 ≠ nσ(n1), n < n1, n < n2, n0 < n1, n0 ≤ n, n0 < n2, nσ(n1) ≤ n3, nσ(n) < n3, stutf(nσ) ⊦ ∃ n4. nσ(n4) = n3 ∧ ¬(∀ m1. n4 < m1 → nσ(m1) = nσ(n4))

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic 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 eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ m + n < m + n0 ↔ n < n0; (s) ⊦ m + 1 ≠ 0; (s)

Proof step 41 from proof leninv-succ (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 42:

  ⊦ stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ(m) → (∃ m0. m0 ≤ m ∧ nσ(m0) = n)

• Proof step 43:

  # nσ = ∞, nσ(n0) = nσ(n), nσ(n + 1) = nσ(n1), nσ(n1) = nσ(n) + 1, nσ(n2) = n3 + 1, n3 + 1 ≠ nσ(n1), n < n2, n < n1, n0 < n2, n0 ≤ n, n0 < n1, nσ(n) < n3, nσ(n1) ≤ n3, stutf(nσ), stutf(nσ) ∧ ⌜n2⌝ < # nσ ∧ n3 ≤ nσ(n2) → (∃ m0. m0 ≤ n2 ∧ nσ(m0) = n3) ⊦ ∃ n4. nσ(n4) = n3 ∧ ¬(∀ m1. n4 < m1 → nσ(m1) = nσ(n4))

Rule argument

Additional information

Used terms: nσ, n2, n3

Proof step 42 from proof leninv-succ (previous step) (back to tree)

Lemma stutf-alltaken

Proof step 43 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 44:

  # nσ = ∞, nσ(n0) = nσ(n), nσ(n + 1) = nσ(n1), nσ(n1) = nσ(n) + 1, nσ(n2) = nσ(n4) + 1, nσ(n2) ≠ nσ(n1), n < n2, n < n1, n0 < n2, n0 ≤ n, n0 < n1, n4 ≤ n2, nσ(n) < nσ(n4), nσ(n1) ≤ nσ(n4), stutf(nσ) ⊦ ∃ n. nσ(n) = nσ(n4) ∧ ¬(∀ m1. n < m1 → nσ(m1) = nσ(n))

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 eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat ⊦ m ≤ m + n; (s) ⊦ ⌜m⌝ < ∞; (s)

Proof step 44 from proof leninv-succ (previous step) (back to tree)

User applied rule exists right

And obtained the following new goals

• Proof step 45:

  # nσ = ∞, nσ(n2) = nσ(n4) + 1, nσ(n1) = nσ(n) + 1, nσ(n + 1) = nσ(n1), nσ(n0) = nσ(n), nσ(n2) ≠ nσ(n1), n < n1, n < n2, n0 < n1, n0 ≤ n, n0 < n2, n4 ≤ n2, nσ(n1) ≤ nσ(n4), nσ(n) < nσ(n4), stutf(nσ) ⊦ nσ(n4) = nσ(n4) ∧ ¬(∀ m1. n4 < m1 → nσ(m1) = nσ(n4))

Rule argument

Additional information

Used terms: n4

Proof step 45 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 46:

  # nσ = ∞, nσ(n0) = nσ(n), nσ(n + 1) = nσ(n1), nσ(n1) = nσ(n) + 1, nσ(n2) = nσ(n4) + 1, nσ(n2) ≠ nσ(n1), n < n2, n < n1, n0 < n2, n0 ≤ n, n0 < n1, n4 ≤ n2, nσ(n) < nσ(n4), nσ(n1) ≤ nσ(n4), stutf(nσ), ∀ m1. n4 < m1 → nσ(m1) = nσ(n4) ⊦

Rule argument

Additional information

Proof step 46 from proof leninv-succ (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 47:

  # nσ = ∞, nσ(n2) = nσ(n4) + 1, nσ(n1) = nσ(n) + 1, nσ(n + 1) = nσ(n1), nσ(n0) = nσ(n), nσ(n2) ≠ nσ(n1), n < n1, n < n2, n0 < n1, n0 ≤ n, n0 < n2, n4 ≤ n2, nσ(n1) ≤ nσ(n4), nσ(n) < nσ(n4), stutf(nσ), n4 < n2 → nσ(n2) = nσ(n4) ⊦

Rule argument

Additional information

Used terms: n2

Proof step 47 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 48:

  # nσ = ∞, nσ(n0) = nσ(n), nσ(n + 1) = nσ(n1), nσ(n1) = nσ(n) + 1, nσ(n2) = nσ(n4) + 1, nσ(n2) ≠ nσ(n1), n < n2, n < n1, n0 < n2, n0 ≤ n, n0 < n1, n4 ≤ n2, nσ(n) < nσ(n4), nσ(n1) ≤ nσ(n4), stutf(nσ), n4 < n2 → nσ(n2) = nσ(n4) ⊦

Rule argument

Additional information

Proof step 48 from proof leninv-succ (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)

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); ⊦ m ≤ n → (n ≤ m ↔ m = n); (s) ⊦ m = n + 1 → ¬ m ≤ n; (s) ⊦ m ≠ n → (m ≤ n ↔ m < n); (s) lf-01: ⊦ m < n ∧ n = n0 → m < n0; (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: nσ, n2, n4

Proof step 49 from proof leninv-succ (previous step) (back to tree)

Lemma monle-leq

Proof step 50 from proof leninv-succ (previous step) (back to tree)

User applied rule weakening

And obtained the following new goals

• Proof step 51:

  # nσ = ∞, nσ(n1) = nσ(n0) + 1, n0 < n1, stutf(nσ) ⊦ ∃ m1. n0 ≤ m1 ∧ m1 < n1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1

Rule argument

Additional information

Proof step 51 from proof leninv-succ (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 52:

  # nσ = ∞, n2.+1 = n1 - n0, nσ(n1) = nσ(n0) + 1, n0 < n1, stutf(nσ), ∀ n1, n0, nσ. n2 = n1 - n0 ∧ # nσ = ∞ ∧ nσ(n1) = nσ(n0) + 1 ∧ n0 < n1 ∧ stutf(nσ) → (∃ m1. n0 ≤ m1 ∧ m1 < n1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1) ⊦ ∃ m1. n0 ≤ m1 ∧ m1 < n1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1

Rule argument

Additional information

Used simplifier rules: Nat freely generated by 0, +1; ⊦ n < m → m - n ≠ 0; (s)

Proof step 52 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 53:

  # nσ = ∞, n2 + 1 = n1 - n0, nσ(n1) = nσ(n0) + 1, n0 < n1, stutf(nσ), ∀ n1, n0, nσ. n2 = n1 - n0 ∧ # nσ = ∞ ∧ nσ(n1) = nσ(n0) + 1 ∧ n0 < n1 ∧ stutf(nσ) → (∃ m1. n0 ≤ m1 ∧ m1 < n1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1) ⊦ ∃ m1. n0 ≤ m1 ∧ m1 < n1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1

Rule argument

Additional information

Used simplifier rules: ⊦ m.+1 = m + 1; (s)

Proof step 53 from proof leninv-succ (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 54:

  # nσ = ∞, n1 = n3 + n0, n2 + 1 = n3, nσ(n1) = nσ(n0) + 1, n0 < n1, n0 ≤ n1, stutf(nσ), ∀ n1, n0, nσ. n2 = n1 - n0 ∧ # nσ = ∞ ∧ nσ(n1) = nσ(n0) + 1 ∧ n0 < n1 ∧ stutf(nσ) → (∃ m1. n0 ≤ m1 ∧ m1 < n1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1) ⊦ ∃ m1. n0 ≤ m1 ∧ m1 < n1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ m < n → ¬ n < m; (s) n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n1, n0, n3

Proof step 54 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 55:

  # nσ = ∞, nσ((n2 + 1) + n0) = nσ(n0) + 1, stutf(nσ), ∀ n1, n0, nσ. n2 = n1 - n0 ∧ # nσ = ∞ ∧ nσ(n1) = nσ(n0) + 1 ∧ n0 < n1 ∧ stutf(nσ) → (∃ m1. n0 ≤ m1 ∧ m1 < n1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1) ⊦ ∃ m1. n0 ≤ m1 ∧ ¬ n2 + n0 < m1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 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 + 1 ↔ ¬ n < m; (s) ⊦ ¬ m + n < m; (s) ⊦ m < n → m ≤ n; (s)

Proof step 55 from proof leninv-succ (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 56:

  # nσ = ∞, nσ((n2 + 1) + n0) = nσ(n0) + 1, stutf(nσ), ∀ n1, n0, nσ. n2 = n1 - n0 ∧ # nσ = ∞ ∧ nσ(n1) = nσ(n0) + 1 ∧ n0 < n1 ∧ stutf(nσ) → (∃ m1. n0 ≤ m1 ∧ m1 < n1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1), n2 = n2 + n0 - n0 ∧ # nσ = ∞ ∧ nσ(n2 + n0) = nσ(n0) + 1 ∧ n0 < n2 + n0 ∧ stutf(nσ) → (∃ m1. n0 ≤ m1 ∧ m1 < n2 + n0 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1) ⊦ ∃ m1. n0 ≤ m1 ∧ ¬ n2 + n0 < m1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1

Rule argument

Additional information

Used terms: n2 + n0, n0, nσ

Proof step 56 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 57:

  # nσ = ∞, nσ((n2 + 1) + n0) = nσ(n0) + 1, stutf(nσ), ∀ n1, n0, nσ. n2 = n1 - n0 ∧ # nσ = ∞ ∧ nσ(n1) = nσ(n0) + 1 ∧ n0 < n1 ∧ stutf(nσ) → (∃ m1. n0 ≤ m1 ∧ m1 < n1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1), nσ(n2 + n0) = nσ(n0) + 1 ∧ n2 ≠ 0 → n0 ≤ n ∧ n < n2 + n0 ∧ nσ(n0) = nσ(n) ∧ nσ(n + 1) = nσ(n) + 1 ⊦ ∃ m1. n0 ≤ m1 ∧ ¬ n2 + n0 < m1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 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 < m + n ↔ n ≠ 0; (s) ⊦ m + n - m = n; (s)

Proof step 57 from proof leninv-succ (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 58:

  # nσ = ∞, nσ((n2 + 1) + n0) = nσ(n0) + 1, stutf(nσ), n2 = (n2 + 1) + n0 - n0 + 1 ∧ # nσ = ∞ ∧ nσ((n2 + 1) + n0) = nσ(n0 + 1) + 1 ∧ n0 + 1 < (n2 + 1) + n0 ∧ stutf(nσ) → (∃ m1. n0 + 1 ≤ m1 ∧ m1 < (n2 + 1) + n0 ∧ nσ(n0 + 1) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1), nσ(n2 + n0) = nσ(n0) + 1 ∧ n2 ≠ 0 → n0 ≤ n ∧ n < n2 + n0 ∧ nσ(n0) = nσ(n) ∧ nσ(n + 1) = nσ(n) + 1 ⊦ ∃ m1. n0 ≤ m1 ∧ ¬ n2 + n0 < m1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1

Rule argument

Additional information

Used terms: (n2 + 1) + n0, n0 + 1, nσ

Proof step 58 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 59:

  # nσ = ∞, nσ((n2 + 1) + n0) = nσ(n0) + 1, stutf(nσ), nσ(n2 + n0) = nσ(n0) + 1 ∧ n2 ≠ 0 → n0 ≤ n ∧ n < n2 + n0 ∧ nσ(n0) = nσ(n) ∧ nσ(n + 1) = nσ(n) + 1, nσ(n0) = nσ(n0 + 1) ∧ n2 ≠ 0 → n0 < n1 ∧ ¬ n2 + n0 < n1 ∧ nσ(n0 + 1) = nσ(n1) ∧ nσ(n1 + 1) = nσ(n1) + 1 ⊦ ∃ m1. n0 ≤ m1 ∧ ¬ n2 + n0 < m1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 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) ⊦ m + 1 ≤ n ↔ m < n; (s) ⊦ 1 = m + 1 ↔ m = 0; (s) ⊦ m + 1 ≠ 0; (s) ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s) ⊦ m + n < m + n0 ↔ n < n0; (s) ⊦ n + 0 = n; (s) ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s) ⊦ m + n - n + n1 = m - n1; (s)

Proof step 59 from proof leninv-succ (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 60:

  # nσ = ∞, nσ((n2 + 1) + n0) = nσ(n0) + 1, nσ(n0) = nσ(n0 + 1), stutf(nσ), nσ(n0) = nσ(n0 + 1) ∧ n2 ≠ 0 → n0 < n1 ∧ ¬ n2 + n0 < n1 ∧ nσ(n0 + 1) = nσ(n1) ∧ nσ(n1 + 1) = nσ(n1) + 1, nσ(n2 + n0) = nσ(n0) + 1 ∧ n2 ≠ 0 → n0 ≤ n ∧ n < n2 + n0 ∧ nσ(n0) = nσ(n) ∧ nσ(n + 1) = nσ(n) + 1 ⊦ ∃ m1. n0 ≤ m1 ∧ ¬ n2 + n0 < m1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1

• Proof step 64:

  # nσ = ∞, nσ((n2 + 1) + n0) = nσ(n0) + 1, nσ(n0) ≠ nσ(n0 + 1), stutf(nσ), nσ(n0) = nσ(n0 + 1) ∧ n2 ≠ 0 → n0 < n1 ∧ ¬ n2 + n0 < n1 ∧ nσ(n0 + 1) = nσ(n1) ∧ nσ(n1 + 1) = nσ(n1) + 1, nσ(n2 + n0) = nσ(n0) + 1 ∧ n2 ≠ 0 → n0 ≤ n ∧ n < n2 + n0 ∧ nσ(n0) = nσ(n) ∧ nσ(n + 1) = nσ(n) + 1 ⊦ ∃ m1. n0 ≤ m1 ∧ ¬ n2 + n0 < m1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1

Rule argument

Additional information

Used formulas: nσ(n0) = nσ(n0 + 1)

Proof step 60 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 61:

  # nσ = ∞, nσ(n0) = nσ(n0 + 1), nσ((n2 + 1) + n0) = nσ(n0) + 1, stutf(nσ), n2 ≠ 0 → n0 < n1 ∧ ¬ n2 + n0 < n1 ∧ nσ(n0) = nσ(n1) ∧ nσ(n1 + 1) = nσ(n1) + 1, nσ(n2 + n0) = nσ(n0) + 1 ∧ n2 ≠ 0 → n0 ≤ n ∧ n < n2 + n0 ∧ nσ(n0) = nσ(n) ∧ nσ(n + 1) = nσ(n) + 1 ⊦ ∃ m1. n0 ≤ m1 ∧ ¬ n2 + n0 < m1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1

Rule argument

Additional information

Proof step 61 from proof leninv-succ (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 62:

  # nσ = ∞, nσ((n2 + 1) + n0) = nσ(n0) + 1, nσ(n0) = nσ(n0 + 1), n2 ≠ 0, stutf(nσ), n2 ≠ 0 → n0 < n1 ∧ ¬ n2 + n0 < n1 ∧ nσ(n0) = nσ(n1) ∧ nσ(n1 + 1) = nσ(n1) + 1, nσ(n2 + n0) = nσ(n0) + 1 ∧ n2 ≠ 0 → n0 ≤ n ∧ n < n2 + n0 ∧ nσ(n0) = nσ(n) ∧ nσ(n + 1) = nσ(n) + 1 ⊦ ∃ m1. n0 ≤ m1 ∧ ¬ n2 + n0 < m1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1

Rule argument

Additional information

Used simplifier rules: ⊦ n + 0 = 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 formulas: n2 ≠ 0

Proof step 62 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 63:

  # nσ = ∞, nσ(n1 + 1) = nσ(n1) + 1, nσ(n0) = nσ(n1), nσ((n2 + 1) + n0) = nσ(n1) + 1, nσ(n1) = nσ(n0 + 1), n2 ≠ 0, n0 < n1, n1 ≤ n2 + n0, stutf(nσ), nσ(n2 + n0) = nσ(n1) + 1 → n0 ≤ n ∧ n < n2 + n0 ∧ nσ(n1) = nσ(n) ∧ nσ(n + 1) = nσ(n) + 1 ⊦ ∃ m1. n0 ≤ m1 ∧ ¬ n2 + n0 < m1 ∧ nσ(n1) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 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 < m + n ↔ n ≠ 0; (s) ⊦ 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)

Proof step 63 from proof leninv-succ (previous step) (back to tree)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s) ⊦ m ≤ n → (n < m ↔ false); (s) Used terms: n1

Proof step 64 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 65:

  # nσ = ∞, nσ((n2 + 1) + n0) = nσ(n0) + 1, nσ(n0) ≠ nσ(n0 + 1), stutf(nσ), nσ(n2 + n0) = nσ(n0) + 1 ∧ n2 ≠ 0 → n0 ≤ n ∧ n < n2 + n0 ∧ nσ(n0) = nσ(n) ∧ nσ(n + 1) = nσ(n) + 1 ⊦ ∃ m1. n0 ≤ m1 ∧ ¬ n2 + n0 < m1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1

Rule argument

Additional information

Proof step 65 from proof leninv-succ (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 66:

  # nσ = ∞, nσ((n2 + 1) + n0) = nσ(n0) + 1, nσ(n0) ≠ nσ(n0 + 1), stutf(nσ) ⊦ ∃ m1. n0 ≤ m1 ∧ ¬ n2 + n0 < m1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1, nσ(n2 + n0) = nσ(n0) + 1 ∧ n2 ≠ 0

• Proof step 72:

  # nσ = ∞, nσ((n2 + 1) + n0) = nσ(n0) + 1, nσ(n0) = nσ(n), nσ(n + 1) = nσ(n) + 1, nσ(n0) ≠ nσ(n0 + 1), n < n2 + n0, n0 ≤ n, stutf(nσ) ⊦ ∃ m1. n0 ≤ m1 ∧ ¬ n2 + n0 < m1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1

Rule argument

Additional information

Proof step 66 from proof leninv-succ (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 67:

  ⊦ stutf(nσ) ↔ nσ ≠ [] ∧ nσ(0) = 0 ∧ (∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ(n + 1) = nσ(n) ∨ nσ(n + 1) = nσ(n) + 1)

• Proof step 68:

  # nσ = ∞, nσ(0) = 0, nσ((n2 + 1) + n0) = nσ(n0) + 1, nσ(n0) ≠ nσ(n0 + 1), nσ ≠ [], ∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ(n + 1) = nσ(n) ∨ nσ(n + 1) = nσ(n) + 1 ⊦ ∃ m1. n0 ≤ m1 ∧ ¬ n2 + n0 < m1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1, nσ(n2 + n0) = nσ(n0) + 1 ∧ n2 ≠ 0

Rule argument

Additional information

Used terms: nσ

Proof step 67 from proof leninv-succ (previous step) (back to tree)

Lemma stutf-def

Proof step 68 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 69:

  # nσ = ∞, nσ(0) = 0, nσ((n2 + 1) + n0) = nσ(n0) + 1, nσ ≠ [], nσ(n0) ≠ nσ(n0 + 1), ∀ n. nσ(n + 1) = nσ(n) ∨ nσ(n + 1) = nσ(n) + 1 ⊦ ∃ m1. n0 ≤ m1 ∧ ¬ n2 + n0 < m1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1, nσ(n2 + n0) = nσ(n0) + 1 ∧ n2 ≠ 0

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)

Proof step 69 from proof leninv-succ (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 70:

  # nσ = ∞, nσ((n2 + 1) + n0) = nσ(n0) + 1, nσ(0) = 0, nσ ≠ [], nσ(n0) ≠ nσ(n0 + 1), nσ(n0 + 1) = nσ(n0) ∨ nσ(n0 + 1) = nσ(n0) + 1 ⊦ ∃ m1. n0 ≤ m1 ∧ ¬ n2 + n0 < m1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1, nσ(n2 + n0) = nσ(n0) + 1 ∧ n2 ≠ 0

Rule argument

Additional information

Used terms: n0

Proof step 70 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 71:

  # nσ = ∞, nσ(n0 + 1) = nσ(n0) + 1, nσ((n2 + 1) + n0) = nσ(n0) + 1, nσ(0) = 0, nσ ≠ [] ⊦ ∃ m1. n0 ≤ m1 ∧ ¬ n2 + n0 < m1 ∧ nσ(n0) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 1, nσ(n2 + n0) = nσ(n0) + 1 ∧ n2 ≠ 0

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 = m + n ↔ n = 0; (s)

Proof step 71 from proof leninv-succ (previous step) (back to tree)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ 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: n0

Proof step 72 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 73:

  # nσ = ∞, nσ((n2 + 1) + n0) = nσ(n) + 1, nσ(n0) = nσ(n), nσ(n + 1) = nσ(n) + 1, n2 ≠ 0, nσ(n) ≠ nσ(n0 + 1), n < n2 + n0, n0 ≤ n, stutf(nσ) ⊦ ∃ m1. n0 ≤ m1 ∧ ¬ n2 + n0 < m1 ∧ nσ(n) = nσ(m1) ∧ nσ(m1 + 1) = nσ(m1) + 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 < m + n ↔ n ≠ 0; (s) ⊦ m < n → n ≠ 0; (s) fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 73 from proof leninv-succ (previous step) (back to tree)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

Additional information

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

Proof step 74 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 75:

  # nσ = ∞, stutf(nσ), ∀ m0. n0 < m0 → nσ(m0) = nσ(n0) + 1 ⊦ (∃ m. nσ(m) = nσ(n0) ∧ ¬(∀ m0. m < m0 → nσ(m0) = nσ(m))) ∧ (nσ(n) = n1 - 1 ∧ ¬(n < n2 → nσ(n2) = n1 - 1) → ¬ nσ(n0) + 1 < n1)

Rule argument

Additional information

Proof step 75 from proof leninv-succ (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 76:

  # nσ = ∞, stutf(nσ), ∀ m0. n0 < m0 → nσ(m0) = nσ(n0) + 1 ⊦ ∃ m. nσ(m) = nσ(n0) ∧ ¬(∀ m0. m < m0 → nσ(m0) = nσ(m))

• Proof step 81:

  # nσ = ∞, nσ(n) = n1 - 1, nσ(n2) ≠ n1 - 1, n < n2, nσ(n0) + 1 < n1, stutf(nσ), ∀ m0. n0 < m0 → nσ(m0) = nσ(n0) + 1 ⊦

Rule argument

Additional information

Proof step 76 from proof leninv-succ (previous step) (back to tree)

User applied rule exists right

And obtained the following new goals

• Proof step 77:

  # nσ = ∞, stutf(nσ), ∀ m0. n0 < m0 → nσ(m0) = nσ(n0) + 1 ⊦ nσ(n0) = nσ(n0) ∧ ¬(∀ m0. n0 < m0 → nσ(m0) = nσ(n0))

Rule argument

Additional information

Used terms: n0

Proof step 77 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 78:

  # nσ = ∞, stutf(nσ), ∀ m0. n0 < m0 → nσ(m0) = nσ(n0) + 1, ∀ m0. n0 < m0 → nσ(m0) = nσ(n0) ⊦

Rule argument

Additional information

Proof step 78 from proof leninv-succ (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 79:

  # nσ = ∞, stutf(nσ), ∀ m0. n0 < m0 → nσ(m0) = nσ(n0) + 1, n0 < n0 + 1 → nσ(n0 + 1) = nσ(n0) ⊦

Rule argument

Additional information

Used terms: n0 + 1

Proof step 79 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 80:

  # nσ = ∞, nσ(n0 + 1) = nσ(n0), stutf(nσ), ∀ 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 ⊦ ¬ n < n; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s)

Proof step 80 from proof leninv-succ (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) ⊦ ¬ 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 + 1

Proof step 81 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 82:

  # nσ = ∞, nσ(n) = n1 - 1, nσ(n2) ≠ n1 - 1, n < n2, nσ(n0) + 1 < n1, 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

Proof step 82 from proof leninv-succ (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 83:

  # nσ = ∞, n1 = n3 + 1, nσ(n) = n3, n1 ≠ 0, nσ(n2) ≠ n3, n < n2, nσ(n0) + 1 < n1, stutf(nσ), ∀ m0. n0 < m0 → nσ(m0) = nσ(n0) + 1 ⊦

Rule argument

Additional information

Used simplifier rules: n0 < n ⊦ ¬ m + (* n) < (* n0); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n1, n3

Proof step 83 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 84:

  # nσ = ∞, nσ(n2) ≠ nσ(n), n < n2, nσ(n0) < nσ(n), 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 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)

Proof step 84 from proof leninv-succ (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 85:

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

• Proof step 86:

  # nσ = ∞, nσ(n2) ≠ nσ(n), n < n2, nσ(n0) < nσ(n), stutf(nσ), ∀ m0. n0 < m0 → nσ(m0) = nσ(n0) + 1, mon≤(nσ) ∧ n ≤ n0 ∧ ⌜n0⌝ < # nσ → nσ(n) ≤ nσ(n0) ⊦

Rule argument

Additional information

Used terms: nσ, n, n0

Proof step 85 from proof leninv-succ (previous step) (back to tree)

Lemma monle-leq

Proof step 86 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 87:

  # nσ = ∞, nσ(n2) ≠ nσ(n), n < n2, n0 < n2, n0 < n, n0 + 1 < n2, nσ(n0) < nσ(n), stutf(nσ), ∀ m0. n0 < m0 → nσ(m0) = nσ(n0) + 1 ⊦

Rule argument

Additional information

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

Proof step 87 from proof leninv-succ (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 88:

  # nσ = ∞, nσ(n2) ≠ nσ(n), n < n2, n0 + 1 < n2, n0 < n, n0 < n2, nσ(n0) < nσ(n), stutf(nσ), ∀ m0. n0 < m0 → nσ(m0) = nσ(n0) + 1, n0 < n → nσ(n) = nσ(n0) + 1 ⊦

Rule argument

Additional information

Used terms: n

Proof step 88 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 89:

  # nσ = ∞, nσ(n) = nσ(n0) + 1, nσ(n2) ≠ nσ(n), n < n2, n0 + 1 < n2, n0 < n, n0 < n2, nσ(n0) < nσ(n), stutf(nσ), ∀ m0. n0 < m0 → nσ(m0) = nσ(n) ⊦

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ ¬ n < n; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ m ≤ m; (s) lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat

Proof step 89 from proof leninv-succ (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n2

Proof step 90 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 91:

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

Rule argument

Additional information

Proof step 91 from proof leninv-succ (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 92:

  ⊦ n∞ = ∞ ↔ ¬(∃ n. n∞ = ⌜n⌝ )

• Proof step 93:

  stutf(nσ), ∃ n. # nσ = ⌜n⌝ ⊦ (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n) ∧ (∀ n0. (∃ n1. ⌜n1⌝ < # nσ ∧ nσ(n1) = n0 - 1) → ¬ n + 1 < n0) ↔ nσ((# nσ).n - 1) = n

Rule argument

Additional information

Used terms: # nσ

Proof step 92 from proof leninv-succ (previous step) (back to tree)

A spec-lemma

Proof step 93 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 94:

  # nσ = ⌜n0⌝ , stutf(nσ) ⊦ (∃ m. m < n0 ∧ nσ(m) = n) ∧ (∀ n0. (∃ n1. ⌜n1⌝ < # nσ ∧ nσ(n1) = n0 - 1) → ¬ n + 1 < n0) ↔ nσ(n0 - 1) = n

Rule argument

Additional information

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

Proof step 94 from proof leninv-succ (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 95:

  # nσ = ⌜n0⌝ , n0 = n2 + 1, n0 ≠ 0, stutf(nσ) ⊦ (∃ m. m < n0 ∧ nσ(m) = n) ∧ (∀ n0. (∃ n1. ⌜n1⌝ < # nσ ∧ nσ(n1) = n0 - 1) → ¬ n + 1 < n0) ↔ nσ(n2) = n

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: n0, n2

Proof step 95 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 96:

  # nσ = ⌜n2⌝ + ⌜1⌝ , stutf(nσ) ⊦ (∃ m. ¬ n2 < m ∧ nσ(m) = n) ∧ (∀ n0. (∃ n1. ⌜n1⌝ < # nσ ∧ nσ(n1) = n0 - 1) → ¬ n + 1 < n0) ↔ nσ(n2) = n

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ m + 1 ≠ 0; (s) ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)

Proof step 96 from proof leninv-succ (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 97:

  # nσ = ⌜n2⌝ + ⌜1⌝ , nσ(n2) ≠ n, stutf(nσ), ∃ m. ¬ n2 < m ∧ nσ(m) = n, ∀ n0. (∃ n1. ⌜n1⌝ < # nσ ∧ nσ(n1) = n0 - 1) → ¬ n + 1 < n0 ⊦

• Proof step 105:

  # nσ = ⌜n2⌝ + ⌜1⌝ , nσ(n2) = n, stutf(nσ) ⊦ (∃ m. ¬ n2 < m ∧ nσ(m) = n) ∧ (∀ n0. (∃ n1. ⌜n1⌝ < # nσ ∧ nσ(n1) = n0 - 1) → ¬ n + 1 < n0)

Rule argument

Additional information

Proof step 97 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 98:

  # nσ = ⌜n2⌝ + ⌜1⌝ , nσ(n2) ≠ nσ(n0), n0 ≤ n2, stutf(nσ), ∀ n1. (∃ n. ⌜n⌝ < # nσ ∧ nσ(n) = n1 - 1) → ¬ nσ(n0) + 1 < n1 ⊦

Rule argument

Additional information

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

Proof step 98 from proof leninv-succ (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 99:

  # nσ = ⌜n2⌝ + ⌜1⌝ , nσ(n2) ≠ nσ(n0), n0 ≤ n2, stutf(nσ), (∃ n. ⌜n⌝ < # nσ ∧ nσ(n) = nσ(n2) + 1 - 1) → ¬ nσ(n0) + 1 < nσ(n2) + 1, ∀ n1. (∃ n. ⌜n⌝ < # nσ ∧ nσ(n) = n1 - 1) → ¬ nσ(n0) + 1 < n1 ⊦

Rule argument

Additional information

Used terms: nσ(n2) + 1

Proof step 99 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 100:

  # nσ = ⌜n2⌝ + ⌜1⌝ , nσ(n2) ≠ nσ(n0), n0 ≤ n2, stutf(nσ), (∃ n. ⌜n⌝ < # nσ ∧ nσ(n) = nσ(n2)) → ¬ nσ(n0) < nσ(n2), ∀ n1. (∃ n. ⌜n⌝ < # nσ ∧ nσ(n) = n1 - 1) → ¬ nσ(n0) + 1 < n1 ⊦

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) ⊦ n + 0 = n; (s) ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)

Proof step 100 from proof leninv-succ (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 101:

  # nσ = ⌜n2⌝ + ⌜1⌝ , nσ(n2) ≠ nσ(n0), n0 ≤ n2, stutf(nσ), ∀ n1. (∃ n. ⌜n⌝ < # nσ ∧ nσ(n) = n1 - 1) → ¬ nσ(n0) + 1 < n1 ⊦ ∃ n. ⌜n⌝ < # nσ ∧ nσ(n) = nσ(n2)

• Proof step 103:

  # nσ = ⌜n2⌝ + ⌜1⌝ , nσ(n2) ≠ nσ(n0), n0 ≤ n2, stutf(nσ), ¬ nσ(n0) < nσ(n2), ∀ n1. (∃ n. ⌜n⌝ < # nσ ∧ nσ(n) = n1 - 1) → ¬ nσ(n0) + 1 < n1 ⊦

Rule argument

Additional information

Proof step 101 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 102:

  # nσ = ⌜n2⌝ + ⌜1⌝ , nσ(n2) ≠ nσ(n0), n0 ≤ n2, stutf(nσ), ∀ n1. (∃ n. ¬ n2 < n ∧ nσ(n) = n1 - 1) → ¬ nσ(n0) + 1 < n1 ⊦ ∃ n. ¬ n2 < n ∧ nσ(n) = nσ(n2)

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-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 102 from proof leninv-succ (previous step) (back to tree)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s) Used terms: n2

Proof step 103 from proof leninv-succ (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 104:

  ⊦ 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 ↔ ¬ n < m); (ss) 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⌝ < n∞ ↔ n < m + n0); (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 → (m ≤ n ↔ m < n); (s) ⊦ m < n → ¬ n < m; (s) Used terms: nσ, n0, n2

Proof step 104 from proof leninv-succ (previous step) (back to tree)

Lemma monle-leq

Proof step 105 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 106:

  # nσ = ⌜n2⌝ + ⌜1⌝ , stutf(nσ) ⊦ (∃ m. ¬ n2 < m ∧ nσ(m) = nσ(n2)) ∧ (⌜n1⌝ < # nσ ∧ nσ(n1) = n0 - 1 → ¬ nσ(n2) + 1 < n0)

Rule argument

Additional information

Proof step 106 from proof leninv-succ (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 107:

  # nσ = ⌜n2⌝ + ⌜1⌝ , stutf(nσ) ⊦ ∃ m. ¬ n2 < m ∧ nσ(m) = nσ(n2)

• Proof step 108:

  # nσ = ⌜n2⌝ + ⌜1⌝ , nσ(n1) = n0 - 1, ⌜n1⌝ < # nσ, nσ(n2) + 1 < n0, stutf(nσ) ⊦

Rule argument

Additional information

Proof step 107 from proof leninv-succ (previous step) (back to tree)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s) Used terms: n2

Proof step 108 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 109:

  # nσ = ⌜n2⌝ + ⌜1⌝ , nσ(n1) = n0 - 1, ⌜n1⌝ < # nσ, nσ(n2) + 1 < n0, 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 109 from proof leninv-succ (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 110:

  # nσ = ⌜n2⌝ + ⌜1⌝ , n0 = n3 + 1, nσ(n1) = n3, n0 ≠ 0, ⌜n1⌝ < # nσ, nσ(n2) + 1 < n0, stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: n0 < n ⊦ ¬ m + (* n) < (* n0); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n0, n3

Proof step 110 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 111:

  # nσ = ⌜n2⌝ + ⌜1⌝ , ⌜n1⌝ < # nσ, nσ(n2) < nσ(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 < m + n0 ↔ n < n0; (s) ⊦ m + 1 ≠ 0; (s)

Proof step 111 from proof leninv-succ (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 112:

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

• Proof step 113:

  # nσ = ⌜n2⌝ + ⌜1⌝ , ⌜n1⌝ < # nσ, nσ(n2) < nσ(n1), stutf(nσ), mon≤(nσ) ∧ n1 ≤ n2 ∧ ⌜n2⌝ < # nσ → nσ(n1) ≤ nσ(n2) ⊦

Rule argument

Additional information

Used terms: nσ, n1, n2

Proof step 112 from proof leninv-succ (previous step) (back to tree)

Lemma monle-leq

Proof step 113 from proof leninv-succ (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); ⊦ m < n → (n ≤ m ↔ false); (s) 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) ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) 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