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Proof step 53 from proof inv-ls-succ (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 54:
nσ[m] = m2, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n + 1) = ∞, n1 = m2 + 1, ⌜m⌝ < # nσ, 1 + n < n1, stutf(nσ), ∞.n ≤ ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to n1)[n], ∀ m. (∃ m0. ⌜m0⌝ < # nσ ∧ nσ[m0] = m - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m1. m0 < m1 → nσ[m1] = m - 1))) → ¬ n1 < m, # nσ ≠ ∞ ∨ ¬(m < m0 → nσ[m0] = m2) ⊦ n1 = 0

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n1, m3]

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, m3