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Proof step 39 from proof inv-get-inverse-last (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 40:
# nσ = ⌜n⌝ , max∞(λ n0. n0 < n ∧ nσ[n0] = nσ[m0]) = ∞, n = m0 + 1, 1 + m + ∞.n < n, stutf(nσ), ∀ n0. n0 < n → nσ[n0] < nσ[m0] ⊦ nσ[∞.n + m + 1] = nσ[m0] + 1, n = 0

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m0]

Additional information

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