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Proof step 94 from proof inv-get-inverse-succ (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 95:
# nσ = ⌜n1⌝ , n1 = m + 1, 1 + max∞(λ n0. n0 < n1 ∧ nσ[n0] = n).n < n1, stutf(nσ), n ≤ nσ[m] ⊦ nσ[max∞(λ n0. n0 < n1 ∧ nσ[n0] = n).n + 1] = n + 1, n1 = 0

Rule argument

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

Additional information

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