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Proof step 93 from proof inv-get-inverse-succ (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 94:
n1 = m + 1, # nσ = ⌜n1⌝ , stutf(nσ), n ≤ nσ[m], 1 + max∞(λ n0. n0 < n1 ∧ nσ[n0] = n).n < n1 ⊦ nσ[max∞(λ n0. n0 < n1 ∧ nσ[n0] = n).n + 1] = n + 1, n1 = 0

Rule argument

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

Additional information

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