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Proof step 34 from proof inv-inverts (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 35:
n1 = m0 + 1, nσ[n + 1] = nσ[n] + 1, # nσ = ⌜n1⌝ , max∞(λ n0. n0 < n1 ∧ nσ[n0] = nσ[n]) = ⌜n⌝ , stutf(nσ), 1 + n < n1 ⊦ ((λ m. max∞(λ n0. n0 < n1 ∧ nσ[n0] = m).n) to nσ[m0] + 1)[nσ[n]] = n, n1 = 0

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n1, m0] flags:

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n1, m0