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Proof step 9 from proof mininv-inverts-inner (previous step)

# nσ = ⌜min(nσ, nσ[n])⌝ , nσ[min(nσ, nσ[n]) - 1] < nσ[n], mon<(nσ), n < min(nσ, nσ[n]) ⊦ nσ[n] = 0, nσ = [], min(nσ, nσ[n]) = n, min(nσ, nσ[n]) = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
min(nσ, nσ[n]) = m + 1, # nσ = ⌜min(nσ, nσ[n])⌝ , nσ[m] < nσ[n], mon<(nσ), n < min(nσ, nσ[n]) ⊦ nσ[n] = 0, nσ = [], min(nσ, nσ[n]) = n, min(nσ, nσ[n]) = 0

Rule argument

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

Additional information

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