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Proof step 10 from proof mininv-inverts-inner (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
# nσ = ⌜min(nσ, nσ[n])⌝ , min(nσ, nσ[n]) = m + 1, nσ[m] < nσ[n], n < min(nσ, nσ[n]), mon<(nσ) ⊦ min(nσ, nσ[n]) = n, min(nσ, nσ[n]) = 0, nσ = []

Rule argument

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

Additional information

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