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Proof step 7 from proof mininv-inverts-inner (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
# nσ = ⌜n0⌝ , min(nσ, nσ[n]) = n0, nσ[min(nσ, nσ[n]) - 1] < nσ[n], ⌜n⌝ < # nσ, mon<(nσ) ⊦ min(nσ, nσ[n]) = n, nσ = [], # nσ = ∞, nσ[n] = 0

Rule argument

spec: nat-inf-basic inst: name: elim seq: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ) subst: [n∞, m] → [# nσ, m] flags:

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: # nσ, m