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Proof step 9 from proof mininv-stays-if-not-taken (previous step)

nσ[min(nσ, n + 1) - 1] = n, min(nσ, n + 1) = (# nσ).n, n ≤ nσ[min(nσ, n)], ⌜min(nσ, n)⌝ < # nσ, mon<(nσ), ∀ m. ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → min(nσ, n) ≤ m, ∀ m. ⌜m⌝ < # nσ → nσ[m] ≠ n ⊦ min(nσ, n + 1) = min(nσ, n), nσ = [], # nσ = ∞

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
# nσ = ⌜n0⌝ , nσ[min(nσ, n + 1) - 1] = n, min(nσ, n + 1) = n0, n ≤ nσ[min(nσ, n)], ⌜min(nσ, n)⌝ < # nσ, mon<(nσ), ∀ m. ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → min(nσ, n) ≤ m, ∀ m. ⌜m⌝ < # nσ → nσ[m] ≠ n ⊦ min(nσ, n + 1) = min(nσ, n), nσ = [], # nσ = ∞

Rule argument

spec: nat-inf-basic inst: name: elim seq: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ) subst: [n∞, m] → [# nσ, m0] flags:

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: # nσ, m0