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Proof step 10 from proof mininv-stays-if-not-taken (previous step)

# 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 = ∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
nσ[min(nσ, n + 1) - 1] = n, # nσ = ⌜min(nσ, n + 1)⌝ , n ≤ nσ[min(nσ, n)], mon<(nσ), min(nσ, n) < min(nσ, n + 1), ∀ m. m < min(nσ, nσ[m] + 1) → nσ[m] ≠ n, ∀ m. m < min(nσ, n + 1) ∧ n ≤ nσ[m] → min(nσ, n) ≤ m ⊦ nσ = [], min(nσ, n + 1) = min(nσ, n), min(nσ, n + 1) = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf%disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic