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Proof step 12 from proof min-stutters (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 ⊦ min(nσ, n + 1) = 1 + min(nσ, n), min(nσ, n + 1) = min(nσ, n), nσ = [], # nσ = ∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
# nσ = ⌜min(nσ, n + 1)⌝ , nσ[min(nσ, n + 1) - 1] = n, min(nσ, n) + 1 < min(nσ, n + 1), min(nσ, n) < min(nσ, n + 1), mon<(nσ), n ≤ nσ[min(nσ, n)], ∀ m. m < min(nσ, n + 1) ∧ n ≤ nσ[m] → min(nσ, n) ≤ m ⊦ min(nσ, n + 1) = 1 + min(nσ, n), min(nσ, n + 1) = min(nσ, n), nσ = []

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; from specification nat-basic1%succ-03: ⊦ m < n ∧ m + 1 ≠ n → m + 1 < n; (forward) from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%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