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Proof step 29 from proof min-stutters (previous step)

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

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 1 flags: rotate, discard

Additional information

Used simplifier rules: ⊦ ⌜m⌝ ≠ ∞; (s)%n∞ ≠ ∞ ⊦ (⌜m⌝ + n∞).n = m + n∞.n; (s)% ⊦ ⌜m⌝ .n = m; (s)%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)% ⊦ 0 + m = m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic