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Proof step 16 from proof aligndown-is-n (previous step)

n0 + n1 = n2 + (n0 + n1) / n * n, n0 + n1 < 2 * n, n ≤ n0 + n1 ⊦ (n0 + n1) / n = 1, n = 0, n2 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: nat-div name: rec seq: n ≠ 0 ⊦ n ≤ m → m / n = (m - n) / n + 1 subst: [n, m] → [n, n0 + n1] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m * (n + 1) = m * n + m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1; (ws)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%m1 ≠ 0 ⊦ m + n < (* m1) * m ↔ n < (m1 - 1) * m; (s)% ⊦ m0 * 1 = m0; (s)% ⊦ m * n + n0 < n ↔ m = 0 ∧ n0 < n; (ws) Used terms: n, n0 + n1