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Proof step 14 from proof aligndown-is-n (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 15:
n0 + n1 = n2 + (n0 + n1) / n * n, n0 + n1 < 2 * n, n ≤ n0 + n1, (n0 + n1) / n * n ≤ n0 + n1 ⊦ (n0 + n1) / n = 1, n = 0, n2 = 0

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [n0 + n1, (n0 + n1) / n * n, n2]

Additional information

Used simplifier rules: n ≠ 0 ⊦ m / n * n ≤ m; (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n0 + n1, (n0 + n1) / n * n, n2