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Proof step 12 from proof aligndown-aligndown-add (previous step)

n0 % n = 0 ⊦ n1 % n = 0, (n1 / n + n0 / n) * n = n0 + n1 / n * n, n = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ m0 * (m + n) = m0 * m + m0 * n

Rule argument

spec: nat-mult name: dist seq: ⊦ m0 * (m + n) = m0 * m + m0 * n subst: [m0, n, m] → [n, n0 / n, n1 / n] flags: <all paths>

Additional information

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