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Proof step 16 from proof le-div-to-mult (previous step)

(n0 + m2) * (m1 + m0) + m2 ≤ m1 * (n0 + m2) ⊦ m2 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 17:
⊦ m0 * (m + n) = m0 * m + m0 * n

Rule argument

spec: nat-mult name: dist seq: ⊦ m0 * (m + n) = m0 * m + m0 * n subst: [n, m, m0] → [m1, m0, n0 + m2] flags: <all paths>

Additional information

Used simplifier rules: mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ m + m0 ≤ m ↔ m0 = 0; (s)% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: m1, m0, n0 + m2