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Proof step 31 from proof mod-by6 (previous step)

3 * n0 * n0 + (n0 * 2) * n0 * n0 + n0 = n * 6 ⊦ (9 * n0 * n0 + n0 * 13 + 2 * n0 * n0 * n0) % 6 = 0, 12 * n0 + 6 * n0 * n0 = 6 * (n0 * n0 + 2 * n0)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: nat-mult name: dist seq: ⊦ m0 * (m + n) = m0 * m + m0 * n subst: [n, m, m0] → [2 * n0, n0 * n0, 6] flags: <all paths>

Additional information

Used simplifier rules: mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ m * n + m0 * n = m1 * n + m2 * n ↔ m + m0 = m1 + m2 ∨ n = 0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic Used terms: 2 * n0, n0 * n0, 6