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Proof step 6 from proof two-neq-02 (previous step)

m = (2 * (m + n) + 2) + n0, n0 < 2 ⊦

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] → [n, m, 2] flags: <all paths>

Additional information

Used simplifier rules: m1 ≠ 0 ⊦ m = (* m1) * m + n0 ↔ 0 = (m1 - 1) * m + n0; (s)% ⊦ m0 * 1 = m0; (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)%n < n0 ⊦ (* n) ≠ m0 + (* n0); (s)%c: ⊦ m + n = n + m; from specification nat-basic Used terms: n, m, 2