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Proof step 6 from proof mult-a (previous step)

∀ m0, m. m0 * m * n0 = (m0 * m) * n0 ⊦ m0 * m * n0 + m0 * m = (m0 * m) * n0 + m0 * m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2