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Proof step 3 from proof mult (previous step)

∀ n0, n. (n0 + m0 * n) / n = n0 / n + m0 ∨ n = 0 ⊦ (n0 + m0 +1 * n) / n = n0 / n + m0 +1, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ n0, n. (n0 + m0 * n) / n = n0 / n + m0 ∨ n = 0 ⊦ (n0 + n * m0) / n = n0 / n + m0, n = 0

Rule argument

Additional information

Used simplifier rules: add: n ≠ 0 ⊦ (n0 + n) / n = n0 / n + 1;% ⊦ m + n = m + n0 ↔ n = n0; (s)%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m * (n + 1) = m * n + m; (s)% ⊦ m +1 = m + 1; (s)