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Proof step 15 from proof eq1 (previous step)

(m + n0) / m * m = n0 + 1, n0 / m * m ≤ n0 ⊦ (n0 - n0 / m * m) + 1 = m, m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
(m + n0) / m * m = n0 + 1 ⊦ n0 + 1 = n0 / m * m + m, m = 0

Rule argument

Additional information

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