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Proof step 8 from proof alignup-add-bounded1 (previous step)

n ≤ m ⊦ n % m = 0, m * n / m + m = m, m = 0, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
n ≤ m ⊦ n % m = 0, n / m = 0, m = 0, n = 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%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0); (s)% ⊦ m = m + n ↔ n = 0; (s)