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Proof step 10 from proof aligndown-is-n (previous step)

n0 + n1 < 2 * n, n ≤ n0 + n1 ⊦ (n0 + n1) % n = 0, (n0 + n1) / n * n = n, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
n0 + n1 < 2 * n, n ≤ n0 + n1 ⊦ (n0 + n1) % n = 0, (n0 + n1) / n = 1, n = 0

Rule argument

Additional information

Used simplifier rules: mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ m * n = m ↔ m = 0 ∨ n = 1; (s)