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Proof step 5 from proof is-aligned-align-up-multiple (previous step)

⊦ (n0 % (n * m) = 0 ⊃ n0;(n * n0 / (n * m) + n) * m) % m = 0, m = 0, n = 0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 6:
n0 % (n * m) = 0 ⊦ n0 % m = 0, m = 0, n = 0

Rule argument

pos: right 1

Additional information

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