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Proof step 10 from proof aligndown-aligndown-add (previous step)

⊦ ((n0 % n = 0 ⊃ n0;n0 / n * n) + n1) % n = 0, ((n0 % n = 0 ⊃ n0;n0 / n * n) + n1) / n * n = (n0 % n = 0 ⊃ n0;n0 / n * n) + (n1 % n = 0 ⊃ n1;n1 / n * n), n = 0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 11:
n0 % n = 0 ⊦ (n0 + n1) % n = 0, ((n0 % n = 0 ⊃ n0;n0 / n * n) + n1) / n * n = (n0 % n = 0 ⊃ n0;n0 / n * n) + (n1 % n = 0 ⊃ n1;n1 / n * n), n = 0

Rule argument

pos: right 1

Additional information

Used simplifier rules: n ≠ 0 ⊦ (n0 + m * n) / n = n0 / n + m; (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%n ≠ 0 ⊦ (m + n0 * n) % n = m % n; (s)% ⊦ m * n + m0 * n = m1 * n ↔ m + m0 = m1 ∨ n = 0; (s)