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Proof step 8 from proof aligndown-aligndown-add (previous step)

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

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: n ≠ 0 ⊦ (m + n0 * n) % n = m % n; (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% ⊦ m0 % n = 0 ∧ 0 ≠ n → (m + m0) % n = m % n; (s)