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Proof step 5 from proof aligndown-is-aligned-add (previous step)

n0 % n = 0 ⊦ (n1 % n = 0 ⊃ n0 + n1;(n1 / n + n0 / n) * n) = n0 + (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 6:
n0 % n = 0 ⊦ n1 % n = 0, (n1 / n + n0 / n) * n = n0 + (n1 % n = 0 ⊃ n1;n1 / n * n), n = 0

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic