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Proof step 6 from proof aligndown-aligndown-add (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 7:
((n0 % n = 0 ⊃ n0;n0 / n * n) + n1) % n = 0 ⊦ n1 % n = 0, n = 0, n1 = n1 / n * n

Rule argument

pos: right 2

Additional information