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Proof step 4 from proof aligndown-is-n (previous step)

n0 + n1 < 2 * n, n ≤ n0 + n1 ⊦ ((n0 + n1) % n = 0 ⊃ n0 + n1;(n0 + n1) / n * n) = n, n = 0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 5:
(n0 + n1) % n = 0, n0 + n1 < 2 * n, n ≤ n0 + n1 ⊦ n0 + n1 = n, n = 0
Proof step 10:
n0 + n1 < 2 * n, n ≤ n0 + n1 ⊦ (n0 + n1) % n = 0, (n0 + n1) / n * n = n, n = 0

Rule argument

pos: right 1

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