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Proof step 4 from proof aligndown-remainder (previous step)

⊦ m = m0 + (m % n = 0 ⊃ m;m / n * n) ↔ m0 = m % n, n = 0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 5:
m = m0 + (m % n = 0 ⊃ m;m / n * n) ⊦ m0 = m % n, n = 0
Proof step 11:
m0 = m % n ⊦ m = m0 + (m % n = 0 ⊃ m;m / n * n), n = 0

Rule argument

pos: right 1

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