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Proof step 5 from proof alignup-le-is-aligned (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 6:
m % n = 0, m < (m0 % n = 0 ⊃ m0;n * m0 / n + n) ⊦ n = 0, m < m0
Proof step 18:
m % n = 0, m < m0 ⊦ n = 0, m < (m0 % n = 0 ⊃ m0;n * m0 / n + n)

Rule argument

pos: right 1

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