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Proof step 19 from proof alignup-le-is-aligned (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 20:
m % n = 0, m < m0, n * m0 / n + n < m0, (m0 % n = 0 ⊃ m0;n * m0 / n + n) ≤ m ⊦ m0 % n = 0, n = 0

Rule argument

pos: left 2

Additional information

Used simplifier rules: ⊦ m < n → (n ≤ m ↔ false); (s)