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Proof step 7 from proof alignup-le-is-aligned (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 8:
m % n = 0, m < (m0 % n = 0 ⊃ m0;n * m0 / n + n), m0 < n * m0 / n + n, m0 ≤ m ⊦ m0 % n = 0, n = 0

Rule argument

pos: left 2

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s)