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Proof step 6 from proof alignup-le (previous step)

m0 % n = 0, m0 < (m1 % n = 0 ⊃ m1;n * m1 / n + n), m1 < (m1 % n = 0 ⊃ m1;n * m1 / n + n), m1 ≤ m0 ⊦ n = 0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 7:
m0 % n = 0, m1 < (m1 % n = 0 ⊃ m1;n * m1 / n + n), m0 < n * m1 / n + n, m1 ≤ m0 ⊦ m1 % n = 0, n = 0

Rule argument

pos: left 2

Additional information

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