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Proof step 5 from proof alignup-upper-bound (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

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

Rule argument

pos: left 2

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s)