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Proof step 5 from proof alignup-add-bounded1 (previous step)

n % m = 0, n ≤ m ⊦ m = 0, n = (n = 0 ⊃ 0;m)

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: right 2

Additional information

Used simplifier rules: ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)% ⊦ m < n → m % n = m; (s)