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Proof step 4 from proof aligndown-add-n (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 5:
n + m / n * n ≤ m ⊦ m % n = 0, n = 0

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ m + m0 ≤ m ↔ m0 = 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic