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Proof step 4 from proof aligndown-is-smaller (previous step)

n < (n % m = 0 ⊃ n;n / m * m), n0 < (n % m = 0 ⊃ n;n / m * m), n ≤ n0 ⊦ m = 0

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: n ≠ 0 ⊦ ¬ m < m / n * n; (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ ¬ n < n; (s)