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Proof step 4 from proof aligndown-add-not-smaller (previous step)

((n0 + n1) % m = 0 ⊃ n0 + n1;(n0 + n1) / m * m) < (n0 % m = 0 ⊃ n0;n0 / m * m) ⊦ m = 0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 5:
(n0 + n1) % m = 0, n0 + n1 < (n0 % m = 0 ⊃ n0;n0 / m * m) ⊦ m = 0
Proof step 8:
(n0 + n1) / m * m < (n0 % m = 0 ⊃ n0;n0 / m * m) ⊦ (n0 + n1) % m = 0, m = 0

Rule argument

pos: left 1

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