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Proof step 15 from proof align-down-add-ls (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 16:
(n0 + n1) % m = 0, n0 % m + n1 < m, n0 ≤ n0 + n1 ⊦ n0 % m = 0
Proof step 20:
n0 % m + n1 < m, n0 ≤ (n0 + n1) / m * m ⊦ n0 % m = 0, (n0 + n1) % m = 0

Rule argument

pos: left 2

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