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Proof step 4 from proof align-down-add-ls (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 5:
((n0 + n1) % m = 0 ⊃ n0 + n1;(n0 + n1) / m * m) < n0 ⊦ m = 0, n0 % m + n1 < m ∧ n0 % m ≠ 0
Proof step 14:
n0 % m + n1 < m ⊦ n0 % m = 0, m = 0, ((n0 + n1) % m = 0 ⊃ n0 + n1;(n0 + n1) / m * m) < n0

Rule argument

pos: right 1

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