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Proof step 5 from proof align-down-add-eq (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 6:
(n0 + n1) / m * m = n0 ⊦ (n0 + n1) % m = 0, m = 0, n1 < m ∧ n0 % m = 0

Rule argument

pos: left 1

Additional information

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