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Proof step 6 from proof align-down-add-ls (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 7:
(n0 + n1) / m * m < n0 ⊦ (n0 + n1) % m = 0, m = 0, n0 % m + n1 < m ∧ n0 % m ≠ 0

Rule argument

pos: left 1

Additional information

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