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Proof step 14 from proof align-down-add-ls (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
n0 % m + n1 < m, n0 ≤ ((n0 + n1) % m = 0 ⊃ n0 + n1;(n0 + n1) / m * m) ⊦ n0 % m = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → n ≠ 0; (s)