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Proof step 2 from proof mod-eq-01 (previous step)

m0 < n ⊦ (m + m0) % n = m0 ↔ m % n = 0, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m0 < n ⊦ (m + m0) % n = m0 ↔ m % n = 0

Rule argument

Additional information

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