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Proof step 1 from proof mod-eq-01 (conclusion)

⊦ m0 < n ∧ n ≠ 0 → ((m + m0) % n = m0 ↔ m % n = 0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
(m + m0) % n = m0, m0 < n ⊦ m % n = 0, n = 0

Rule argument

Additional information

Used simplifier rules: mod-add: ⊦ m0 % n = 0 ∧ 0 ≠ n → (m + m0) % n = m % n;%mod-eq: ⊦ m < n → m % n = m;%mod-ls: n ≠ 0 ⊦ m % n < n;%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2