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Proof step 4 from proof mod-eq-01 (previous step)

m = m1 * n + n0 ∧ n0 < n, (m + m0) % n = m0, m0 < n ⊦ n0 = 0, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
((m1 * n + n0) + m0) % n = m0, m0 < n, n0 < n ⊦ n = 0, n0 = 0

Rule argument

Additional information

Used simplifier rules: mod-ls: n ≠ 0 ⊦ m % n < n;%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1