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Proof step 3 from proof mod-mult (previous step)

∀ m, n. (m + n1 * n) % n = m % n ∨ n = 0 ⊦ (m + n1 +1 * n) % n = m % n, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ m, n. (m + n1 * n) % n = m % n ∨ n = 0 ⊦ n = 0, (m + n * n1) % n = m % n

Rule argument

Additional information

Used simplifier rules: mod-once: n ≠ 0 ⊦ (m + n) % n = m % n;%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%rsucc-02: ⊦ m * (n + 1) = m * n + m; from specification nat-mult%succ: ⊦ m +1 = m + 1; from specification nat-basic2