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Proof step 2 from proof div-add (previous step)

m0 % n = 0 ⊦ (m + m0) / n = m / n + m0 / n, 0 = n

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: nat-div inst: name: modex seq: ⊦ n ≠ 0 → (m % n = 0 ↔ (∃ n0. m = n0 * n)) subst: [n, m] → [n, m0] flags: rewrite, current

Additional information

Used simplifier rules: mult: n ≠ 0 ⊦ (n0 + m * n) / n = n0 / n + m;%div-mult: ⊦ m ≠ 0 → (m * m0) / m = m0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2 Used terms: n, m0