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Proof step 2 from proof mod-mult-02 (previous step)

⊦ (i0 * i) % (i0 * j) = i % j * i0, i0 = 0, 0 = j

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-div inst: name: mod-rew seq: j ≠ 0 ⊦ i % j = i - i / j * j subst: [j, i] → [i0 * j, i0 * i] flags: rewrite, current

Additional information

Used simplifier rules: div-mult-01: j ≠ 0, i0 ≠ 0 ⊦ (i * i0) / (j * i0) = i / j;%divmod-rew: j ≠ 0 ⊦ i / j * j + i % j = i;%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%dist-sub-merge: ⊦ i * j - i * j0 = i * (j - j0); from specification int-mult%eq-nonzero: ⊦ i ≠ 0 → (i * j = i * j0 ↔ j = j0); from specification int-mult%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%zero-eq: ⊦ i * j = 0 ↔ i = 0 ∨ j = 0; from specification int-mult Used terms: i0 * j, i0 * i