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Proof step 14 from proof modmod-mult-02 (previous step)

i0 < j, i0 < 0, 0 ≤ i, 0 < j, i0 + 1 < j ⊦ (i % j * i0 % j) % j = (i * i0) % j

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: int-div inst: name: modmod-mult seq: 0 < j ⊦ 0 ≤ i ∧ 0 ≤ i0 → (i % j * i0 % j) % j = (i * i0) % j subst: [j, i, i0] → [j, i, ~ i0] flags: current, []

Additional information

Used simplifier rules: mod-neg-01: j ≠ 0 ⊦ ~ i % j = ~(i % j);%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%negle-07: ⊦ i ≤ ~ j ↔ i + j ≤ 0; from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%zero-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%minus-left: ⊦ ~ i * j = ~(i * j); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%neg-12: ⊦ ~ i = ~ j ↔ i = j; from specification int-abs Used terms: j, i, ~ i0