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Proof step 8 from proof div-mult-01 (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: int-div inst: name: div-mult-02 seq: 0 < j, 0 < i0, 0 ≤ i ⊦ (i * i0) / (j * i0) = i / j subst: [j, i0, i] → [j, ~ i0, ~ i] flags: current, []

Additional information

Used simplifier rules: neg: j ≠ 0 ⊦ i / ~ j = ~(i / j);%neg-01: j ≠ 0 ⊦ ~ i / j = ~(i / j);%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-02: ⊦ ~ ~ i = i; from specification int-abs%zero-neq-01: ⊦ 0 < i → (i * j = 0 ↔ j = 0); from specification int-mult%zero-left: ⊦ 0 * i = 0; from specification int-mult%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%neg-12: ⊦ ~ i = ~ j ↔ i = j; from specification int-abs%neg: ⊦ 0 < ~ i ↔ i < 0; 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 Used terms: j, ~ i0, ~ i