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Proof step 24 from proof div-div (previous step)

j < 0, (i / ~ j) / abs(j0) + ~ i / (~ j * abs(j0)) = 0, i < 0 ⊦ (i / j) / j0 = i / (j * j0), j0 = 0, j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: Div-right: ⊦ j ≠ 0 → i / ~ j = ~(i / j);%neg-01: j ≠ 0 ⊦ ~ i / j = ~(i / j);%neg: j ≠ 0 ⊦ i / ~ j = ~(i / j);%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%neg-15: ⊦ i + ~ j = i - j; from specification int-abs%neg-02: ⊦ ~ ~ i = i; from specification int-abs%zero-neq: ⊦ i < 0 → (i * j = 0 ↔ j = 0); from specification int-mult%neg-10: ⊦ ~ i = 0 ↔ i = 0; from specification int-abs%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%minus-left: ⊦ ~ i * j = ~(i * j); from specification int-mult%abs-07: ⊦ abs(i) = 0 ↔ i = 0; from specification int-abs