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Proof step 16 from proof div-div (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
(i / abs(j)) / abs(j0) + ~ i / (abs(j) * abs(j0)) = 0, i < 0 ⊦ (i / j) / j0 = i / (j * j0), j0 = 0, j = 0

Rule argument

Additional information

Used simplifier rules: neg-01: j ≠ 0 ⊦ ~ i / j = ~(i / j);%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%abs-07: ⊦ abs(i) = 0 ↔ i = 0; from specification int-abs