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Proof step 7 from proof abs (previous step)

¬ ¬ j < 0, i < 0 ⊦ j = 0, i / abs(j) + abs(i / j) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
i < 0, j < 0, i / j < 0 ⊦ j = 0

Rule argument

Additional information

Used simplifier rules: Div-right: ⊦ j ≠ 0 → i / ~ j = ~(i / j);%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%abs-18: ⊦ abs(i) = i ↔ 0 ≤ i; 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%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs