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Proof step 4 from proof abs (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
i < 0 ⊦ j = 0, i / abs(j) + abs(i / j) = 0

Rule argument

Additional information

Used simplifier rules: neg-01: j ≠ 0 ⊦ ~ i / j = ~(i / j);%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%abs-07: ⊦ abs(i) = 0 ↔ i = 0; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs