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Proof step 6 from proof div-zero-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
i + j < 0, j < 0, 0 ≤ i ⊦ i / j = 0

Rule argument

Additional information

Used simplifier rules: ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1