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Proof step 6 from proof div-eq-zero-01 (previous step)

i < 0 ⊦ i / j = 0 ↔ abs(i) < abs(j), 0 = j, 0 < i ∧ 0 < j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
i < 0 ⊦ 0 = j, i / j = 0 ↔ 0 < abs(j) + i

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%neg-13: ⊦ ~ i < j ↔ 0 < j + i; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs