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Proof step 2 from proof div-ngr-zero (previous step)

i < 0, 0 < j ⊦ 0 ≤ i / j ↔ abs(i) < j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
0 < j, i < 0 ⊦ 0 ≤ i / j ↔ 0 < j + i

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ ~ i < j ↔ 0 < j + i; (s)% ⊦ i < 0 → abs(i) = ~ i; (s)