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Proof step 1 from proof div-zero-ngr (conclusion)

⊦ i < 0 ∧ 0 < j → ¬ 0 < i / j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
i < 0, 0 < i / j, i + 1 < j, i + 1 < i / j, i < j, 1 < j, 0 < j ⊦

Rule argument

Additional information

Used simplifier rules: div-ls-same: ⊦ 0 < j → (i < i / j ↔ i < 0 ∧ 1 < j);%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs