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Proof step 4 from proof zero-05 (previous step)

i < 0 ∧ j < 0 → 0 ≤ i / j, i / j < 0, i < 0, i / j < i, i / j < ~1 ⊦ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
i < j, i / j < j, i / j < ~1, i / j < i, i < 0, i / j < 0, 0 < j, i + 1 < j ⊦

Rule argument

Additional information

Used simplifier rules: lst-04: ⊦ i + j < i0 ∧ j0 < j → i + j0 < i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%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%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs