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Proof step 3 from proof div2-notzero (previous step)

j = i0 + 1, i0 / 2 = 0, j < 0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: div-ls-same: ⊦ 0 < j → (i < i / j ↔ i < 0 ∧ 1 < j);%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%numlsf: ⊦ i < (* j) ∧ j < 0 → i < 0; (forward) from specification int-basic1%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs