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Proof step 12 from proof div-eq-zero-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - 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%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%sub-le: ⊦ i - j ≤ i0 ↔ i ≤ j + i0; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%neg-15: ⊦ i + ~ j = i - j; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs