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Proof step 15 from proof div-eq-zero-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
1 < abs(j), i < 0, 0 < abs(j) + i ⊦ i / j = 0, 0 = j

Rule argument

Additional information

Used simplifier rules: forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%abs-06: ⊦ 0 < abs(i) ↔ i ≠ 0; from specification int-abs%ls-02: ⊦ i < i + j ↔ 0 < j; 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%ls-01: ⊦ i + j < i + i0 ↔ j < i0; from specification int-basic2%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (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