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Proof step 19 from proof div-eq-zero-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: numlelsdel: j0 ≠ 0, j ≤ j0 ⊦ i < (* j) → i < (* j0); from specification int-abs%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%neg-07: ⊦ i < ~ j ↔ i + j < 0; from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j; 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%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs