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Proof step 13 from proof div-nearer-zero (previous step)

i < 0, 0 - i < (0 - i) / abs(j) * abs(j) ⊦ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
i / abs(j) * abs(j) < 0, i < 0, i / abs(j) * abs(j) < i, i / abs(j) * abs(j) < ~1 ⊦ j = 0

Rule argument

Additional information

Used simplifier rules: neg-01: j ≠ 0 ⊦ ~ i / j = ~(i / j);%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%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (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%neg-01: ⊦ ~ j < ~ i ↔ i < j; from specification int-abs%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%minus-left: ⊦ ~ i * j = ~(i * j); from specification int-mult%abs-07: ⊦ abs(i) = 0 ↔ i = 0; from specification int-abs%neg-06: ⊦ 0 - i = ~ i; from specification int-abs