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Proof step 10 from proof div-nearer-zero (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
i / j < 0, j < 0, 0 ≤ i, i < j * i / j ⊦ j = 0

Rule argument

Additional information

Used simplifier rules: Div-right: ⊦ j ≠ 0 → i / ~ j = ~(i / j);%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%zero-gr-01: ⊦ i < 0 → (0 < i * j ↔ j < 0); from specification int-mult%neg-02: ⊦ ~ ~ i = i; 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%neg-06: ⊦ 0 - i = ~ i; from specification int-abs