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Proof step 26 from proof div-eq-zero-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: div-zero-05: ⊦ i < j ∧ 0 < j ∧ 0 ≤ i → i / j = 0;%div-eq-zero: ⊦ 0 < i ∧ 0 < j → (i / j = 0 ↔ i < j);%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; 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%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%neg-07: ⊦ i < ~ j ↔ i + j < 0; 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%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-le: ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward) from specification int-abs%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%Abs-pos: ⊦ ¬ j < 0 → abs(j) = j; from specification int-abs