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Proof step 4 from proof div-nearer-zero (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
i < i / abs(j) * abs(j), 0 ≤ i, 0 < i / abs(j) * abs(j) ⊦ j = 0

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) 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%Abs-pos: ⊦ ¬ j < 0 → abs(j) = j; from specification int-abs