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Proof step 27 from proof div-eq-zero-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
0 ≤ i ⊦ i / j = 0 ↔ i < abs(j), 0 = j, 0 < i ∧ 0 < j

Rule argument

Additional information

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