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Proof step 2 from proof le-abs-numerator (previous step)

⊦ j = 0, ~ abs(i) ≤ i / j ∧ i / j ≤ abs(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ j = 0, 0 ≤ abs(i) + i / j ∧ i / j ≤ abs(i)

Rule argument

Additional information

Used simplifier rules: ⊦ ~ i ≤ j ↔ 0 ≤ i + j; (s)