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Proof step 1 from proof le-abs (conclusion)

⊦ i * abs(j) ≤ i0 * abs(j) ↔ ¬ ¬(j = 0 ∨ i ≤ i0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ i * abs(j) ≤ i0 * abs(j) ↔ j = 0 ∨ i ≤ i0

Rule argument

Additional information