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Proof step 1 from proof abs-ls-same-01 (conclusion)

⊦ abs(i) * j < i ↔ 0 < i ∧ j ≤ 0 ∨ i < 0 ∧ j < ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ abs(i) * j < i ↔ 0 < i ∧ j ≤ 0 ∨ i < 0 ∧ j < ~1

Rule argument

Additional information