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Proof step 1 from proof absls-02 (conclusion)

⊦ i ≤ 0 ∧ j ≤ 0 → (abs(i) < abs(j) ↔ j < i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ abs(i) < abs(j) ↔ j < i, 0 < i, 0 < j

Rule argument

Additional information

Used simplifier rules: le-def: ⊦ i ≤ j ↔ ¬ j < i;