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Proof step 1 from proof absle-01 (conclusion)

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

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: le-def: ⊦ i ≤ j ↔ ¬ j < i;%absls-02: ⊦ i ≤ 0 ∧ j ≤ 0 → (abs(i) < abs(j) ↔ j < i);