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Proof step 2 from proof absls-02 (
previous step
)
i ≤ 0, j ≤ 0 ⊦ abs(i) < abs(j) ↔ j < i
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 3
:
⊦ abs(i) < abs(j) ↔ j < i, 0 < j, 0 < i
Rule argument
Additional information
Used simplifier rules: le-def: ⊦ i ≤ j ↔ ¬ j < i;