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Proof step 1 from proof count-i2j-empty-02 (conclusion)

⊦ count-i2j(i, j) = [] ↔ j < i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
count-i2j(i, j) = [], i ≤ j ⊦

Rule argument

Additional information

Used simplifier rules: counti2j-empty: ⊦ j < i → count-i2j(i, j) = [];%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs