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Proof step 2 from proof sr (previous step)

⊦ sgn(i) = ~1, sgn(i) = 1

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: sgn-02: ⊦ sgn(i) = ~1 ↔ i < 0;%sgn+1: ⊦ 0 ≤ j → sgn(j) = 1;% ⊦ i < j ↔ ¬ j ≤ i; (ss)