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Proof step 2 from proof sgn-03 (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-01: ⊦ sgn(i) = 1 ↔ 0 ≤ i;%Sign-neg: ⊦ j < 0 → sgn(j) = 0 -1;% ⊦ i ≤ j ↔ ¬ j < i; (ss)