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Proof step 4 from proof uint2bits-sgn (previous step)

0 ≤ i → ¬ uint2bits(i).first, 0 < i ⊦ sgn(uint2bits(i)) = uint2bits(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 < i ⊦ sgn(uint2bits(i)) = uint2bits(i), uint2bits(i).first

Rule argument

Additional information

Used simplifier rules: ⊦ i < j → ¬ j < i;% ⊦ ¬ i ≤ j ↔ j < i;