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

uint2bits(j) ≠ [] ∧ (uint2bits(j).first ↔ false), 0 ≤ j ⊦ sgn(uint2bits(j)) = uint2bits(j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 ≤ j ⊦ sgn(uint2bits(j)) = uint2bits(j)

Rule argument

Additional information

Used simplifier rules: uint2bits-first: ⊦ 0 ≤ i → ¬ uint2bits(i).first;%uint2bits-empty: 0 ≤ i ⊦ uint2bits(i) ≠ [];% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);