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Proof step 4 from proof int2uint-neg (previous step)

j < 0, j < 0 ⊦ bits2uint(sgn(bitsinvert(uint2bits(abs(j))) +1)) = 2 ^ n→i(# sgn(bitsinvert(uint2bits(abs(j))) +1)) + j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
j < 0 ⊦ bits2uint(sgn(bitsinvert(uint2bits(~ j)) +1)) = 2 ^ n→i(# sgn(bitsinvert(uint2bits(~ j)) +1)) + j

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 → abs(i) = ~ i;