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Proof step 10 from proof int2uint-neg (previous step)

∃ bits. uint2bits(~ j) = false ' + true ' + bits, j < 0 ⊦ bits2uint(sgn(bitsinvert(uint2bits(~ j)) +1)) = 2 ^ n→i(# sgn(bitsinvert(uint2bits(~ j)) +1)) + j, true ∈ uint2bits(~ j).rest.rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
uint2bits(~ j) = false ' + true ' + bits, false ∈ uint2bits(~ j), j < 0 ⊦ bits2uint(sgn(bitsinvert(uint2bits(~ j)) +1)) = 2 ^ n→i(# sgn(bitsinvert(uint2bits(~ j)) +1)) + j, true ∈ uint2bits(~ j).rest.rest

Rule argument

Additional information

Used simplifier rules: uint2bits-onein-01: ⊦ j < 0 → true ∈ uint2bits(~ j);% ⊦ bits = bits0 + boolvar ' + bits1 → boolvar ∈ bits; (forward)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)