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Proof step 10 from proof int2bits-truein (previous step)

true ∈ uint2bits(~ j) ↔ (∃ bits0, bits1. bitsinvert(uint2bits(~ j)) +1 = bitsinvert(bits0) + true ' + bits1 ∧ ¬ true ∈ bits1 ∧ uint2bits(~ j) = bits0 + true ' + bits1), j < 0 ⊦ j = 0, true ∈ sgn(bitsinvert(uint2bits(~ j)) +1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
uint2bits(~ j) = bits + true ' + bits0, bitsinvert(uint2bits(~ j)) +1 = bitsinvert(bits) + true ' + bits0, j < 0 ⊦ j = 0, true ∈ bits0, true ∈ sgn(bitsinvert(bits) + true ' + bits0)

Rule argument

Additional information

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