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Proof step 3 from proof bits2int-sign (previous step)

bits = boolvar + bits1 ⊦ bits = [], bits2int(bits0 + bits) = bits2int(bits), (¬ boolvar) ∈ bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ bits2int(bits0 + boolvar ' + bits1) = bits2int(boolvar ' + bits1), (¬ boolvar) ∈ bits0

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits ≠ [];% ⊦ boolvar + bits = boolvar ' + bits;