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Proof step 3 from proof sgn-append-06 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
(¬ boolvar) ∈ bits ⊦ sgn(bits + boolvar ' + bits1) = sgn(bits) + boolvar ' + bits1

Rule argument

Additional information

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