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Proof step 5 from proof sgn-sgn-01 (previous step)

¬(bits0 ≠ [] ∧ (bits0.first ↔ boolvar)), ∀ bits1. sgn(sgn(bits0) + bits1) = sgn(bits0 + bits1) ⊦ sgn(sgn(boolvar ' + bits0) + bits1) = sgn((boolvar ' + bits0) + bits1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ bits1. sgn(sgn(bits0) + bits1) = sgn(bits0 + bits1) ⊦ sgn(sgn(boolvar ' + bits0) + bits1) = sgn((boolvar ' + bits0) + bits1), bits0.first ↔ boolvar, bits0 = []

Rule argument

Additional information

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