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Proof step 11 from proof sgn-append-03 (previous step)

boolvar0, bits0 = [] ∨ (∀ bits1, boolvar. sgn(bits0 + bits1) = sgn((¬ boolvar)' + bits1) ∨ boolvar ∈ bits0) ⊦ boolvar, (bits0 + bits1).first ↔ boolvar, sgn((boolvar ' + bits0) + bits1) = sgn((¬ boolvar0)' + bits1), boolvar0 ∈ bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
bits0.first, ∀ bits1, boolvar. sgn(bits0 + bits1) = sgn((¬ boolvar)' + bits1) ∨ boolvar ∈ bits0 ⊦ sgn((false ' + bits0) + bits1) = sgn(false ' + bits1), bits0 = [], true ∈ bits0

Rule argument

Additional information

Used simplifier rules: ⊦ bits ≠ [] → ((bits + bits0).first ↔ bits.first);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ [] + bits = bits;