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Proof step 10 from proof sgn-append-03 (previous step)

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

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 3

Additional information