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Proof step 4 from proof sgn-append (previous step)

∀ boolvar, bits1. sgn(boolvar ' + bits0 + bits1) = sgn(boolvar ' + bits1) ∨ (¬ boolvar) ∈ bits0 ⊦ sgn(boolvar0 ' + (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 2

Additional information