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Proof step 3 from proof sgn-last (previous step)

(¬ boolvar1) ∈ bits0, ∀ boolvar, boolvar0. (¬ boolvar) ∈ bits0 → sgn(boolvar ' + bits0 + boolvar0 ') = sgn(boolvar ' + bits0) + boolvar0 ' ⊦ sgn(boolvar1 ' + (boolvar ' + bits0) + boolvar0 ') = sgn(boolvar1 ' + boolvar ' + bits0) + boolvar0 ', ¬ boolvar1 ↔ boolvar

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 2

Additional information