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

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

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 2

Additional information