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Proof step 42 from proof sgn-last-04 (previous step)

¬ boolvar1 ↔ boolvar1, ∀ boolvar. (¬ boolvar ↔ boolvar1) ∨ (¬ boolvar) ∈ bits1 → sgn((boolvar1 ' + bits1) + boolvar ') = sgn(boolvar1 ' + bits1) + boolvar ' ⊦ sgn(boolvar1 ' + bits1 + boolvar0 ') = boolvar1 ' + (boolvar1 ' + bits1) + boolvar0 ', ¬(¬ boolvar0 ↔ boolvar1) ∧ ¬ (¬ boolvar0) ∈ bits1 ∧ ¬(¬ boolvar0 ↔ boolvar1)

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

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