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Proof step 6 from proof sgn-first (previous step)

∀ boolvar. sgn(boolvar ' + bits0).first ↔ boolvar ⊦ sgn(boolvar0 ' + boolvar ' + bits0).first ↔ boolvar0, boolvar ↔ boolvar0

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 2

Additional information

Used simplifier rules: sgn-base-02: ⊦ sgn(false ' + true ' + bits) = false ' + true ' + bits;%sgn-base-01: ⊦ sgn(true ' + false ' + bits) = true ' + false ' + bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (boolvar ' + bits).first ↔ boolvar;