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Proof step 8 from proof sgn-sgn-01 (previous step)

∀ bits0. sgn(sgn(boolvar0 ' + bits) + bits0) = sgn((boolvar0 ' + bits) + bits0) ⊦ boolvar0 ↔ boolvar, sgn(sgn(boolvar ' + boolvar0 ' + bits) + bits1) = sgn(boolvar ' + (boolvar0 ' + bits) + bits1)

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

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;% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;