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Proof step 5 from proof sgn-sgn (previous step)

¬(bits0 ≠ [] ∧ (bits0.first ↔ boolvar)), sgn(sgn(bits0)) = sgn(bits0) ⊦ sgn(sgn(boolvar ' + bits0)) = sgn(boolvar ' + bits0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: sgn-empty: ⊦ sgn([]) = [];%sgn-one: ⊦ sgn(boolvar ') = boolvar ';% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;