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Proof step 7 from proof sgn-sgn-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
∀ bits0. sgn(sgn(boolvar0 ' + bits) + bits0) = sgn((boolvar0 ' + bits) + bits0) ⊦ boolvar0 ↔ boolvar, sgn(sgn(boolvar ' + boolvar0 ' + bits) + bits1) = sgn(boolvar ' + (boolvar0 ' + bits) + bits1)

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits ≠ [];% ⊦ boolvar + bits = boolvar ' + bits;