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Proof step 8 from proof sgn-append-03 (previous step)

bits0 = [] ∨ (∀ bits1, boolvar. sgn(bits0 + bits1) = sgn((¬ boolvar)' + bits1) ∨ boolvar ∈ bits0) ⊦ bits0 ≠ [] ∨ bits1 ≠ [], sgn((boolvar ' + bits0) + bits1) = sgn((¬ boolvar0)' + bits1), boolvar0 ↔ boolvar, boolvar0 ∈ bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
⊦ boolvar0 ↔ boolvar, boolvar ↔ ¬ boolvar0

Rule argument

Additional information

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