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Proof step 3 from proof sgn-last-04 (previous step)

∀ boolvar. (¬ boolvar) ∈ bits0 → sgn(bits0 + boolvar ') = sgn(bits0) + boolvar ' ⊦ (¬ boolvar0) ∈ boolvar + bits0 → sgn((boolvar + bits0) + boolvar0 ') = sgn(boolvar + bits0) + boolvar0 '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ boolvar + bits = boolvar ' + bits;