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Proof step 2 from proof sgn-last (previous step)

∀ boolvar, boolvar0. (¬ boolvar) ∈ bits0 → sgn(boolvar ' + bits0 + boolvar0 ') = sgn(boolvar ' + bits0) + boolvar0 ' ⊦ (¬ boolvar1) ∈ boolvar + bits0 → sgn(boolvar1 ' + (boolvar + bits0) + boolvar0 ') = sgn(boolvar1 ' + (boolvar + bits0)) + boolvar0 '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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