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Proof step 16 from proof sgn-last-04 (previous step)

bits0 + boolvar0 ' ≠ [] ∧ ((bits0 + boolvar0 ').first ↔ boolvar), ∀ boolvar. (¬ boolvar) ∈ bits0 → sgn(bits0 + boolvar ') = sgn(bits0) + boolvar ' ⊦ sgn(bits0 + boolvar0 ') = sgn(boolvar ' + bits0) + boolvar0 ', ¬ (¬ boolvar0) ∈ bits0 ∧ ¬(¬ boolvar0 ↔ boolvar)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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