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Proof step 18 from proof sgn-last-diff (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
⊦ (¬ boolvar)' = sgn((¬ boolvar)' + bits0), boolvar ∈ bits0

Rule argument

Additional information

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