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Proof step 11 from proof bitsinvert-size (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
bitsinvert(bits0) = boolvar0 ' + bits, # sgn(bitsinvert(bits0)) = # sgn(bits0), (¬ boolvar0) ∈ bits0 ⊦ boolvar0 ↔ ¬ boolvar, bits0 = [], # sgn((¬ boolvar)' + bitsinvert(bits0)) = # sgn(boolvar ' + bits0), bitsinvert(bits0).first ↔ boolvar

Rule argument

Additional information

Used simplifier rules: inv-empty: ⊦ bitsinvert(bits) = [] ↔ bits = [];