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Proof step 18 from proof bitsinvert-size (previous step)

¬(bits0 ≠ [] ∧ (bits0.first ↔ ¬ boolvar)), bitsinvert(bits0) = boolvar ' + bits1, # sgn(bitsinvert(bits0)) = # sgn(bits0), (¬ boolvar) ∈ bits0 ⊦ bits0 = [], 1 + # bits0 = # sgn(boolvar ' + bits0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
bitsinvert(bits0) = boolvar ' + bits1, # sgn(bitsinvert(bits0)) = # sgn(bits0), (¬ boolvar) ∈ bits0 ⊦ bits0 = [], 1 + # bits0 = # sgn(boolvar ' + bits0), bits0.first ↔ ¬ boolvar

Proof step 18 from proof bitsinvert-size (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information