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Proof step 3 from proof inv-in (previous step)

∀ boolvar. boolvar ∈ bitsinvert(bits0) ↔ (¬ boolvar) ∈ bits0 ⊦ boolvar0 ∈ bitsinvert(bits0) ∨ (boolvar0 ↔ ¬ boolvar) ↔ (¬ boolvar0 ↔ boolvar) ∨ (¬ boolvar0) ∈ bits0

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information