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Proof step 6 from proof inv-in (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
(¬ boolvar0) ∈ bits0, ∀ boolvar. boolvar ∈ bitsinvert(bits0) ↔ (¬ boolvar) ∈ bits0 ⊦ boolvar0 ↔ ¬ boolvar, ¬ boolvar0 ↔ boolvar, boolvar0 ∈ bitsinvert(bits0)

Proof step 6 from proof inv-in (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information