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Proof step 4 from proof inv-in (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
boolvar0 ∈ bitsinvert(bits0), ∀ boolvar. boolvar ∈ bitsinvert(bits0) ↔ (¬ boolvar) ∈ bits0 ⊦ ¬ boolvar0 ↔ boolvar, boolvar0 ↔ ¬ boolvar, (¬ boolvar0) ∈ bits0

Proof step 4 from proof inv-in (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information