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Proof step 1 from proof inv-in (conclusion)

⊦ boolvar ∈ bitsinvert(bits) ↔ (¬ boolvar) ∈ bits

User applied rule structural induction

And obtained the following new goals

Proof step 2:
∀ boolvar. boolvar ∈ bitsinvert(bits0) ↔ (¬ boolvar) ∈ bits0 ⊦ boolvar0 ∈ bitsinvert(boolvar + bits0) ↔ (¬ boolvar0) ∈ boolvar + bits0

Rule argument

indvar: bits

Additional information

Used simplifier rules: invert-base: ⊦ bitsinvert([]) = [];%bitlist freely generated by [], +;;% ⊦ ¬ boolvar ∈ [];