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Proof step 22 from proof and-onesin (previous step)

bits ≠ [] ∧ bits0 = true ', ∀ bits1, bits0, boolvar. # bits1 < # bits → true → bits1 = [] ∨ bits0 = [] ∨ (boolvar ∈ bitsand(bits1, bits0) ↔ boolvar ∈ bits0) ∨ false ∈ bits1 ⊦ boolvar ∈ bits ↔ boolvar ∈ bits0, bits0 = [], bits = [], false ∈ bits, bits ≠ bits.first ' ∧ bits0 ≠ bits0.first '

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ true ∈ bits ∧ bits ≠ [] → (¬ false ∈ bits ↔ false);% ⊦ boolvar ' ≠ [];% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;