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Proof step 1 from proof bitsand-onein (conclusion)

⊦ ¬ true ∈ bits ∧ bits1 ≠ [] → (true ∈ bitsand(bits + false ' + bits0, bits1) ↔ true ∈ bitsand(false ' + bits0, bits1))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information