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Proof step 1 from proof and-first (conclusion)

⊦ bits ≠ [] ∧ bits0 ≠ [] → (bitsand(bits, bits0).first ↔ bits.first ∧ bits0.first)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ bits = [], bits0 = [], bitsand(bits, bits0).first ↔ bits.first ∧ bits0.first

Rule argument

Additional information