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Proof step 1 from proof and-pos (conclusion)

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

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: and-first: ⊦ bits ≠ [] ∧ bits0 ≠ [] → (bitsand(bits, bits0).first ↔ bits.first ∧ bits0.first);