Here is one of my favorite examples of a bad proof for what turns out to be a correct theorem. **Theorem:**If NP is in BPP then the whole polynomial-time hierarchy is in BPP.Let's focus on simply showing Σ

_{2}is in BPP if NP is in BPP. The rest is straightforward induction. Here is our first proof:Σ Do you see the problem with this proof?_{2}=NP^{NP}⊆ NP^{BPP}⊆BPP^{BPP}=BPP.To get a correct proof (first due to Zachos) we need to use Arthur Merlin games. Consider a Σ

_{2}language L as an ∃∀ expression. Since NP is in BPP, we can replace the ∀ with a probabilistic test. This gives us what is known as MA or a Merlin-Arthur game where the powerful Merlin sends a message that Arthur can probabilistically verify. A now classic result shows that MA is contained in AM, where Arthur provides a random string to Merlin who must then provide a proof based on that string. Once again we apply the NP in BPP assumption to allow Arthur to simulate Merlin probabilistically and now we have a BPP algorithm for L.The problem in the first proof is in the second "⊆". The assumption NP in BPP does not imply NP

^{A}in BPP^{A}for all A.

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Posted by Lance Fortnow to My Computational Complexity Web Log at 10/24/2003 08:59:53 AM

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