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Let $$\Sigma_i\text{SAT} = \{\phi(u_1,\cdots, u_i): \exists{u_1}\forall{u_2}\exists\cdots Q_iu_i \phi(u_1,\cdots, u_i) = 1\}$$, where here $\phi$ is a Boolean formula (not necessarily in CNF form, although this does not make much difference), each $u_i$ is a vector of Boolean variables, and $Q_i$ is either $\exists$ or $\forall$ depending on whether $i$ is odd or even.

My question is how can I prove that $\Sigma_i\text{SAT}$ is hard for $\Pi_{i−1}^P$? We know that $\Sigma_i\text{SAT}$ is $\Sigma_{i}^P$ complete, is it help to prove the hardness for $\Pi_{i−1}^P$?

Adam Finch
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1 Answers1

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$\Pi^P_{i - 1} \subseteq \Delta^P_{i} \subseteq \Sigma_i^P$ follows from definition of the polynomial hierarchy. As you said, $\Sigma_i SAT$ is $\Sigma_i^P$-complete, so your result follows immediately.

anonymousBeaver
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