This question already has an answer here:

Assuming $\mathsf{P} \ne \mathsf{NP}$ can we show $\mathsf{P} \ne \mathsf{P/poly} \cap \mathsf{NP}$? Obviously this would be the case if $\mathsf{P} \ne \mathsf{NP}$ and $\mathsf{P/poly} \supset \mathsf{NP}$ but this is unlikely. This is also true under the assumption $\mathsf{P} \ne \mathsf{BPP}$, however AFAIK it is commonly believed $\mathsf{P} = \mathsf{BPP}$. Can we prove $\mathsf{P} \ne \mathsf{P/poly} \cap \mathsf{NP}$ under a weaker assumption? What evidence can be provided for/against? Is there a specific language which is believed to be in $\mathsf{P/poly} \cap \mathsf{NP}$ but not in $\mathsf{P}$?


marked as duplicate by Kaveh Feb 11 '13 at 20:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.