I have seen this statement before, but I haven't really seen a proof of it:If $SAT\in PCP_{1,2^{q}}[\log(n),q]$, for some constant $q$, then $P = NP$.Now, if $SAT\in PCP_{1,2^{q}}[\log(n),q]$, then SAT reduces to $GAP-q-CSP[1,2^{-q}]$. So to show $P = NP$, how do we show then that $GAP-q-CSP[1,2^{-q}]$ is in $P$?
If SAT is in PCP, for some constant q, then P = NP
cc.complexity theory;sat;pcp;p vs np