Prove that N P$^$N P $\backslash$cap coNP$=$N P$.$ Be sure you understand what this means. A language is in N P$^$N P $\backslash$cap c o N P if it can be written in the form $\{$x $|$ w: M$^$A$($x$,$ w$)=1\},$ where M is a poly$-$time oracle TM$,$ and A is a language in NP $\backslash$cap coNP. In other words, A has both an $"$NP$-$style" algorithm and a "coNP$-$style" one. More formally, there are polytime TMs M$\_1$ and M$\_2$ such that A$=\{$a $|$ w$\_1$: M$\_1($a$,$ w$\_1)=1\}=\{$a $|$ w$\_2$: M$\_2($a$,$ w$\_2)=1\}.$ Hint: compare the situation to N P$^$N P or N P$^$coNP, both of which are equal to $\backslash$Sigma $\_2.$ Think about where the additional quantifier comes from in these cases, and why you might not need one in our problem.