Complete the logical proof for the following argument.

EExP$(x)$

AAx$(P(x)Q(x))$

:$.$EExQ$(x)$

$\backslash$table$[[$Step$,$Proposition,Justification$],[1,$EExP$($x$),],[2,$c is an element in the domain $^^$P$($c$),],[3,$c is an element in the domain,$],[4,,$Hypothesis$],[5,$P$($c$)->$Q$($c$),],[6,,$Simplification, $2],[7,$Q$($c$),],[8,$Q$($x$),]]$

Note that the justification for each step is either hypothesis or it would include both the name of the law or rule and the step$($s$)$ to which it is applied to$.$

Copy and paste the logical operators when filling in the blanks: $vv,{?}^{{?}^{?}},,$not,$AA,EE.$