Match each schema with the corresponding rule from predicate logic to prove it$,$ or

indicate why it is not a theorem

$p|--(AAx)(pvvx=y)$

$((AAx)xz{)}^{{?}^{?}}x=y|--(AAy)x=y$

$(AAx)(hgx=y^{{?}^{?}}TT)|--hgx=y^{{?}^{?}}TT$

$*$Note there are more options to choose from than needed, and you can use any

option as many times as you wish $($or not at all$)$