Consider the following argument:

All wizards can do magic. Muggles can't do magic. Peter is a muggle.

Therefore, Peter is not a wizard.

Several students in Hogwarts formalized this argument $($without using magic$).$ Here is Luna's attempt:

Luna: Let's define the predicates: W$="$ is a wizard", M$="$ is a muggle", A$="$ can do magic".

Universe of discourse: all people $($magical and non$-$magical$)$

$1.$xW$($x$)->$ A$($x$)($assumption$)$

$2.$xM$($x$)->$A$($x$)($assumption$)$

$3.$ M$($Peter$)($assumption$)$

$4.$ M$($Peter$)->$A$($Peter$)($Universal instantiation of $2)$

$5.$A$($Peter$)($Modus ponens from $3$ and $4)$

$6.$ W$($Peter$)->$ A$($Peter$)($Universal instantiation of $1)$

$7.$W$($Peter$)($Modus tollens from $5$ and $6)$

If you think, Luna's argument is valid, check the "valid" box

If you think, Luna's argument is invalid, check the "invalid" box and check all the reasons why you think that Luna's argument is invalid.

Group of answer choices

Luna' s argument is valid

Luna's argument is invalid.

Luna's argument is invalid, because in $1$ the quantifier only refers to W$($x$)$ and not to A$($x$)$ A$($x$)$ is not bound to any quantifier. Thus, $1$ is not a proposition.

Luna's argument is invalid, because in $2$ the quantifier only refers to M$($x$)$ and not to $$A$($x$)$A$($x$)$ is not bound to any quantifier. Thus, $2$ is not a proposition.

Luna's argument is invalid, because his definitions of the predicates are incorrect. They should have been defined as W$($x$)="$x is a wizard", M$($x$)="$x is a muggle" and A$($x$)="$x can do magic".