# Basic second-order logic example contains a mistake?

I'm reading the following course on second-order logic, by Péter Mekis : http://phil.elte.hu/mekis/sol.pdf . The course seems excellent, but I'm stuck on one of his first examples for showing the power of second-order logic w.r.t first-order logic.

In this example, he works on the following statement : "Santa Claus has all characteristic properties of a pedophile." This is translated into the second-order logic sentence : ∀X ( R(X,P)→P(a) ), where a is santa, P is the property of being a pedophile, X is a property and R tells us whether a property is characteristic of another property.
I understand this as : for all properties, if a property is characteristic of being a pedophile, then santa is a pedophile. This is obviously not what the statement says, so am I misunderstanding or should the second-order logic sentence instead be : ∀X ( R(X,P)→X(a) ), effectively meaning : for all properties, if the property is characteristic of being a pedophile, then santa has this property.

• Wow, that example is in spectacularly bad taste. – David Richerby Apr 23 at 14:14

Yes, the example contains a mistake. The statement $$\forall X (\mathcal{R}(X,P)\rightarrow P(a))$$ is in fact equivalent to $$P(a)$$, under the assumption that $$\mathcal{R}(X,P)$$ is true for at least one $$X$$.
So indeed, the example should be $$\forall X (\mathcal{R}(X,P)\rightarrow X(a))$$, as you state. Perhaps the author would not have made this mistake if they had chosen a class of people that would be described by another letter than $$P$$.