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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.

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    $\begingroup$ Wow, that example is in spectacularly bad taste. $\endgroup$ – David Richerby Apr 23 at 14:14
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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 $P$.

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