Is the following propositional function well defined?

Question

My question is fairly simple:

Is $(P \wedge Q)(x)$ equivalent to $P(x)\wedge Q(x)$?

Reason I'm asking, is that when I asked my tutor he said the statements weren't equivalent because if $P(x) = "x\ is\ beautiful"$ then $P = "is\ beautiful"$ which to his mind meant something like $(P \wedge Q)(x)$ wasn't well-defined, which to me didn't make much sense.

So, is it well-defined? And if not, why?

Answer 1

The question is a little bit "weird" ...

If we work in propositional logic, propositional variables like $P,Q, \ldots$ stands for sentences, i.e. something with a definite truth-value, like : "Plato is a philosopher" (which is true) and "Napoleon is a painter" (which is false).

Thus, we can build up "complex" sentences with the conncetives, like $\land$ ("and") that are again true or false according to the truth-conditions for the connectives (their truth tables).

In my (silly) example :

"Plato is a philosopher and Napoleon is a painter"

is false, because $TRUE \land FALSE$ is $FALSE$.

In this context, a formula like $(P \land Q)(x)$ is not well-formed, i.e. it is meaningless, because individual variables, like $x$, are not allowed in the syntax of propositional logic.

---

If instead we are working in predicate logic, where we have prdicate symbols, like $P,Q, \ldots$ and variables, like $x,y, \ldots$, we can use $P(x)$ to symbolize the expression "$x$ is a philosopher" and $Q(x)$ for "$x$ is a painter".

Thus, we can build complex expressions like $P(x) \land Q(x)$ meaning, in my example :

"$x$ is a philosopher and $x$ is a painter".

Usually, the syntax of predicate logic does not allow for expressions like $(P \land Q)(x)$, but if we want use it to translate the expression :

"$x$ is a philosopher and a painter"

it seems quite clear that it must be equivalent to (the more correct) : $P(x) \land Q(x)$.

---

Your tutor's concern is not very clear ...

According to an "old" point of view (see Gottlob Frege), the variable $x$ in the expression for a predicate is only a "place-holder".

We can use, instead of "$x$ is a philosopher", the expression :

"__ is a philospher".

In this case, it is more clear that the expression is incomplete ("unsaturated") and that it must be completed with a "name" in the place indicated by the place-holder in order to get a meningful expression (a sentence), like the (true) sentence "Plato is a philospher" or the (false) sentence "Napoleon is a philospher".

The weak point of this convention is that it needs some care to handle predicates expressing $n$-ary relations ($n \ge 2$); writing ; "__ loves __" is ambiguous between "John loves John" and "John loves Mary".

This is why we prefer to use : $loves(x,y)$.

Answer 2

Yes, the propositions are equivalent, by definition.