Confusions about FOL existential quantifier with conditional statement

Question

My question is that, from my textbook it states that:

The FOL $\exists x(Doctor(x) \Rightarrow Smart(x))$ translates to the english sentence "Some doctors are smart"

But I think the correct way to interpret the FOL is "There exists a person $x$ such that if $x$ is a doctor then this person is smart". The reason this is a better way (IMO) to express the FOL statement is because of the implication "If $x$ is a doctor then ...", which means we can have a world where there are no doctors at all, and this statement would still be true vacuously. However, if we consider the sentence "Some Doctors are smart", which tells or gives us the impression that there must be at least one person who is a Doctor. But this would be false in a world without doctors. Thus the two interpretations are completely different.

The textbook (from chapters before the above claim) mentioned that:

we express "Some doctors are smart" with $\exists x(Doctor(x) \wedge Smart(x))$ instead of $\exists x(Doctor(x) \Rightarrow Smart(x))$, and this further asserts my reasoning above.

But in the later sections of the textbook where we encountered the statement $\exists x(Cube(x) \Rightarrow Small(x))$ the textbook again used the sentence "Some cubes are small", this is purely miss leading and a contradiction to what the textbook mentioned.

I will appreciate it very much if anyone can verify that the two statements:

"There exists a person $x$ such that if $x$ is a doctor then this person is smart" and

"Some doctors are smart" are completely two different meanings, and that the first sentence is the correct English translation to the FOL $\exists x(Doctor(x) \Rightarrow Smart(x))$.

Answer 1

You are exactly correct. The claim

The FOL $\exists x(Doctor(x) \Rightarrow Smart(x))$ translates to the english sentence "Some doctors are smart"

is completely false, and if that's what the text really says that's an embarrassing mistake (made even worse by the fact that they apparently double down on it with the cubes example).

Your analysis of the mistake is completely correct as well: "$\exists x(P(x)\implies Q(x))$" is automatically true in any world where there is some object not satisfying $P$, whereas "$\exists x(P(x)\wedge Q(x))$" isn't, so they're clearly not equivalent.

(In fact, the only way for "$\exists x(P(x)\implies Q(x))$" to be false is if every object $x$ satisfies $P$ and no object $x$ satisfies $Q$. So it's really, really far from "some $P$ are $Q$.")

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Out of morbid curiosity, what textbook is this?

Answer 2

One way to think about

$$\exists x (Doctor(x) \rightarrow Smart(x))$$

is as its equivalent:

$$\exists x (\neg Doctor(x) \lor Smart(x))$$

i.e. It says that there is something that is either not a doctor or that is smart ... which is clearly different from there being something that is a doctor and that is smart, whcih indeed correctly translates to:

$$\exists x (Doctor(x) \land Smart(x))$$