# Using existential quantifier to express “There is a $P$ that is $Q$” [duplicate]

The textbook says that to express "There is a $$P$$ that is $$Q$$", one can write the following preposition:

$$\exists x [P(x) \land Q(x)\ ]$$

However, I am unsure why I can't write like this:

$$\exists x [P(x) \rightarrow Q(x)]$$

For the case of universial quantifier, I have the opposite question. The textbook says "All $$P$$ is $$Q$$" can be written as below:

$$\forall x [P(x) \rightarrow Q(x)]$$

and I am unsure why I can't write like this:

$$\forall x [P(x) \land Q(x)]$$

I do understand that $$P \land Q$$ and $$P \rightarrow Q$$ are different in that $$P \rightarrow Q$$ also accepts vacuously true. But I am curious why $$\forall$$ can accept vacously true while $$\exists$$ cannot.

For your first $$\exists x [P(x) \rightarrow Q(x)]$$ is true if there is some $$x$$ for which $$P(x)$$ is false. It is still true if $$P(x)$$ is always false. Then there is not a $$P$$, so there cannot be a $$P$$ that is $$Q$$. Similarly for the second $$\forall x [P(x) \land Q(x)]$$ requires that $$P(x)$$ and $$Q(x)$$ are true for all $$x$$. You only want to require $$Q(x)$$ when $$P(x)$$ is true. If our universe is all naturals and $$P(x)$$ is $$x=1$$ while $$Q(x)$$ is $$x \lt 10$$ it is true that all $$P$$ are $$Q$$, but your translation is false because $$x$$ could be $$11$$