# How can this statement be false? "If $\forall x \in D$, $P(x)$ then $\exists x \in D$ such that $P(x)$."

I'm a college student taking a discrete mathematic course for summer. I took midterm last monday and got back grades and solutions for the exam, but I'm still confused with this specific question.

The question is:

Let $$D$$ represent a set and $$P(x)$$ represent a predicate where $$x\in D$$. Is this a true or false statement? Explain briefly.

"If $$\forall x \in D$$, $$P(x)$$ then $$\exists x \in D$$ such that $$P(x)$$."

And the answer is: This is a false statement.

How can this statement be false?

( It's late night so I don't wanna bother my professor)

• Hint: what if $D$ is empty? Jun 21, 2020 at 1:15
• "If every student who bothers the professor late at night fails the course, then there exists a student who has failed."
– Blue
Jun 21, 2020 at 1:15

$$\forall x\in D,\,P(x)$$ can be vacuously true. For example, $$\forall x\in\varnothing\,(x>0)$$ is vacuously true, since $$\varnothing$$ has no elements, but $$\exists x\in\varnothing (x>0)$$ is false, since clearly no such element exists.
• Is this whole thing only false for $D=\emptyset$? Jun 21, 2020 at 1:45
• @gen-zreadytoperish If $D$ is nonempty, then $\exists x_0\in D$, and if $\forall x\in D:P(x)$, then $P(x_0)$, so $\exists x\in D:P(x)$. So I believe the only fringe case is when $D=\varnothing$. Jun 21, 2020 at 4:15
• Yes - if $D$ is nonempty then the statement is true, but unfortunately that wasn’t stated as an assumption. Jun 22, 2020 at 18:22