Well formed formula in predicate logic

Question

Considering the following definition of well formed formula (wff) in predicate logic:

https://en.wikipedia.org/wiki/Well-formed_formula#Predicate_logic

is $\exists x \exists x \varphi(x)$ a wff?

Answer 1

Yes, it is.

If we apply the definition, if $\varphi(x)$ is a wff, then $\exists x \varphi(x)$ is a wff.

And thus, being $\exists x \varphi(x)$ a wff, also $\exists x \exists x \varphi(x)$ is a wff.