How to prove this is a tautology $\forall x : (p(x) \wedge \neg q(x)) \vee \exists x: \neg p(x) \vee \neg \exists x: q(x)$
The problem is I know that this is a tautology, but I don't know how to prove it. Can anyone can help me?

Edit
Here is a way without EI
(a lot quicker,too):
$\forall x : (p(x) \wedge \neg q(x)) \vee \exists x: \neg p(x) \vee \neg \exists x: q(x) \Rightarrow$
$\forall x : \neg(\neg p(x) \vee  q(x)) \vee \exists x: \neg p(x) \vee \neg \exists x: q(x)\Rightarrow$
$\exists x : (\neg p(x) \vee  q(x)) \vee \exists x: \neg p(x) \vee \neg \exists x: q(x)\Rightarrow$
$\exists x : \neg p(x) \vee  \exists x: q(x) \vee \exists x: \neg p(x) \vee \neg \exists x: q(x)\Rightarrow$
$\exists x : \neg p(x) \vee  (\exists x: q(x) \vee \neg \exists x: q(x))\vee \exists x: \neg p(x)\Rightarrow$
$\exists x : \neg p(x) \vee  T \vee \exists x: \neg p(x)\Rightarrow $
$\exists x : \neg p(x) \vee  T \Rightarrow  T$
