Implication with a there exists quantifier When I negate 
$ \forall x \in \mathbb R, T(x) \Rightarrow G(x) $ 
I get $ \exists x \in \mathbb R, T(x) \wedge \neg G(x) $ 
and NOT
$ \exists x \in \mathbb R, T(x) \Rightarrow \neg G(x) $ 
right?
What would it mean if I said $ \exists x \in \mathbb R, T(x) \Rightarrow \neg G(x) $ ? I know in symbolic logic a statement like $ \forall x \in \mathbb R, T(x) \Rightarrow G(x) $ means every T is a G, but what claim am I making between T & G with $ \exists x \in \mathbb R, T(x) \Rightarrow G(x) $ in simple everyday english if you can?
Thanks,

You're correct that the negation of $\forall x (T(x) \rightarrow G(x))$ is $\exists x (T(x) \wedge \neg G(x))$.
The short answer is that $\exists x (\varphi(x) \rightarrow \psi(x))$ doesn't really have a good English translation. You could try turning this into a disjunction, so that $\exists x (\varphi(x) \rightarrow \psi(x))$ becomes $\exists x (\neg \varphi(x) \vee \psi(x))$. But this is equivalent to $\exists x \neg\varphi(x) \vee \exists x \psi(x)$, which just says "There either exists something which is not $\varphi$ or there exists something which is $\psi$. That's the best you're going to get though. $\exists x (\varphi(x) \rightarrow \psi(x))$ is just something that doesn't have a good translation because it's a rather weak statement.
Contrast this with the dual problem $\forall x (\varphi(x) \wedge \psi(x))$, which is a rather strong statement, saying "Everything is both a $\varphi$ and a $\psi$."
