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,

Hint:$\exists x :s(x)\equiv \neg(\forall x :\neg s(x))$
$$\exists b \in \mathbb R, T(x) \wedge \neg G(x) \equiv\exists b \in \mathbb R, \neg (\neg T(x)\lor  G(x)) \equiv\exists b \in \mathbb R, \neg ( T(x)\to  G(x))\equiv\neg(\forall b \in \mathbb R,  ( T(x)\to  G(x)))$$ then $$\color{red}{\neg(\exists b \in \mathbb R, T(x) \wedge \neg G(x)) }\equiv\neg(\neg(\forall b \in \mathbb R,  ( T(x)\to  G(x)))\equiv\color{red}{\forall b \in \mathbb R,  ( T(x)\to  G(x)))}$$
