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,

In all of your expressions $\forall b\in\Bbb R$ and $\exists b\in\Bbb R$ should be $\forall x\in\Bbb R$ and $\exists x\in\Bbb R$, respectively, to match the variable in $T(x)$ and $G(x)$. Once that correction is made, you’re right: the negation of $$\forall x\in\Bbb R\big(T(x)\to G(x)\big)$$ is $$\exists x\in\Bbb R\big(T(x)\land\neg G(x)\big)\;.$$
To interpret the statement
$$\exists x\in\Bbb R\big(T(x)\to\neg G(x)\big)\tag{1}$$
most easily, remember that $P\to Q$ is exactly equivalent to $\neg P\lor Q$, so $(1)$ can be rewritten as
$$\exists x\in\Bbb R\big(\neg T(x)\lor\neg G(x)\big)\;;$$
in words, there is some real number $x$ that fails to have at least one of the properties $T$ and $G$. It’s clearly the negation of the assertion that every real number has both properties, so it’s
$$\neg\forall x\in\Bbb R\big(T(x)\land G(x)\big)\;.$$
