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,
 A: 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 $\psi$."
A: 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
$$\neg\forall x\in\Bbb R\big(T(x)\land G(x)\big)\;.$$
A: 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 $$\color{red}{\neg(\exists b \in \mathbb R, T(x) \wedge \neg G(x)) }\equiv\neg(\neg(\forall b \in \mathbb R,  ( T(x)\to  G(x)))\equiv\color{red}{\forall b \in \mathbb R,  ( T(x)\to  G(x)))}$$
