# 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,

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Why are you quantifying over b when your statements are about x? – David H Jun 28 '13 at 19:49
thats an oversight on my part. thanks for telling me! – tuba09 Jun 28 '13 at 19:50

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 $\psi$."

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aah, I see. Thinking of it in terms of OR makes a lot more sense – tuba09 Jun 28 '13 at 19:56
The problem is in the notion that "implication" in logic actual means what we intuit "implication" means. It doesn't. In logic, $A\implies B$ is not really a statement about $B$ following logically from $A$. For example, "If I had a banana for lunch, my name is Thomas," is a true statement, even it it sounds ridiculous - my name has nothing to do with whether or not I ate a banana. – Thomas Andrews Jun 28 '13 at 20:04
yup, i agree. that's still a habit i'm trying to break out of. thanks again for all the help. – tuba09 Jun 28 '13 at 20:15

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

$$\neg\forall x\in\Bbb R\big(T(x)\land G(x)\big)\;.$$

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

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