# Show that there is a false statement of the form:

Show that there is a false statement of the form:

$$\big(\exists xG(x)\land\exists xH(x)\big)\to\exists x\big(G(x)\land H(x)\big)$$

my question is , is the $x$ in $H(x)$ must be the same $x$ in $G(x)$ ? or it's not necessary that they are equal ??

if they can be diffrent i can show this easily , but if the must be the same element in the doman - the same individual - i think that any statement in this form must be true

this problem is in " first order mathematical logic by angelo margaris , 1990 ed "

page 30

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Think about the "scope" of the existential quantifiers. This determines which $x$ has to be "the same" as another $x$. – hardmath Feb 22 '13 at 20:01

$(\exists x G(x))\land \exists x H(x) )\rightarrow (\exists x ( G(x) \land H(x) ) )$

The left-hand side could equivalently be expressed as:

$$\exists x G(x) \land \exists y H(y)$$

This makes clear that all the left-hand side claims is the existence of some x such that G(x), and the existence of something, say y, such that $H(y)$

Whereas on the right hand side, there is only one quantified variable: $x$, in the right hand side, so the scope of the existent $x$ on the right-hand side is "over" all of $(G(x) \land H(x))$: there is some $x$ such that both $G(x)$ and $H(x)$ hold, for that given $x$.

Can you see how the left hand side does not necessarily imply the right hand side, but that the right-hand side implies the left hand side of the implication?

Suppose the domain of x is all people.
Suppose $G(x)$ means "x is a woman"
Suppose $H(x)$ means "x is a man".

Then, we have the left hand side asserting:

There exists someone that is a woman, and there exists someone who is a man.

That is certainly true: men exist, and women exist.

But now let's see how the right-hand-side translates:

There is someone who (is both a woman and a man).

I hope this example helps make clear that we can easily construct an example in which $\exists x G(x) \land \exists x H(x)$ is true, but $\exists x (G(x) \land H(x)$ is clearly false. And because of this, the implication is false.

There exists

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thanx alot :) , it's obivous now for me :) thanx again :) – Maths Lover Feb 22 '13 at 21:25
Your welcome, MrWhy! – amWhy Feb 22 '13 at 21:26

In the expression

$$\big(\exists xG(x)\land\exists xH(x)\big)\to\exists x\big(G(x)\land H(x)\big)\;,\tag{1}$$

the scope of the first existential quantifier is just $G(x)$, the scope of the second is just $H(x)$, and the scope of the third is $G(x)\land H(x)$. This means that the lefthand expression says that there is some $x$ such that $G(x)$ is true, and there is some $x$ such that $H(x)$ is true; there is no connection between the two things whose existence is asserted. In fact, the meaning of $(1)$ would be unchanged if we rewrote it as

$$\big(\exists xG(x)\land\exists yH(y)\big)\to\exists x\big(G(x)\land H(x)\big)\;.$$

The righthand side, however, is asserting the existence of a single $x$ that satisfies both $G(x)$ and $H(x)$.

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M.Scott , thanx very much :)) .. – Maths Lover Feb 22 '13 at 21:26
@MrWhy: You’re welcome! – Brian M. Scott Feb 22 '13 at 21:28