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
$(\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?
Added example:
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