1. ## logical quantification

Hello everyone,
I'm having problems with the following type of tutorial questions (yes, there are more of them):

Use logic notation to express the following:
"There is at most one x with P(x)"

I understand that to mean, "if there exists an x such that P(x), then for all other x not P(x)."

$\displaystyle (\exists x)[P(x)] \rightarrow (\forall y)[\neg P(x)], x \neq y$

but this is wrong, or at least badly expressed. Would the following be right?

$\displaystyle (\exists x, \forall y; x \neq y)[P(x)]$

2. Originally Posted by bmp05
Hello everyone,
I'm having problems with the following type of tutorial questions (yes, there are more of them):

Use logic notation to express the following:
"There is at most one x with P(x)"

I understand that to mean, "if there exists an x such that P(x), then for all other x not P(x)."

$\displaystyle (\exists x)[P(x)] \rightarrow (\forall y)[\neg P(x)], x \neq y$

but this is wrong, or at least badly expressed. Would the following be right?

$\displaystyle (\exists x, \forall y; x \neq y)[P(x)]$
Hmm... I would have expressed this as a conditional. If there two x-values x1 and x2 such that P(x1) and P(x2), then x1 = x2. Does this make sense?

3. Thanks for the reply apcalculus, that was a good explanation.

so you mean something like:

$\displaystyle (\exists x,y)[(P(x) \leftrightarrow P(y)) \rightarrow (x = y)]$

can you write this as follows?

$\displaystyle (\exists x,y; x=y)[P(x) \leftrightarrow P(y)]$

4. Apcalculus gave the right hint.

Translating sentences of the form "If there exists $\displaystyle x$ such that $\displaystyle P(x)$, then $\displaystyle Q(x)$" is a little counterintuitive. One could try

$\displaystyle \left(\exists x\,P(x)\right)\to Q(x)$

but this would not be right because the scope of $\displaystyle \exists x$ does not extend to $\displaystyle Q$. Another attempt is

$\displaystyle \exists x.\,(P(x)\to Q(x))$

This is also wrong because this proposition affirms the existence of something, while the original plain language sentence did not claim this.

The right way to translate "If there exists $\displaystyle x$ such that $\displaystyle P(x)$, then $\displaystyle Q(x)$" is using the universal quantifier:

$\displaystyle \forall x.\,(P(x)\to Q(x))$

5. That may well be, but you're making me cry!!

6. Originally Posted by bmp05

I understand that to mean, "if there exists an x such that P(x), then for all other x not P(x)."
It is not a conditional and the right expression is:

There exist an x with P(x) and for all other x not P(x)
This in logic in quantifier form is expressed as follows:

$\displaystyle \exists xP(x)\wedge\forall x\forall y( P(x)\wedge P(y)\Longrightarrow x = y)$

7. Well, look: you need to translate the following statement.
If there two x-values x1 and x2 such that P(x1) and P(x2), then x1 = x2
Instead of this, I'll translate a not-very-meaningful statement just to give an example:

If there is an $\displaystyle x$ such that $\displaystyle P(x)$ and $\displaystyle P(x)$, then $\displaystyle x=x$.

This would be

$\displaystyle \forall x.\,\big(P(x)\land P(x)\to x=x\big)$

8. Of course, to really solve the problem, besides writing the actual formula, you need to understand very well why
If there two x-values x1 and x2 such that P(x1) and P(x2), then x1 = x2.
is equivalent to your original statement
There is at most one x with P(x)

9. Originally Posted by emakarov
Well, look: you need to translate the following statement.
Instead of this, I'll translate a not-very-meaningful statement just to give an example:

If there is an $\displaystyle x$ such that $\displaystyle P(x)$ and $\displaystyle P(x)$, then $\displaystyle x=x$.

This would be

$\displaystyle \forall x.\,\big(P(x)\land P(x)\to x=x\big)$
Are you referring to my formula

10. Originally Posted by xalk
Are you referring to my formula
No, I am referring to apcalculus's post.

I don't fully agree with your version.
Originally Posted by xalk
It is not a conditional and the right expression is:

There exist an x with P(x) and for all other x not P(x)
The right expression of the original statement ("There is at most one...")? This is not correct because the original statement does not assert the existence of anything while your sentence does.

This in logic in quantifier form is expressed as follows:

$\displaystyle \forall xP(x)\wedge\forall x\forall y( P(x)\wedge P(y)\Longrightarrow x = y)$
This is also strange because now instead of "There exist an x with P(x)" you write $\displaystyle \forall xP(x)$.

11. Originally Posted by emakarov
Of course, to really solve the problem, besides writing the actual formula, you need to understand very well why
is equivalent to your original statement
Yes, I understand the reason. That makes sense, but the quantifiers and the logical operators do not make sense. The problem is the overloading of the x-variable. That hurts my head. It's the formula-writing which is the problem, not the problem itself.

I have to admit, I find this very hard. Thank you all for your help, nevertheless.

12. Originally Posted by emakarov
No, I am referring to apcalculus's post.

I don't fully agree with your version.

The right expression of the original statement ("There is at most one...")? This is not correct because the original statement does not assert the existence of anything while your sentence does.

This is also strange because now instead of "There exist an x with P(x)" you write $\displaystyle \forall xP(x)$.
Yes thank a typo mistake,i corrected my post

13. Thanks emakarov, it took a while for this to sink in. Really. Anyway, thanks again.