logical quantification

**bmp05** · November 25th 2009, 03:56 AM

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)."

$ (\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?

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

**apcalculus** · November 25th 2009, 03:59 AM

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)."

$ (\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?

$ (\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?

**bmp05** · November 25th 2009, 04:08 AM

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

so you mean something like:

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

can you write this as follows?

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

**emakarov** · November 25th 2009, 04:29 AM

Apcalculus gave the right hint.

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

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

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

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

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 $x$ such that $P(x)$ , then $Q(x)$ " is using the universal quantifier:

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

**bmp05** · November 25th 2009, 04:42 AM

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

**xalk** · November 25th 2009, 04:47 AM

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:

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

**emakarov** · November 25th 2009, 04:50 AM

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 $x$ such that $P(x)$ and $P(x)$ , then $x=x$ .

This would be

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

**emakarov** · November 25th 2009, 04:54 AM

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)

**xalk** · November 25th 2009, 04:58 AM

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 $x$ such that $P(x)$ and $P(x)$ , then $x=x$ .

This would be

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

Are you referring to my formula

**emakarov** · November 25th 2009, 05:04 AM

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:

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

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

**bmp05** · November 25th 2009, 05:04 AM

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.

**xalk** · November 25th 2009, 05:13 AM

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 $\forall xP(x)$ .

Yes thank a typo mistake,i corrected my post

**bmp05** · November 25th 2009, 10:47 AM

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

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