There Exists a Unique Solution?

[microtopian]

Let's say you have a proposition E! P(x) which means that there is only one unique x such that P(x) is true. How do you rewrite E!P(x) in terms of only the E quantifier (there exists at least one x) and another one in terms of only the A quantifier (for all x)?

Thanks

 

[honestrosewater]

I hope someone answers this question.
$$\exists ! x (Px) \equiv_{df} \exists x (Px \wedge \forall y (Py \implies x=y))$$
It seems you could try to get the same quantifiers by using some clever negations and substitutions, especially:
$$\neg (\exists x (Px)) \equiv \forall x (\neg Px)$$ and $$\neg (\forall x (Px)) \equiv \exists x (\neg Px)$$
For instance, negating the whole statement twice:
$$\neg (\neg (\exists x (Px \wedge \forall y (Py \implies x=y))))$$
and distributing the inner negation, but I don't know how to do that.

Okay, I've done some searching (and discovered I was right the first time I tried to answer the question- minor hooray).
$$\exists ! x (Px) \equiv (\exists x (Px)) \wedge (\forall x \forall y ((Px \wedge Py) \implies x=y))$$

Negate the whole statement twice $$(\neg (\neg p) \equiv p)$$:
$$\neg (\neg (\exists x (Px)) \wedge (\forall x \forall y ((Px \wedge Py) \implies x=y))))$$

Distribute the inner negation $$(\neg (p \wedge q) \equiv \neg p \vee \neg q)$$:
$$\neg (\neg (\exists x (Px)) \vee \neg(\forall x \forall y ((Px \wedge Py) \implies x=y)))$$

Substitute the negated existential $$(\neg (\exists x (Px)) \equiv \forall x (\neg Px))$$:
$$\neg (\forall x (\neg Px) \vee \neg(\forall x \forall y ((Px \wedge Py) \implies x=y)))$$

I think that's correct. Eh, the universal case is left to you.

Note that $$(\forall x \forall y ((Px \wedge Py) \implies x=y))$$ leaves open the possibility that Px is empty, IOW that no x such that P exists, IOW $\forall x (\neg Px)$.

 

[microtopian]

ok i was trying some stuff and this is what i got
$$\exists x (Px) \wedge (\exists y (Py) \wedge x != y)$$

i think ^ is right too :\

 

[gnome]

I don't think that looks right, but
how's about

$$\exists x P(x)\; \wedge \; \forall x,y ((P(x)\; \wedge \; P(y)) \implies (x = y))$$

 

[honestrosewater]

(Clue: when someone says, "such and such is left to the reader/student," it means they don't know how to do such and such )
You could just distribute the third negation here:
$$\neg (\neg (\exists x (Px)) \vee \neg(\forall x \forall y ((Px \wedge Py) \implies x=y)))$$,
but I don't know how to do that- I haven't gotten that far, and I can't see how to do it using the rules I know. However, Quine may have saved the day by only using the existential quantifier.
If $$\neg (\forall x (Px)) \equiv \exists x (\neg Px)$$, then $$\neg (\neg (\forall x (Px))) \equiv \neg (\exists x (\neg Px))$$.
So you can go back to the definition:
$$\exists ! x (Px) \equiv_{df} \exists x (Px \wedge \forall y (Py \implies x=y))$$ and rewrite it using $$\forall x (Px) \equiv \neg (\exists x (\neg Px))$$. The following is my best guess on how to proceed. Given:
$$\exists x (Px \wedge \forall y (Py \implies x=y))$$
Replace the conditional:
$$\exists x (Px \wedge \forall y (\neg Py \vee x=y))$$
Replace the universal quantifier (I'm making up a rule here):
$$\exists x (Px \wedge \neg (\exists y (\neg (\neg Py \vee x=y))))$$
Distribute the second negation:
$$\exists x (Px \wedge \neg (\exists y (Py \wedge x\not= y)))$$.
This reads, "There exists some x such that Px, and it is not the case that there exists some y such that Py and x does not equal y." That certainly sounds correct, even if I arrived at it incorrectly. Hopefully someone will check this.

 

[honestrosewater]

ok i was trying some stuff and this is what i got
$$\exists x (Px) \wedge (\exists y (Py) \wedge x != y)$$

i think ^ is right too :\

What does "x!" mean?

 

[gnome]

I assume microtopian's

$$x != y$$

means $$x \neq y$$


I see I misread the original question. You're looking for two different expressions that express $E!P(x)$, but each one using only one type of quantifier.

So:
$$\exists x P(x)\; \wedge \; \neg \exists x,y ( P(x) \; \wedge \; P(y) \; \wedge \; (x \neq y))$$

and

$$\neg \forall x ( \neg P(x))\; \wedge \; \forall x,y ((P(x)\; \wedge \; P(y)) \implies (x = y))$$

honest, I don't understand why you're going through all that work. microtopian didn't ask how to derive the expressions, just how to write them.

 

[honestrosewater]

honest, I don't understand why you're going through all that work. microtopian didn't ask how to derive the expressions, just how to write them.

It doesn't hurt, and it's good practice.

 

[microtopian]

ooo... okay i get it now
Thanks a bunch!

yea sorry about the \neg.. i couldn't find the latex code for "not equal to"
(I'm a noob at latex)

 

[Owen Holden]

Let's say you have a proposition E! P(x) which means that there is only one unique x such that P(x) is true. How do you rewrite E!P(x) in terms of only the E quantifier (there exists at least one x) and another one in terms of only the A quantifier (for all x)?

Thanks

There is one and only one x such that (Px), is written E!x(Px), and,

E!x(Px) <-> Ey(Ax(x=y <-> Px)).

 

[honestrosewater]

Welcome to PF, Owen!

Hope you like it here- it's a great group of people. (BTW, you can click on any LaTex graphic like $$\mbox{this one}$$ to see the code and a link to the crash course.)

 

[Owen Holden]

Welcome to PF, Owen!

Hope you like it here- it's a great group of people. (BTW, you can click on any LaTex graphic like $$\mbox{this one}$$ to see the code and a link to the crash course.)

Thanks for the welcome and for the LaTex link, I will try to use it but I am not very computer literate.

I have noticed many interesting threads here, and I hope I can contribute in a positive way.

Owen