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There exists a unique

  1. Feb 11, 2005 #1
    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)?

  2. jcsd
  3. Feb 11, 2005 #2


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    I hope someone answers this question.
    [tex]\exists ! x (Px) \equiv_{df} \exists x (Px \wedge \forall y (Py \implies x=y))[/tex]
    It seems you could try to get the same quantifiers by using some clever negations and substitutions, especially:
    [tex]\neg (\exists x (Px)) \equiv \forall x (\neg Px)[/tex] and [tex]\neg (\forall x (Px)) \equiv \exists x (\neg Px)[/tex]
    For instance, negating the whole statement twice:
    [tex]\neg (\neg (\exists x (Px \wedge \forall y (Py \implies x=y))))[/tex]
    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).
    [tex]\exists ! x (Px) \equiv (\exists x (Px)) \wedge (\forall x \forall y ((Px \wedge Py) \implies x=y))[/tex]

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

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

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

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

    Note that [tex](\forall x \forall y ((Px \wedge Py) \implies x=y))[/tex] leaves open the possibility that Px is empty, IOW that no x such that P exists, IOW [itex]\forall x (\neg Px)[/itex].
    Last edited: Feb 11, 2005
  4. Feb 11, 2005 #3
    ok i was trying some stuff and this is what i got
    [tex]\exists x (Px) \wedge (\exists y (Py) \wedge x != y) [/tex]

    i think ^ is right too :\
  5. Feb 11, 2005 #4
    I don't think that looks right, but
    how's about

    [tex]\exists x P(x)\; \wedge \; \forall x,y ((P(x)\; \wedge \; P(y)) \implies (x = y))[/tex]
  6. Feb 12, 2005 #5


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    (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 :wink: )
    You could just distribute the third negation here:
    [tex]\neg (\neg (\exists x (Px)) \vee \neg(\forall x \forall y ((Px \wedge Py) \implies x=y)))[/tex],
    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 [tex]\neg (\forall x (Px)) \equiv \exists x (\neg Px)[/tex], then [tex]\neg (\neg (\forall x (Px))) \equiv \neg (\exists x (\neg Px))[/tex].
    So you can go back to the definition:
    [tex]\exists ! x (Px) \equiv_{df} \exists x (Px \wedge \forall y (Py \implies x=y))[/tex] and rewrite it using [tex]\forall x (Px) \equiv \neg (\exists x (\neg Px))[/tex]. The following is my best guess on how to proceed. Given:
    [tex]\exists x (Px \wedge \forall y (Py \implies x=y))[/tex]
    Replace the conditional:
    [tex]\exists x (Px \wedge \forall y (\neg Py \vee x=y))[/tex]
    Replace the universal quantifier (I'm making up a rule here):
    [tex]\exists x (Px \wedge \neg (\exists y (\neg (\neg Py \vee x=y))))[/tex]
    Distribute the second negation:
    [tex]\exists x (Px \wedge \neg (\exists y (Py \wedge x\not= y)))[/tex].
    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.
  7. Feb 12, 2005 #6


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    What does "x!" mean?
  8. Feb 12, 2005 #7
    I assume microtopian's
    means [tex]x \neq y[/tex]

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

    [tex]\exists x P(x)\; \wedge \; \neg \exists x,y ( P(x) \; \wedge \; P(y) \; \wedge \; (x \neq y))[/tex]


    [tex]\neg \forall x ( \neg P(x))\; \wedge \; \forall x,y ((P(x)\; \wedge \; P(y)) \implies (x = y))[/tex]

    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.
  9. Feb 12, 2005 #8


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    It doesn't hurt, and it's good practice. :biggrin:
  10. Feb 12, 2005 #9
    ooo... okay i get it now
    Thanks a bunch!!

    yea sorry about the \neg.. i couldnt find the latex code for "not equal to"
    (I'm a noob at latex)
  11. Feb 22, 2005 #10
    There is one and only one x such that (Px), is written E!x(Px), and,

    E!x(Px) <-> Ey(Ax(x=y <-> Px)).
  12. Feb 22, 2005 #11


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    Welcome to PF, Owen!! :biggrin:

    Hope you like it here- it's a great group of people. (BTW, you can click on any LaTex graphic like [tex]\mbox{this one}[/tex] to see the code and a link to the crash course.)
  13. Feb 23, 2005 #12
    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.

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