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Math Help - Predicate calculus - propositions

  1. #1
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    Predicate calculus - propositions

    Let p(x,y) denote the predicate "x\leq y" and let q(x,y) denote "x<y" both with domain of definition N x N.

    \forall x \forall y \forall z  ((p(x,y) \wedge p(y,z)) \rightarrow (p(x,z))

    \forall x \forall y   (\exists z(q(x,z) \wedge q(z,y)) \rightarrow (q(x,y))

    \forall x \forall y   (\exists z(q(x,z) \wedge q(z,y)) \leftrightarrow (q(x,y))

    \exists x \forall y p(x,y)

    \exists y \forall x p(x,y)

    This is a question I've been given. The question asks for me to translate them all into good English phrases. This I believe I have done correctly. It also asks me to say whether the proposition is true for each.
    I have found that each is true, but I find this a bit suspicious, assuming at least one should be false, or maybe they're trying to throw me off
    So I was wondering, does anyone disagree with my finding that all are true?
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  2. #2
    Senior Member Danneedshelp's Avatar
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    \exists x \forall y p(x,y)

    \exists y \forall x p(x,y)

    For these, try fixing x and y and seeing if they hold.

    Suppose, for the second one, we choose some really large number; such as, y=10^10 . By doing so, I am saying I found a y such that every number x is less than or equal to this number. But, what do we know about the natural numbers? Can this be true?
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  3. #3
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    <br /> <br />
\exists x \forall y p(x,y)<br />



    This one is saying...that there is a natural number x, such that all natural numbers y are greater than or equal to x?
    if I fix 1 as x, then this is true. (assuming 1 is the lowest natural number)

    and

    <br /> <br />
\exists y \forall x p(x,y)<br />

    That one is saying, that there is a natural number y, such that all natural numbers x, are less than y?
    no matter what value you fix for y, this cannot be true..

    Soo, first is true and second is false?
    Also... in the instance given (Domain of definition N x N) or even in general, is 0 a natural number?

    Thanks so much Dan =]
    Last edited by yoonsi; August 2nd 2009 at 08:57 PM. Reason: typos =p
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  4. #4
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by yoonsi View Post
    <br /> <br />
\exists x \forall y p(x,y)<br />



    This one is saying...that there is a natural number x, such that all natural numbers y are greater than or equal to x?
    if I fix 1 as x, then this is true. (assuming 1 is the lowest natural number)

    and

    <br /> <br />
\exists y \forall x p(x,y)<br />

    That one is saying, that there is a natural number y, such that all natural numbers x, are less than y?
    no matter what value you fix for y, this cannot be true..

    Soo, first is true and second is false?
    Also... in the instance given (Domain of definition N x N) or even in general, is 0 a natural number?

    Thanks so much Dan =]
    Yes, you got it. As for the natural numbers, the definition does not include zero (1,2,3…).
    Last edited by Danneedshelp; August 2nd 2009 at 10:22 PM.
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