Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Predicate Logic Problem

  1. Nov 10, 2006 #1
    I have the sentence: No American who hasn't met any Canadian's knows Canada. The teacher gave the correct answer as being:

    Vx-Ex((Ax ^ Cy ^ Mxy) -> -Kxc)

    Would this version also work?:

    -ExEy(Ax ^ Kxc ^ Cy ^ Mxy)

    or is it supposed to be:

    -ExEy(Ax ^ Kxc ^ Cy ^ -Mxy)

    After thinking about it, I think it may be the second. Which means I got it wrong on the assignment.
     
  2. jcsd
  3. Nov 10, 2006 #2

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    A correct answer must be logically equivalent to:

    [tex]\neg (\exists x)(Ax\ \wedge \ (\forall y)(Cy\ \rightarrow \ \neg Mxy)\ \wedge \ Kxc)[/tex]

    The answer you said your teacher gave is:

    [tex](\forall x)\neg (\exists y)[(Ax \wedge Cy \wedge Mxy) \rightarrow \neg Kxc][/tex]

    which is logically equivalent to:

    [tex](\forall x)(\forall y)(Ax \wedge Cy \wedge Mxy \wedge Kxc)[/tex]

    which says: "Everybody is American, everybody is Canadian, everybody has met everyone, and everyone knows Canada" which is obviously not what the original English sentence says. Anyways, neither of your answers are logically equivalent to the answer your teacher supposedly gave, nor the answer I gave at the start of this post.
     
  4. Nov 13, 2006 #3
    Let U = the set of all Americans (presumably U.S. citizens).
    Let M(x) denote: x has never met a Canadian.
    Let K(x) denote: x does not know Canada.

    For any x [C(x)) -> K(x)] <=>
    There does not exit an x [C(x) and ~K(x)]
     
  5. Nov 13, 2006 #4
    Correction: Post #3 should read, C(x) instead of C(x)).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?