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Math Help - Universal Quantifier in Predicate Logic

  1. #1
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    Universal Quantifier in Predicate Logic

    Hi,

    I have the following:
    B(x) stands for "x is a baker",
    F(x) stands for "x is a farmer",
    K(x,y) stands for "x knows y".


    Ax: for all x
    Ex: there exists an x

    (a) Express "Everybody knows a farmer" in terms of predicate logic.

    For this I have:

    Ax(Ey (F(y) -> K(x,y))


    Which I believe translates to "For all X, there exists a farmer Y, such that X knows a Y". Which in turn implies that everybody knows a farmer.

    Now, the second question is:

    (b) "Everybody knows someone who knows a farmer"

    This one confuses me, as the same relation is involved twice.

    So far, I have:

    Ax(Ey K(x,y) -> (K(y,z) && Ez(F(z))

    But this does not seem right. How do I approach this question?
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  2. #2
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    Re: Universal Quantifier in Predicate Logic

    Quote Originally Posted by aprilrocks92 View Post
    I have the following:
    B(x) stands for "x is a baker",
    F(x) stands for "x is a farmer",
    K(x,y) stands for "x knows y".

    (a) Express "Everybody knows a farmer" in terms of predicate logic.
    Here is that one: \left( {\forall x} \right)\left( {\exists y} \right)\left[ {F(y) \wedge K(x,y)} \right]

    Now you post your effort on the other parts.
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  3. #3
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    Re: Universal Quantifier in Predicate Logic

    For all x, there exists a y,z such that if z is a farmer, and y knows z, then x knows y.

    (Ax)(Ey,z) [F(z) & K(z, y)]->K(x,y).

    As stated above, I'm rather confused as to how this is put.
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  4. #4
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    Re: Universal Quantifier in Predicate Logic

    Quote Originally Posted by aprilrocks92 View Post
    For all x, there exists a y,z such that if z is a farmer, and y knows z, then x knows y.
    (Ax)(Ey,z) [F(z) & K(z, y)]->K(x,y).
    What the devil are you on about?

    Look at the statement ""Everybody knows someone who knows a farmer""
    Where do you see an implication (an if...then) there?

    \left( {\forall x} \right)\left( {\exists y} \right)\left( {\exists z} \right)\left[ {K(x,y) \wedge F(z) \wedge K(y,z)} \right]
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