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Math Help - quantifier

  1. #1
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    quantifier

    T(x) : x is in this class
    B(x) : x has a motor bike
    C(x) : x has a car

    how do I state:
    Only one person in this class has a motor bike

    my current answer is:

    \exists x\forall y[(x \neq y) \wedge T(x) \wedge B(x)]

    is this right?

    how about if there is at least two people in this class have a car

     \exists x\exists y[T(x) \wedge T(y) \wedge C(x) \wedge C(y)]
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  2. #2
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    Quote Originally Posted by TheRekz View Post
    T(x) : x is in this class
    B(x) : x has a motor bike
    C(x) : x has a car
    Only one person in this class has a motor bike
    There is at least two people in this class have a car
    For the first on I would expect:
     \left( {\exists x} \right)\left( {\forall y} \right)\left[ {\left( {T(x) \wedge B(x)} \right) \wedge \left( {\left( {T(y) \wedge B(y)} \right) \Rightarrow \left( {x = y} \right)} \right)} \right].

    The second one could be:
    \left( {\exists x} \right)\left( {\exists y} \right)\left[ {\left( {T(x) \wedge C(x)} \right) \wedge \left( {\left( {T(y) \wedge C(y)} \right) \wedge \left( {x \not= y} \right)} \right)} \right]
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  3. #3
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    Quote Originally Posted by Plato View Post
    For the first on I would expect:
     \left( {\exists x} \right)\left( {\forall y} \right)\left[ {\left( {T(x) \wedge B(x)} \right) \wedge \left( {\left( {T(y) \wedge B(y)} \right) \Rightarrow \left( {x = y} \right)} \right)} \right].

    The second one could be:
    \left( {\exists x} \right)\left( {\exists y} \right)\left[ {\left( {T(x) \wedge C(x)} \right) \wedge \left( {\left( {T(y) \wedge C(y)} \right) \wedge \left( {x \not= y} \right)} \right)} \right]
    for the first one would it be the same if I say y = x, rather than x = y??
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  4. #4
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    WELL, of course!
    It is always true that \left( {x = y} \right) \equiv \left( {y = x} \right)
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  5. #5
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    Quote Originally Posted by Plato View Post
    WELL, of course!
    It is always true that \left( {x = y} \right) \equiv \left( {y = x} \right)
    thanks Plato, it really helps.. what about the second one?? is my answer considered fine?
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  6. #6
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    Look at the answer I gave for the second one:
    \left( {\exists x} \right)\left( {\exists y} \right)\left[ {\left( {T(x) \wedge C(x)} \right) \wedge \left( {\left( {T(y) \wedge C(y)} \right) \wedge \left( {x \not= y} \right)} \right)} \right]
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  7. #7
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    Quote Originally Posted by Plato View Post
    Look at the answer I gave for the second one:
    \left( {\exists x} \right)\left( {\exists y} \right)\left[ {\left( {T(x) \wedge C(x)} \right) \wedge \left( {\left( {T(y) \wedge C(y)} \right) \wedge \left( {x \not= y} \right)} \right)} \right]
    yeah but I think it's almost similar from what I did
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  8. #8
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    But in what you did it is possible that x=y.
    In that case we don't have two, but maybe only one.
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