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Thread: Reading predicates and counterexample

  1. #1
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    Reading predicates and counterexample

    How do I read these two predicates:

    D = are natural numbers
    P(x) = x is even number
    Q(x) = x is odd number

    <br />
$\forall x (P(x)  \lor  Q(x))$<br /> <br />
$\forall x P(x)  \lor  \forall x Q(x) $<br /> <br />

    And the first predicate has a True value, the second has False. How?
    Last edited by Nforce; Nov 14th 2010 at 08:40 AM.
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  2. #2
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    because the first one reads that for all possible values of x (in the natural numbers), whatever value you take could be either odd or even. obviously true!

    the second reads that for all x, it is an even number and for all x (in another case!) it is an odd number. which is quite clearly false!

    the difference between the two statements is that the first gives one case being either odd or even (which is true, as any natural number you choose can only be odd or even) where the second statement gives two separate cases each claiming something is false..

    do you see?
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  3. #3
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    Yes I see, thank you.
    <br /> <br />
$\forall x (P(x)  \land  Q(x))$<br />

    so, this is also true.

    For all x in natural numbers there are odd and even. Is it right?
    Last edited by Nforce; Nov 14th 2010 at 08:44 AM.
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  4. #4
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    Quote Originally Posted by Nforce View Post
    Yes I see, thank you.
    <br /> <br />
$\forall x (P(x)  \land  Q(x))$<br />

    so, this is also true.

    For all x in natural numbers there are odd and even. Is it right?
    no unfortunately that is false! you have said there that x can be both odd and even. which is untrue
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