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Math Help - logic sets predicates calculus

  1. #1
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    logic sets predicates calculus

    i need help on this question
    Attached Thumbnails Attached Thumbnails logic sets predicates calculus-6.jpg  
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  2. #2
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    Quote Originally Posted by srk619 View Post
    i need help on this question
    \forall x\forall y[ O(x) +O(y) = E(x+y)]

    Now the negation of :

    .......... \exists m[ m\in Z\wedge O(2m)] is.............

    ............ \forall m[\neg( m\in Z\wedge O(2m)]......................

    Now drop \forall,m and we are left with:

    ............~  (m\in Z\wedge O(2m)).................................................. .....................................1

    .But (1) is equivalent to:


    ....................... [m\in Z\rightarrow\neg O(2m).................................................. .......................................2


    And introducing back the for all quantifier we get:


    .................. \forall m[ m\in z\rightarrow\neg O(2m)].................................................. ........................................3


    FOR all integers 2m is not odd............................................... ...........
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  3. #3
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    can anyone help to fill in the blanks
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  4. #4
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    The negation is correct and well detailled, no blank to fill in!

    For a), O and E are predicates, so O(x)+O(y)=E(x+y) has no sense. But the answer is not so far from the formula proposed, only two symbols to change (and maybe a pair of parenthesis to add).
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