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Quantifiers problem

  1. Nov 11, 2007 #1
    use of quantifiers

    1. The problem statement, all variables and given/known data


    a) Let P(x), Q(x) be open statements in the variable x with a given universe. prove that

    AxP(x)\/AxQ(x)=>Ax[P(x)\/Q(x)]
    [that is, prove that if the statement AxP(x)\/AxQ(x) is true then Ax[P(x)\/Q(x)] is true.

    b) find a counter example for the converse in part a, that is, find an open statment for P(x) and Q(x) and a universe such that Ax[P(x)\/Q(x)] is true while AxP(x)\/AxQ(x) is false



    2. Relevant equations

    (i will use \/ for "or", /\ for "and", Ax for "all x in a universe", Ex for (some x in a universe) )


    3. The attempt at a solution



    no idea how to start rreally...

    for a i thought maybe

    AxP(x)\/AxQ(x)
    P(c)\/Q(c)
    therefore Ax[P(x)\/Q(x)]

    but i totaly think that is a lame wrong asnwer haha

    and even if thats correct i have NO idea how to do b...as far as im concerned they are the same?!
     
  2. jcsd
  3. Nov 11, 2007 #2
  4. Nov 11, 2007 #3
    1. The problem statement, all variables and given/known data


    a) Let P(x), Q(x) be open statements in the variable x with a given universe. prove that

    AxP(x)\/AxQ(x)=>Ax[P(x)\/Q(x)]
    [that is, prove that if the statement AxP(x)\/AxQ(x) is true then Ax[P(x)\/Q(x)] is true.

    b) find a counter example for the converse in part a, that is, find an open statment for P(x) and Q(x) and a universe such that Ax[P(x)\/Q(x)] is true while AxP(x)\/AxQ(x) is false



    2. Relevant equations

    (i will use \/ for "or", /\ for "and", Ax for "all x in a universe", Ex for (some x in a universe) )


    3. The attempt at a solution



    no idea how to start rreally...

    for a i thought maybe

    AxP(x)\/AxQ(x)
    P(c)\/Q(c)
    therefore Ax[P(x)\/Q(x)]

    but i totaly think that is a lame wrong asnwer haha

    and even if thats correct i have NO idea how to do b...as far as im concerned they are the same?!
     
  5. Nov 12, 2007 #4

    HallsofIvy

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    ?? What is "c"? You haven't mentioned it before or defined it!
    Think about what AXP(x) means. If that's true what does it tell you about Ax(Px\/Q(x)).
    What does AxQ(x) mean? If that's true, what does it tell you about Ax(Px\/Q(x))?

    First, AxP(x)\/AxQ(x) is NOT the same as Ax(P(x)\/Q(x)). If you don't understand exactly what each side means, you have no hope of doing this problem! Second, what is the "converse" of this statement? What is the "converse" of any statement?

    For both (a) and (b) consider this example: the "universe" is the set of all people. P(x) is "x is a female" and Q(x) is "x is a male". Now what do both sides of your statement mean?
     
  6. Nov 12, 2007 #5
    C was from the rule of generalization (i forgot exact name)
    it states that if Axg(x) is true there exists a G(c) that is true, thats just how were supposed to start it whenevr it has Ax(of a function)



    i don't know the difference, that's half of what i was asking
    does AxP(x)\/AxQ(x) mean all of P(x) OR all of q(x) but not both....aka....all boys OR all girls..

    and then Ax(P(x)\/Q(x)) would be all the boys and girls combined

    or compare it to a set...lets P(x) be {1,2,3} Q(x) be {4,5,6}

    Ax(P(x)\/Q(x)). is {1,2,3,4,5,6} and AxP(x)\/AxQ(x) is {1,2,3} OR {4,5,6}

    is that the differencE?

    converse? don't know?
     
  7. Nov 12, 2007 #6
    Hah, I'm taking a similar class.

    I have no clue on how to do (a). But (b) is really easy. I'll give you a hint and say that the universe is the natural numbers or even the integers if you wish. Find two properties of numbers such that if one is true, then the other isn't. (take advantage of the "for all")

    edit: BTW, you may or maynot find this useful: linky
    The username and password is 250 ;)
     
  8. Nov 12, 2007 #7
    I think c is some specialization constant....
    I posted in the other thread you created.
     
  9. Nov 12, 2007 #8
    wow that formual sheet is like a god! haha
    we get to bring a double sided formula sheet in so that is like heaven sent!
    !!
    thanks
     
  10. Nov 13, 2007 #9

    HallsofIvy

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    There's something seriously wrong here! You apparently are asked to do problems in which you say you have no idea what the symbols or the words mean! I see three possiblities: (1)You are in a course with pre-requisites you have not mastered. (2) Your teacher is a sadist who enjoys giving students problems they can't possibly do! (3) You are simply not reading the text book carefully enough and/or are not listening carefully enough to the lectures. If the first is true, you should withdraw from the course. If the second is true you should withdraw from the course and complain to the dept. chair (I would recommend you check with other students to make sure you are right before doing that!). If the third is true, you need to work on your study skills!

    In your original post you quoted the question as "b) find a counter example for the converse in part a, that is, find an open statment for P(x) and Q(x) and a universe such that Ax[P(x)\/Q(x)] is true while AxP(x)\/AxQ(x) is false". The word "converse is used in that question. Since your problem asked you about the converse, I'm surprised you didn't immediately look up "converse". How can you expect to answer a question in which you don't know the meaning of some of the words?
    (If a statement is "If P then Q" then its "converse" is "If Q then P".)

    The words "that is" imply that the rest of the sentence is explaining the first part. You are asked to find an example that shows Ax[P(x)\/Q(x)] can be true while AxP(x)\/AxQ(x). Look at the "grouping" in Ax[P(x)\/Q(x)] the "Ax" ("for all x") is outside the braces while P(x)\/Q(x) is inside: "for all x" applies to everything inside the braces: that says "for all x, either P(x) is true or Q(x) is true". Now consider my example: P(x) means "x is a male", Q(x) means "x is a female". Ax[P(x)\/Q(x)] means "for every person x, either x is a male or x is a female". Is that true or false?

    On the other hand, AxP(x)\/AxQ(x) has two different "Ax"s. AxP(x) means "for every x, P(x) is true" or, in terms of my example, "for every person x, x is a male". AxQ(x) means "for every x, Q(x) is true" or, in terms of my example, "for every person x, x is a female". AxP(x)\/Ax Q(x) means "either, for every person x, x is a male, or, for every person x, x is a female". Is that true or false.
     
  11. Nov 13, 2007 #10

    HallsofIvy

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    Since this has been posted in both "Precalculus" and "Calculus and beyond", I am merging the two threads.
     
  12. Nov 13, 2007 #11

    i was super sick with an ear infection for 2.5 weeks and missed just as many lectures and im trying desperatly to catch up, but my ta/teacher had no office hours over the weekend, so i came here for help. ive bene attempting tocatch up in 6 hard courses and this just happens to tbe the last one im catching up in ( how it went)


    true i think? because if your male your not female, and visa versa, there are no other choices.
    false? i think this one states that everyonein he world is male, OR every one in the world is female, but if there exists a female there is no males. which is false.

    it said "that is..blah blah...so i assumed that was an example, or the converse, and so i didnt look it up
     
  13. Nov 13, 2007 #12
    Formally, let [tex] I = (A, \beta)[/tex] be an interpretation under arbitrary assignment mapping [tex]\beta[/tex]. Assume that [tex] I \vDash (\forall x P(x) \vee \forall x Q(x) ) [/tex]. This is true if and only if

    for all [tex] a \in A , \; I^{\frac{a}{x}}\vDash P(x)[/tex] or for all [tex] b \in A, \; I^{\frac{b}{x}}\vDash Q(x) [/tex]

    If it is the case that for all [tex]a\in A, \; I^{\frac{a}{x}}\vDash P(x) [/tex] then certainly [tex]I\vDash \forall x (P(x) \vee Q(x)) [/tex]

    Similarily, if for all [tex] b \in A, \; I^{\frac{b}{x}}\vDash Q(x) [/tex] then [tex]I\vDash \forall x ( P(x) \vee Q(x) )[/tex].

    Thus in either case we have [tex] I\vDash (\forall x P(x) \vee \forall x Q(x) ) \rightarrow \forall x ( P(x) \vee Q(x) ) \[/tex]

    Thus [tex] (\forall x P(x) \vee \forall x Q(x) ) \vDash \forall x ( P(x) \vee Q(x) ) [/tex] and the result follows.
     
    Last edited: Nov 13, 2007
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