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Thread: Questions and more questions

  1. #1
    Oct 2008

    Questions and more questions

    Consider the following wff (∀y)(∃x)Q(x,y)→(∃x)(∀y)Q(x,y)

    Find an interpretation to prove this wff is not valid

    Find the flaw in the following "proof" of this wff.

    I am not sure how to disprove this or where the flaw is.

    This problem I know is invalid, again, I am not sure how to prove it is invalid.

    (∀x)P(x) V (∃x) Q(x)→(∀x)[P(x) V Q(x)]

    My textbook does not have answers to exercises in the back. Therefore, I have no way of knowing whether I am doing these problems correctly. I would like to see, hopefully correct solutions, so I know what I am doing and gain confidence.
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  2. #2
    Aug 2008

    In (∀y)(∃x)Q(x,y), you choose x according to a given y, whereas in (∃x)(∀y)Q(x,y) you can't change x once you fix it. For example, when Q(x,y) means "x is wiser than y", (∀y)(∃x)Q(x,y) means that any y can be beaten by some x, whereas (∃x)(∀y)Q(x,y) means that there is some kind of god x(the wisest one) who is superior to every y. Try to interpret this into some mathematical example. To discover the flaw, interpret each line of the "proof " to know which line fails.

    (∃x) Q(x) (and hence (∀x)P(x) V (∃x) Q(x)) happens even if Q is something rare, occuring for only one instance say, a. So, Q(a) occurs but Q(x) is false for all x except the case x=a. Then, it is obvious that you can't expect [P(x) V Q(x)] to occur for all x (since nothing is assumed for x except for the case x=a.)

    There are many ways to disprove a wff (showing counterexamples, or using tableaux,...). I don't know which ones are used in your book.

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