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Is this statement true?

  1. Mar 15, 2015 #1
    I am starting to learn propositional calculus and am trying to make sense of the notation. I am trying to express the idea that sets A and B are equivalent. I want to know if the following statement is true and if it shows three equally valid ways of saying that A and B are the same set.

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    Thank you for your time. Any help and/or recommendations would be greatly appreciated.

    Edit : Looking back at it, I think the first part does not imply that there are no elements of B that are not also in A. It does not eliminate the possibility that A is a subset of B. Should I write :

    gif.gif

    ?
     
    Last edited: Mar 15, 2015
  2. jcsd
  3. Mar 16, 2015 #2

    Bacle2

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    Maybe a more direct way would be : ## x \in A ## iff ## x \in B ##.
     
  4. Mar 16, 2015 #3
    Doesn't that only say that all elements of A are also elements of B, making A a subset of B, and not necessarily equivalent to B? Or does using ''iff'' imply that ## x \in B ## iff ## x \in A ## ?
    Also, I understand that the way I put it isn't the most direct way of doing it, but I want to know if my usage of these symbols and operators makes sense.

    Thank you for your time.
     
  5. Mar 16, 2015 #4
    If you move the negations inside of
    you get the axiom of extensionality of Zermelo-Fraenkel. That is, this "iff" is valid.
    But it is unclear what you mean by "equivalent". Equivalence requires a relation. Do you mean "equivalent under the relation of equality"? Then that "iff" would be (trivially) valid. But if you mean, say, equinumerability as your equivalence relation, then the implication only goes in one direction. So, what do you mean by "equivalent"?
     
  6. Mar 16, 2015 #5
    I mean ''equivalent under the relation of equality'' as in ''A and B are the same object''. Because A and B share not only the same cardinality, but also the same elements.
    So if A = {p, q, r, t}, then B = {p, q, r, t} also, and thus A=B.

    Edit : I'm currently looking at the Zermelo-Fraenkel axioms. That's very helpful, thank you!
     
    Last edited: Mar 16, 2015
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