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Confusing Axiomatic Set Theory Proof

  1. Aug 30, 2013 #1
    This proof makes no sense to me.

    The theorem to be proved is

    Theorem 44. {x,y} = {u,v} → (x = u & y = v) V (x = v & y = u)

    where {x,y} and {u,v} are sets with exactly two members, which can be either sets or individuals. The proof relies on:

    Theorem 43. z [itex]\in[/itex] {x,y} z = x V z = y.


    The given proof is:

    "By virtue of Theorem 43
    u [itex]\in[/itex] {u,v},​
    and thus by the hypothesis of the theorem
    u [itex]\in[/itex] {x,y}.​
    Hence, by virtue of Th. 43 again
    (1)
    u = x V u = y.​
    By exactly similar arguments
    (2)
    v = x V v = y,​
    (3)
    x = u V x = v,​
    (4)
    y = u V y = v.​
    We may now consider two cases.
    Case 1: x = y. Then by virtue of (1), x = u, and by virtue of (2), y = v."

    This is where I got lost. Couldn't I just as easily argue that x = v by virtue of (2) and y = u by virtue of (1)? Or by virtue of (3) and (4)? What's the rationale behind the assumed values of x and y, and couldn't any of the four propositions support it? And if x = y, how could I justify the argument that x and y were equal to two ostensibly different variables without showing that u = v? On one, hand I can sort of see that the assumption x = y and the conditions (1) - (4) would necessarily make it true that u = v...but, given that then either x or y could be said to be equal to either v or u, would there be any need for this part of the proof at all?

    In the interest of completeness, the rest of the proof is:

    Case 2: x ≠ y. In view of (1), either x = u or y = u. Suppose x ≠ u. Then y = u and by (3) x = v. On the other hand, suppose y ≠ u. Then x = u and by (4), y = v.
     
  2. jcsd
  3. Sep 1, 2013 #2

    CompuChip

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    Science Advisor
    Homework Helper

    Yes, if x = y then basically the four propositions read (replace x and y by (x = y) everywhere):
    (1)' u = x = y V u = x = y
    (2)' v = x = y V v = x = y
    (3)' x = y = u V x = y = v
    (4)' x = y = u V x = y = v

    Which reduces to
    (1)'' u = x = y
    (2)'' v = x = y
    (3)'' x = y = u V x = y = v

    Which reduces to
    (1,2) u = x = y = v
    (3)'' x = y = u V x = y = v

    Which reduces to
    (1-3) x = y = u = v

    So indeed, you can use any of them to draw that conclusion.

    Actually, x = y is a pretty boring case, but you have to handle it :-)
     
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