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Logic: demonstration

  1. Aug 23, 2004 #1
    How do you prove this is a function?

    g: P(A) > P(A)
    g(X) = X^-1

    X^-1 means inverse of X

    anyway, does anyone know how to do a formal demonstration for this?
  2. jcsd
  3. Aug 23, 2004 #2


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    You need to be a bit clearer. If this were NOT
  4. Aug 23, 2004 #3


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    You need to be a bit clearer. If this were NOT a "logic" question, I would assume you meant that g is a relation on the Power set of A. However, that would still leave open the question of what is meant by "inverse of X"- a set does not have an inverse.
  5. Aug 24, 2004 #4

    No, I do MEAN PROVE THAT IT IS A FUNCTION. somehow you first have to prove that for the domain there is a solution for every element.. and that there is Unity in the solutions for every element that x contains ;/

    Now what I mean that X^c... it´s hard to put the friggin´ notation on that thing but it´s defined something like this

    if xRy

    then xR^cy means that yRx
    that´s all :wink:

    (x,y) belongs R
    then (x,y) belongs R^-1 means (y,x) belongs R
  6. Aug 26, 2004 #5


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    You still haven't told us:

    What A is.

    What P(A) means.

    What X is!
  7. Aug 27, 2004 #6
    I though you were to know that ;\

    but here we go anyway, I think I'm getting it solved from another source, so when I'll get it, I'll post it =d

    P(A) - means Parts of set A

    so when I say P(A) -> P(A) I simply mean that the function goes from a domain of P(A) to a range of P(A)

    g(x) = X^c

    That's the function that we're talking about. I believe I explained that one... X is a set by the way

    What you have to prove????

    1. That under the function X^c any entry and exit belongs to P(A)

    2. That there's Unicity that is .....

    for all x1 x2(g(x1) != g(x2) -> x1 != x2)
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