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Math Help - A contradiction about sets!

  1. #1
    Junior Member Researcher's Avatar
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    Exclamation A contradiction about sets!

    Hello all.
    I know the last expresion is not allways true(by making examples), but it can be proved that is true!

    x \in ( A \cap B) \equiv [( x \in A ) \wedge ( x \in B )]

    We know that to being the x \in ( A \cap B) a true expression, both the expressions right to the (first) equalency sign must be true. So we have:

    x \in ( A \cap B) \equiv [ ( x \in A ) \wedge ( x \in B) ] \equiv [ [ ( x \in A ) \wedge T ] \equiv ( x \in A ) ] \Rightarrow ( A \cap B ) = A

    It can be proved by a similar way that: ( A \cap B ) = B

    -------------------------
    What do examples contradict to this theorem?
    Does the proof have a problem?

    Thanks.
    Last edited by Researcher; September 8th 2009 at 03:09 PM.
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  2. #2
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    Quote Originally Posted by Researcher View Post
    Hello all.
    I know the last expresion is not allways true(by making examples), but it can be proved that is true!

    x \in a ( A \cap B) \equiv [( x \in A ) \wedge ( x \in B )]

    We know that to being the x \in ( A \cap B) a true expression, both the expressions right to the (first) equalency sign must be true. So we have:

    x \in ( A \cap B) \equiv [ ( x \in A ) \wedge ( x \in B) ] \equiv [ [ ( x \in A ) \wedge T ] \equiv ( x \in A ) ] \Rightarrow ( A \cap B ) = A

    It can be proved by a similar way that: ( A \cap B ) = B

    -------------------------
    What do examples contradict to this theorem?
    Does the proof have a problem?

    Thanks.
    Yes, there is a problem with the "theorem"! You have proved that "if x\in A\cap B then x\in A". That only proves A\cap B\subset A. In order to prove A\cap B= A, you would have to prove "if x\in A then x\in A\cap B" which you can't do.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    Yes, there is a problem with the "theorem"! You have proved that "if x\in A\cap B then x\in A". That only proves A\cap B\subset A. In order to prove A\cap B= A, you would have to prove "if x\in A then x\in A\cap B" which you can't do.
    No you are wrong!
    Because by propositional calculus we know that ( P \wedge T ) \equiv P. And We know that P \equiv Q is the same (P => Q and Q => P).
    (So i have proved that [( x\in A\cap B then x\in A) And ( x\in A then x\in A\cap B)]
    And we know from The Axiom of Extensoinality:
    If for every x, [( x\in A ) \equiv ( x\in B )] \Rightarrow  A = B

    You can see that i have proved that [x \in ( A \cap B) \equiv (  x \in A )] \Rightarrow ( A \cap B ) = A
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  4. #4
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    Quote Originally Posted by Researcher View Post
    No you are wrong!
    Because by propositional calculus we know that ( P \wedge T ) \equiv P. And We know that P \equiv Q is the same (P => Q and Q => P).
    (So i have proved that [( x\in A\cap B then x\in A) And ( x\in A then x\in A\cap B)]
    And we know from The Axiom of Extensoinality:
    If for every x, [( x\in A ) \equiv ( x\in B )] \Rightarrow  A = B

    You can see that i have proved that [x \in ( A \cap B) \equiv (  x \in A )] \Rightarrow ( A \cap B ) = A

    Take A = { 1 } ,B = { 2} ,THEN  A\cap B\neq A

    So your theorem is not correct
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  5. #5
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    I said in my first post that there are examples wich contradict with this theorem.
    But my question is: Why by using the same valid logical expresion(wich are also used to buld all the bases of set theory)does this theorem appear?
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  6. #6
    Super Member Matt Westwood's Avatar
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    By propositional calculus you indeed have that P \land Q \implies P but it is NOT true that P \implies P \land Q.

    So where you have (in your original expression): [[x \in A \land T] \equiv x \in A] the equivalence does not hold because you do not have [x \in A \implies [x \in A \land T]].


    That's where you have gone wrong.
    Last edited by Matt Westwood; September 8th 2009 at 11:42 PM. Reason: amplifying
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  7. #7
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    Quote Originally Posted by Matt Westwood View Post
    By propositional calculus you indeed have that P \land Q \implies P but it is NOT true that P \implies P \land Q.

    So where you have (in your original expression): [[x \in A \land T] \equiv x \in A] the equivalence does not hold because you do not have [x \in A \implies [x \in A \land T]].


    That's where you have gone wrong.
    No, you are wrong!!
    Your statement ( P \implies P \land Q is not true) holds whenever we don't know about validity of Q.(i mean we don't know it is true or false).
    But i've said clearly that Q = T here (wich T is a true statement, and here it is x \in B ).
    So the equvalency sign is i have chosen is TRUE.
    Because P \land T \implies P and P \implies P \land T.
    Last edited by Researcher; September 9th 2009 at 12:18 AM.
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    Quote Originally Posted by Researcher View Post
    Hello all.
    I know the last expresion is not allways true(by making examples), but it can be proved that is true!

    x \in ( A \cap B) \equiv [( x \in A ) \wedge ( x \in B )]

    We know that to being the x \in ( A \cap B) a true expression, both the expressions right to the (first) equalency sign must be true. So we have:

    x \in ( A \cap B) \equiv [ ( x \in A ) \wedge ( x \in B) ] \equiv [ [ ( x \in A ) \wedge T ] \equiv ( x \in A ) ] \Rightarrow ( A \cap B ) = A

    It can be proved by a similar way that: ( A \cap B ) = B

    -------------------------
    What do examples contradict to this theorem?
    Does the proof have a problem?

    Thanks.
    First of all we cannot mix a syntactical with semantical proof ,a thing that you have done in your proof.

    In a syntactical proof one uses the rules of logic without using true or false values for parts of formulas.

    In a semantical proof one uses the true tables .

    Mixing the two the results are wrong conclusions.

    Besides that, to have: [ ( x \in A ) \wedge ( x \in B) ] \equiv [ [ ( x \in A ) \wedge T ] we must have that :

    (x\in A)\wedge T) logicaly implies  (x\in A)\wedge (x\in B)

    ................................................OR.......................

    The conditional : [(x\in A)\wedge T]\Longrightarrow [(x\in A)\wedge (x\in B)] is a tautology.


    But it is not
    .

    Hence : [ ( x \in A ) \wedge ( x \in B) ] \equiv [ [ ( x \in A ) \wedge T ] does not hold .

    Thus your proof is wrong

    AND another more obvious thing: In the backwards proof ( the one starting with xεΑ ) When you substitute T = xεΒ ,this is wrong because you do not know now the value of xεΒ
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  9. #9
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by Matt Westwood View Post
    By propositional calculus you indeed have that P \land Q \implies P but it is NOT true that P \implies P \land Q.

    So where you have (in your original expression): [[x \in A \land T] \equiv x \in A] the equivalence does not hold because you do not have [x \in A \implies [x \in A \land T]].


    That's where you have gone wrong.
    I think I understand now (of course, "T" is "True" = "tautology")

    IF it is the case that x \in B \equiv T, which it only is if x is ALWAYS an element of B, then the equivalence indeed holds. But then that means A \subseteq B which is indeed equivalent to A \cup B = A.

    But first you have to take as a truth that x \in B ALL THE TIME, WHATEVER x and B may be.

    WHICH IS RUBBISH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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