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.

2. Originally Posted by Researcher
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.

3. Originally Posted by HallsofIvy
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 $[x \in ( A \cap B) \equiv ( x \in A )] \Rightarrow ( A \cap B ) = A$

4. Originally Posted by Researcher
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

5. 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?

6. 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.

7. Originally Posted by Matt Westwood
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 $P \implies P \land T$.

8. Originally Posted by Researcher
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 .

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

9. Originally Posted by Matt Westwood
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!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!