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!

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

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

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

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

-------------------------

What do examples contradict to this theorem?

Does the proof have a problem?

Thanks.