Random variables identities - how to make a formal proof.

Let $X, Y, Z$ be three random discrete variables. Consider the below random variables:

$A = X\vert Y\vert Z$ ,$B= X\vert Y,Z$

Question: Can I conclude that $A$ and $B$ are the same random variable? And if yes, how do I construct a formal proof?

Clarification of the notation:

We can define the random variable $X \vert Y$ as the random discrete variable which can take the value $X=x \vert Y = y$ with the probability of this outcome denoted by $P(X=x \vert Y = y)$.

We let $T = X \vert Y$. Now we introduce a new variable $Z$. From $Z, T$ we define two new variables:

The first one we denote by $A$ and define it as:

$A = T \vert Z$

The second one we denote by $B$ and define it as the multivariable

$B = (T, Z)$

My attempt

I have tried to show that $A$ and $B$ is the same variable. I have first deduce that $A$ is the same variable as $B$ if the following statement is true:

$P(X\vert Y \vert Z) = P(X \vert Y , Z)$

If $X_{1}, X_{2}$ are discrete random variables we have the below result:

$P( X_{1} \vert X_{2} ) = \frac{P(X_{1}, X_{2})}{P(X_{2})} \Leftrightarrow P(X_{1},X_{2}) = P(X_{2})P(X_{1}\vert X_{2})$

In this case we can obtain the following identity:

$P(X\vert Y \vert Z ) = \frac{P(X,Y\vert Z)}{P(Y\vert Z)} = \frac{P(X,Y\vert Z)P(Z)}{P(Y,Z)} = \frac{P(X,Y,Z)}{P(Y,Z)} = P(X\vert Y, Z)$

And hence we are done.

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Are you sure that A = X|Y|Z ? I've never seen that notation before. – TakeS Feb 20 '13 at 3:09
Please note that $X|Y,Z$ is $\sigma(Y,Z)-$measurable,while $X|Y|Z$ is only $\sigma(Z)-$measurable. Are you sure the conclusion is correct? – Coiacy Feb 20 '13 at 9:24
If you define the variable $W = Y\vert Z$ what is wrong with the definition of the variable $A = X \vert W$? I have not seen this notation, but I think it makes sense! – guestfromthepast Feb 24 '13 at 14:59
Neither X|Y|Z nor X|Y,Z make any sense to me. To ask a question involving these notions, you might want to first define them as precisely as possible. – Did Feb 25 '13 at 21:56
@guestfromthepast I think you're confusing the notation: $X\vert Y$ is not a random variable, rather, $P(\cdot \vert \cdot)$ is the notation for conditional probability, so far as I know. Admittedly, I don't really dabble in probability, but... – tomasz Feb 28 '13 at 12:24