Bayes' Theorem with multiple random variables I'm reviewing some notes regarding probability, and the section regarding Conditional Probability gives the following example:
$P(X,Y|Z)=\frac{P(Z|X,Y)P(X,Y)}{P(Z)}=\frac{P(Y,Z|X)P(X)}{P(Z)}$
The middle expression is clearly just the application of Bayes' Theorem, but I can't see how the third expression is equal to the second.  Can someone please clarify how the two are equal?
 A: We know
$$P(X,Y)=P(X)P(Y|X)$$
and
$$P(Y,Z|X)=P(Y|X)P(Z|X,Y)$$
(to understand this, note that if you ignore the fact that everything is conditioned on $X$ then it is just like the first example). 
Therefore
\begin{align*}
P(Z|X,Y)P(X,Y)&=P(Z|X,Y)P(X)P(Y|X)\\
&=P(Y,Z|X)P(X)
\end{align*}
Which derives the third expression from the second.
(However I don't have any good intuition for what the third expression means. Does anyone else?)
A: We have
$$P(X,Y\mid Z) \tag1$$
Considering X and Y as a single event, we call them A. Now we have
$$P(A\mid Z) = P(X,Y\mid Z) \tag2$$
Using the Joint Probabilities Rule, we have
$$P(A,Z) = P(A\mid Z)\times P(Z) \tag3$$
So we can say that
$$P(A\mid Z) = \frac{P(A,Z)}{P(Z)} \tag4$$
We know that
$$P(A,Z) = P(Z,A) \tag5$$
Again using the Joint Probabilities Rule, we have
$$P(Z,A) = P(Z \mid A)\times P(A) \tag6$$
We defined $P(A)$ as the following
$$P(A) = P(X,Y) \tag7$$
Again using the Joint Probabilities Rule, we have
$$P(X,Y) = P(X\mid Y)\times P(Y) \tag8$$
Plugging $(8)$ into $(7)$, we have
$$P(A) = P(X\mid Y)\times P(Y) \tag9$$
Plugging $(9)$ into $(6)$, we have
$$P(Z,A) = P(Z\mid A)\times P(X\mid Y)\times P(Y) \tag{10}$$
Plugging $(10)$ into $(5)$ we have
$$P(A,Z) = P(Z\mid A)\times P(X\mid Y)\times P(Y) \tag{11}$$
Plugging $(11)$ into $(4)$, we have
$$P(A\mid Z) = \frac{P(Z\mid A)\times P(X\mid Y)\times P(Y)}{P(Z)} \tag{12}$$
Plugging $(12)$ into $(2)$, we have
$$P(X,Y\mid Z) = \frac{P(Z\mid A)\times P(X\mid Y)\times P(Y)}{P(Z)} \tag{13}$$
Observe that in $(13)$, using the Joint Probabilities Rule, we have
$$P(X,Y) = P(X\mid Y)\times P(Y) \tag{14}$$
Since we defined $P(A)$ is $P(X,Y)$, we have
$$P(A) = P(X\mid Y)\times P(Y) \tag{15}$$
Plugging $(15)$ into $(13)$, we have
$$P(X,Y\mid Z) = \frac{P(Z\mid A)\times P(A)}{P(Z)} \tag{16}$$
Observe that in $(16)$, using the Joint Probabilities Rule, we have
$$P(Z\mid A) = \frac{P(Z,A)}{P(A)} \tag{17}$$
Plugging $(17)$ into $(16)$, we have
$$P(X,Y\mid Z) = \frac{P(Z,A)}{P(Z)} \tag{18}$$
Now observe the following
$$P(Z,A) = P(Z,X,Y) = P(Y,Z,X) \tag{19}$$
Similar to what we did at the beginning, treating $Y$ and $Z$ as a single event and using the Joint Probabilities Rule, we have
$$P(Y,Z,X) = P(Y,Z\mid X)\times P(X) \tag{20}$$
Plugging $(20)$ into $(19)$, we have
$$P(Z,A) = P(Y,Z\mid X)\times P(X) \tag{21}$$
Plugging $(21)$ into $(18)$, we have
$$P(X,Y\mid Z) = \frac{P(Y,Z\mid X)\times P(X)}{P(Z)} \tag{22}$$
I don't know if this clarifies or complicates things more but nevertheless I wanted to include this here as well.
Right now, I can't prove why treating multiple joint events as if they were a single event is "legal".
A: It is easy to follow the following argumentation
$$P(Z|X,Y)P(X,Y) = \frac{P(X,Y,Z)}{P(X,Y)}P(X,Y) = P(X,Y,Z)=\frac{P(X,Y,Z)}{P(X)}P(X)=P(Y,Z|X)P(X).$$
