Why wrong or unacceptable to write Pipes on their own, without any probability? I see mixed messages. heropup commented $(C \mid B) \mid (A \mid B)$, but Michael Hardy chided that  "$\color{Red}{\text{There's no such thing as A∣B.}}$ When one writes Pr(A∣B), one is NOT writing about the probability of something that's called A∣B". Tom Loredo answered

Now suppose our information about the problem tells us that $A$ and $B$ are conditionally independent given $C$.  A conventional notation for this is:
$$
A \perp\!\!\!\perp B \,|\, C,
$$
which means (among other implications),
$
P(A|B,C) = P(A|C).$

But why is there $\color{Red}{\text{"no such thing as A∣B"}}$? A|B makes perfect sense to me. $A \mid B$ would simply mean the event A, given that and after the event B happened and anteceded A. A|B would meaningfully and gainfully differ from $A \cap B$, because $A \cap B$ reveals nothing about the echelon/order/tier  of A and B! But A|B and B|A do!
I retort Bey's rebuttal. $(A|B)|(C|D)$ can be construed  meaningfully! $(A|B)|(C|D)$  would simply mean A, given B, given C, given D. So D happened first. Then C happened after D. Then B happened after C, D. Finally A happened after B, C, D.
 A: There's nothing about timing in the notation $A\mid B.$
For example: an urn contain five red marbles and five blue marbles. Two marbles are to be chosen without replacement.
Let $A$ be the event that the first one chosen is red.
Let $B$ be the event that the second is red.
Then $\Pr(B\mid A) = \dfrac 4 9 = \Pr(A\mid B).$
One can choose the two marbles at the same time, and the "first" may then be, not the first one chosen, but the first one whose color is revealed. That does not alter the statement about the two conditional probabilities above.
A: The pipe cannot really be “chained” in a coherent way — the right hand side fixes the “universe” we are considering when thinking about events.
Another way to think about this is how would you represent the event $A|B|C$ using a Venm diagram? This notation is hopelessly ambiguous.
A more precise version of this is $ A|(B\cap C)$. However, what if we had $(A|B) | (C|D)$?
Note that $C|D$ is effectively telling us the same thing as $C \cap D$ in terms of events. The only difference is that the conditional expressions is telling us what serves as $\Omega$ in any probability calculation.
A: I think there is a sense in which A|B can be thought of as an event, but nothing to do with timing.  Rather, it relies on the construction of a new kind of algebra:
https://en.wikipedia.org/wiki/Goodman%E2%80%93Nguyen%E2%80%93Van_Fraassen_algebra
