TITLE: Why wrong or unacceptable to write Pipes on their own, without any probability?
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: 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