Subtree definition confusion Let $A$ be a finite prefix-closed set with operation $\cdot$ (so $A$ is a semigroup, "prefix-closed" means that if $x\cdot y\in A$, then $x\in A$ as well, it is not necessarily about strings, btw).
Consider the following set, which we call the subtree with root at the node $a$:
$B|_a=\{b\mid a\cdot b\in A\}$.
My question is: does this definition imply that:
$\forall a'=a\cdot b\in A\Rightarrow b\in B|_a$?
UPD, in other words, let us consider the following example:
let $A=\{1, 1\cdot 1,1\cdot 2,1\cdot 3\}$, consider $B|_1$. Will it be necessarily $\{1,2,3\}$, or it can be $\{1,2\}$ or $\{2,3\}$?
UPD-2: I expected that the answer is yes. But in this case, how I should properly describe/define the case when $B|_1$ can be $\{1,2\}$ as well?
 A: As I understand it, this is partially a notation question about set
builder notation. If there is no specified set which the
expression of the notation is iterating over (such as
the positive integers in $\{x \in \mathbb Z_{>0}: x < 10\}$)
the notation is frankly ambiguous.
So supposing there is some universal set $S$, this is how I read your definition of $B|_a$.
For a fixed $a$,
for all $b$ in $S$,
$b \in B|_a$ if
and only if  $a \cdot b \in A$.
So if $A=\{1, 1\cdot 1,1\cdot 2,1\cdot 3\}$, and $S = \{1,2,3\}$,
then $B|_1$ will be precisely $\{1,2,3\}$.
The other part of your question, if you wish to describe a set $B'$ such that
For a fixed $a$, if $b \in B'$ then
$a \cdot b \in A$.
You are then not defining a single set but a collection of sets.
The property that defines them is $B' \subseteq B|_a$.
A: The definition of $B|_a$ is the set $\{b \mid a \cdot b \in A\}$. This is just set-builder notation which means that if $a \cdot b \in A$, then we have $b \in B|_a$, so the answer to your question is yes.
Hence, if $A = \{1, 1 \cdot 1, 1 \cdot 2, 1 \cdot 3\}$, then $B|_1 = \{1, 2, 3\}$.
