I am reading Baader's paper "The Theory of Idempotent Semigroups is of Unification Type Zero" and I have a doubt about a property of idempotent semigroups.
Let $AI = \{(xy)z = x(yz), x^2 = x \}$.
Here is the part I didn't understand:
Some facts about Idempotent Subgroups
For a word $s$ let $\alpha(s)$ denote the alphabet of $s$, i.e., the set of symbols occuring in $s$. The following facts about $=_{AI}$ are well-known (see e.g. [1, 4, 5, 6]).
(2.1) - (Not related)
(2.2) - For any word $s$ let $l(s)$ be the shortest prefix of $s$ satisfying $\alpha(l(s)) = \alpha(s)$ and $r(s)$ be the shortest suffix of $s$ satisfying $\alpha(r(s)) = \alpha(s)$. Then $s =_{AI} t \leftrightarrow l(s) =_{AI} l(t) \text{ and } r(s) =_{AI} r(t)$.
(2.3) - (Not related)
(2.4) - (Not related)
I was able to understand properties (2.1), (2.3) and (2.4) [which I am not showing], but I was not able to prove the $\leftarrow$ part of (2.2).
I thought about a counterexample (that I don't know if it's correct) for it:
Let s = (xyz)x(xyz) and t = (xyz)y(xyz). Then: \begin{align} l(s) = xyz = l(t)\\ r(s) = xyz = r(t) \end{align} But I was not able to prove that $s = (xyz)x(xyz)$ is equal modulo AI to $(xyz)y(xyz) = t$.
All I noticed is that $s = (xyz)x(xyz) =_{AI} xyz$.
Can anyone tell me if my counterexample is correct or what is the mistake in it?
If it's wrong, can anyone help me with the proof of (2.2)?
Thanks in advance.
EDIT: The references [1, 4, 5, 6] are shown below. I could not access reference 1 (because it's in German) and 6 (couldn't find). I could not find the information in the references 4 and 5, but I don't know a lot of algebra so there may be something I am missing.
1 - (German reference)
4 - Howie, J.M (1976) "An Introduction to Semigroup Theory", Academic Press
5 - Kimura, N. (1958) "The structure of idempotent semigroups (I)", Pacific Journal of Math
6 - McLean, D. (1954) "Idempotent Semigroups", Am. Math. Mon 61.
EDIT2: My counterexample is not correct:
As mentioned before, $s =_{AI} xyz$. But, the chain that follows shows that $t =_{AI} xyz$ too. $$t = (xyz)y(xyz) = (xy)zyxyz = (xyxy)zyxyz = x \ (yxyz)(yxyz) = x (yxyz) = (xyxy)z = xyz $$