Anticommutative Operation with Left Cancellative Element is Right Operation -- why? Seth Warner's "Modern Algebra" (1965) once again: Exercise 7.17.
We are given a semigroup $(S, \circ)$ such that:

*

*$(1): \quad \forall x, y \in S: x \circ y = y \circ x \implies x = y$


*$(2): \quad \exists z \in S: \forall x, y \in S: z \circ x = z \circ y \implies x = y$
That is:

*

*$\circ$ is anticommutative


*$S$ has a left cancellative element.
The object of the exercise is:
Prove that $\circ$ is what Warner calls the "right operation", that is:
$\forall x, y \in S: x \circ y = y$
(which seems to me the same thing as saying that $x \circ y$ is the second projection on $(x, y)$, although this is probably going to be excoriated as appalling abuse of notation.)
I have tried various approaches, like:

*

*assuming $x \circ z = z \circ x$ and seeing where that takes me (to $z = x$ and where from there?)

*starting from the end: $x \circ y = y$, and working backwards via $z \circ (x \circ y) = z \circ y$ which is derived from $(z \circ x) \circ y = z \circ y$, but trying to get there from the axioms escapes me.

Again, a hint will probably be all I need.
 A: First, note that anti-commutativity implies each element is idempotent, as
$$(x \circ x) \circ x = x \circ (x \circ x) \implies x = x \circ x.$$
Next, we show $z \circ y = y$. We have,
$$z \circ y = y \impliedby z \circ y \circ y = y \circ z \circ y \impliedby z \circ z \circ y \circ y = z\circ y \circ z \circ y \impliedby z \circ y = z \circ y,$$
the rightmost implication being true by idempotency. We also have
$$x \circ z = z \impliedby z \circ x \circ z = x \circ z \circ z \impliedby x \circ z = x \circ z.$$
We can therefore conclude from this that any $x$ is left-cancellative:
$$x \circ v = x \circ w \implies x \circ z \circ v = x \circ z \circ w \implies z \circ v = z \circ w \implies v = w.$$
Thus, the proof that $z \circ y = y$ works for any $x$, i.e. $x \circ y = y$ for all $x, y$.
A: Since $a$ and $a\circ a$ commute, we can conclude $a=a\circ a$, i.e. all $a\in S$ are idempotent.
(i) $z$ idempotent implies $z\circ z\circ x=z\circ x$, and left-cancelling $z$ gives $z\circ x=x$ for all $x\in S$.
(ii) If ever $x=x\circ z$ then we could replace the $x$ on the left side with $z\circ x$ to get $z\circ x=x\circ z$ which would imply that $x=z$. Therefore, $x\ne x\circ z$ for all $x\in S\setminus\{z\}$, i.e. the only fixed point of right-multiplication by $z$ is $z$ itself. But $x\circ z=x\circ z\circ z$ implies $x\circ z$ is a fixed point, so $x\circ z=z$ for all $x\in S$.
Finally, $x\circ y=x\circ (z\circ y)=(x\circ z)\circ y=z\circ y=y$. That is, $x\circ y=y$.
A: Here is my own attempt:
By associativity:

*

*$\forall a \in S: (a \circ a) \circ a = a \circ (a \circ a)$
By anticommutativity:

*

*$\forall a \in S: a \circ a = a$
that is idempotence.

Then by idempotence:

*

*$\forall a, b \in S: (a \circ a) \circ b \circ a = a \circ b \circ (a \circ a)$
By associativity:

*

*$a \circ (a \circ b \circ a) = (a \circ b \circ a) \circ a$
By anticommutativity:

*

*$(1): \quad a \circ b \circ a = a$
Now we have from $(1)$:

*

*$a \circ (c \circ b) \circ a \circ b = a \circ b$
and also from $(1)$:

*

*$a \circ c \circ (b \circ a \circ b) = a \circ c \circ b$
So:

*

*$\forall a, b, c \in S: a \circ c \circ b = a \circ b$
In particular, setting $a = z, c = x, b = y$:

*

*$\forall x, y, z \in S: z \circ x \circ y = z \circ y$
As we have asserted that $z$ is left cancellative:

*

*$\forall x, y \in S: x \circ y = y$
Ta-da.
The nudge that sent me down this route was the obvious fact of idempotence of an anticommutative associative operation (the first 2 lines).
Then I remembered those two crucial identities (included and proved in the above) that Warner had us prove back in Exercise $2.17$ of "Modern Algebra" where (now I look more closely) he actually refers us back to for that definition of anticommutativity:

*

*$\quad a \circ b \circ a = a$
and in particular as a result of this:

*

*$\forall x, y, z \in S: x \circ y \circ z = x \circ z$
