I have a problem with this definition of anticommutative operators. I found the following: An operation $\circledast$ is called anticommutative if it satisfies the following:
(i) There is a right identity element $r:=r_x$, that is, $\exists r \in \mathbb{X}: x \circledast r = x, x \in \mathbb{X}$.
(ii) $x \circledast y = r \Leftrightarrow (x\circledast y)\circledast (y \circledast x) = r \Leftrightarrow x = y $ for all $x,y \in X$.
Now, my problem is, if you consider a set $\mathbb{X}$ with more than one element, and you choose $x \neq r, y=r$, the following happens:
None of the equivalent equations are true (because of $x \neq y$), so equation 2 says: $(x\circledast r)\circledast (r \circledast x) \neq r$ which is $x \circledast x \neq r$, but this is a contradiction to the equivalence of equation 1 ($x \circledast y = r$) and equation 3 ($x = y $ for all $x,y \in X$).
Is the definition correct?
Thx in advance.
In saying $(x \circledast r) \circledast (r \circledast x) \neq r$ is equivalent to $x \circledast x \neq r$, you are assuming the operator is associative:
$$(x \circledast r) \circledast (r \circledast x) = x \circledast (r \circledast x) = (x \circledast r) \circledast x = x \circledast x$$
Take as an example $x \circledast y = x-y$ on the (say) real numbers. Then the axioms are satisfied with $r = 0$.