Element under Left Operation is Right Identity

Theorem
Let $$\left({S, \leftarrow}\right)$$ be an algebraic structure in which the operation $$\leftarrow$$ is the left operation.

Then no matter what $$S$$ is, $$\left({S, \leftarrow}\right)$$ is a semigroup all of whose elements are right identities.

Proof

 * It has been established that $$\leftarrow$$ is associative.


 * It is also immediately apparent that $$\left({S, \leftarrow}\right)$$ is closed, from the nature of the left operation:
 * $$\forall x, y \in S: x \leftarrow y = x \in S$$

whatever $$S$$ may be.

So $$\left({S, \leftarrow}\right)$$ is definitely a semigroup.


 * From the definition of left operation:
 * $$\forall x, y \in S: x \leftarrow y = x$$

from which it can immediately be seen that all elements of $$S$$ are indeed right identities.

From More than One Left Identity then No Right Identity, it also follows that there is no left identity.

Also see

 * Right Operation All Elements Left Identities