Element under Right Operation is Left Identity

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

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

Proof

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


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

whatever $$S$$ may be.

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


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

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

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

Also see

 * Left Operation All Elements Right Identities