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# interchange law

Let $S$ be a set and $\circ$ and $\bullet$ two partial binary operations on $S$. Then $\circ$ and $\bullet$ are said to satisfy the *interchange law* if

$(a\circ b)\bullet(c\circ d)=(a\bullet c)\circ(b\bullet d),$ |

provided that the operations are defined on both sides of the equation.

An element $e\in S$ is a $\circ$-identity, or an identity with respect to $\circ$, if $e\circ a=a\circ e=a$ provided the operations are defined.

###### Proposition 1.

If $\circ$ is a total function (defined for all of $S\times S$), then there is at most one $\circ$-identity.

###### Proof.

If $e$ and $f$ are both $\circ$-idenities, then $e=e\circ f=f$. ∎

###### Proposition 2.

If both $\circ$ and $\bullet$ are total functions, and identities exist and the same with respect to both operations, then $\circ=\bullet$ and is commutative.

###### Proof.

Suppose that $e$ is both the $\circ$-identity and the $\bullet$-identity. Then, according to the interchange law, $a\bullet d=(a\circ e)\bullet(e\circ d)=(a\bullet e)\circ(e\bullet d)=a\circ d$, showing that $\bullet=\circ$. Again, using the interchange law, $a\bullet d=(e\circ a)\bullet(d\circ e)=(e\bullet d)\circ(a\bullet e)=d\circ a=% d\bullet a$, showing that $\bullet$ is commutative. ∎

## Mathematics Subject Classification

18A99*no label found*08A99

*no label found*

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