# Is there a name for this set identity?

I proved that the following identity is true. It seems relatively simple and somewhat applicable, so I was wondering if there is a name for it (maybe something like De Morgan's Laws, for example).

$$(A \setminus B) \cup (B \setminus A) = (A \cup B) \setminus (A \cap B)$$.

• There isn't a name for every set identity... – MathematicsStudent1122 Feb 23 '17 at 5:36
• @MathematicsStudent1122 I am aware, but it just seemed possible to me that this one might have a name (and if so, I would like to know what it is). – pzp Feb 23 '17 at 5:38
• The wiki page on the symmetric difference gives no name. en.wikipedia.org/wiki/Symmetric_difference – Thomas Andrews Feb 23 '17 at 6:00
• I don't think the identity itself has a name, but it is describing two equivalent definitions of the symmetric difference of two sets. – Dan Simon Feb 23 '17 at 6:02
• Well, I was going to suggest we call it pzp's law.... – fleablood Feb 24 '17 at 0:31

## 2 Answers

Since you always can combine identities to come up with new identities they can't all have names. Of course you can come up with names as easy as you can come up with identities, but those don't get widely accepted as easily. This means that it's normally the most basic identities that get a name.

It's also the usefulness of identities that makes them get a name.

I would say that this identity fails in both categories, it's too complex and not that widely used.

One should note that in set algebra and boolean algebra at a more advanced level one can rely on truth tables to determine if an identity is valid so there is no actual need to manipulate expressions. One simple moves from one expression to the form that is convenient. This removes much of the need of names for identities.

There isn't a name, it is describing two equivalent definitions of the symmetric difference of two sets and you can prove the equivalence:

Definition: symmetric difference $$A \bigtriangleup B := (A \setminus B) \cup (B\setminus A)$$ Proof: $(A \cup B) \setminus (A \cap B)=(A \setminus B) \cup (B\setminus A)$ \begin{align*} (A \cup B)\setminus (A\cap B) &= ((A \cup B) \setminus A) \cup ((A \cup B) \setminus B)\\ &=(B \cup A) \setminus A) \cup ((A \cup B) \setminus B) \\ &= (B \setminus A) \cup (A \setminus B) \\ &=(A \setminus B) \cup (B\setminus A) \end{align*}