List of set identities and relations

This article lists mathematical properties and laws of sets, involving the set-theoretic operations of union, intersection, and complementation and the relations of set equality and set inclusion. It also provides systematic procedures for evaluating expressions, and performing calculations, involving these operations and relations.

The binary operations of set union (${\displaystyle \cup }$) and intersection (${\displaystyle \cap }$) satisfy many identities. Several of these identities or "laws" have well established names.

Throughout this article, capital letters (such as ${\displaystyle A,B,C,L,M,R,S,}$ and ${\displaystyle X}$) will denote sets. On the left hand side of an identity, typically,

• ${\displaystyle L}$ will be the leftmost set,
• ${\displaystyle M}$ will be the middle set, and
• ${\displaystyle R}$ will be the rightmost set.

This is to facilitate applying identities to expressions that are complicated or use the same symbols as the identity.^{[note 1]} For example, the identity $${\displaystyle (L\,\setminus \,M)\,\setminus \,R~=~(L\,\setminus \,R)\,\setminus \,(M\,\setminus \,R)}$$ may be read as: $${\displaystyle ({\text{Left set}}\,\setminus \,{\text{Middle set}})\,\setminus \,{\text{Right set}}~=~({\text{Left set}}\,\setminus \,{\text{Right set}})\,\setminus \,({\text{Middle set}}\,\setminus \,{\text{Right set}}).}$$

For sets ${\displaystyle L}$ and ${\displaystyle R,}$ define: $${\displaystyle {\begin{alignedat}{4}L\cup R&&~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\{~x~:~x\in L\;&&{\text{ or }}\;\,&&\;x\in R~\}\\L\cap R&&~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\{~x~:~x\in L\;&&{\text{ and }}&&\;x\in R~\}\\L\setminus R&&~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\{~x~:~x\in L\;&&{\text{ and }}&&\;x\notin R~\}\\\end{alignedat}}}$$ and $${\displaystyle L\triangle R~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\{~x~:~x{\text{ belongs to exactly one of }}L{\text{ and }}R~\}}$$ where the symmetric difference ${\displaystyle L\triangle R}$ is sometimes denoted by ${\displaystyle L\ominus R}$ and equals:^[1][2] $${\displaystyle {\begin{alignedat}{4}L\;\triangle \;R~&=~(L~\setminus ~&&R)~\cup ~&&(R~\setminus ~&&L)\\~&=~(L~\cup ~&&R)~\setminus ~&&(L~\cap ~&&R).\end{alignedat}}}$$

One set ${\displaystyle L}$ is said to intersect another set ${\displaystyle R}$ if ${\displaystyle L\cap R\neq \varnothing .}$ Sets that do not intersect are said to be disjoint.

The power set of ${\displaystyle X}$ is the set of all subsets of ${\displaystyle X}$ and will be denoted by $${\displaystyle \wp (X)~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\{~L~:~L\subseteq X~\}.}$$

Universe set and complement notation

The notation $${\displaystyle L^{\complement }~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~X\setminus L.}$$ may be used if ${\displaystyle L}$ is a subset of some set ${\displaystyle X}$ that is understood (say from context, or because it is clearly stated what the superset ${\displaystyle X}$ is). It is emphasized that the definition of ${\displaystyle L^{\complement }}$ depends on context. For instance, had ${\displaystyle L}$ been declared as a subset of ${\displaystyle Y,}$ with the sets ${\displaystyle Y}$ and ${\displaystyle X}$ not necessarily related to each other in any way, then ${\displaystyle L^{\complement }}$ would likely mean ${\displaystyle Y\setminus L}$ instead of ${\displaystyle X\setminus L.}$

If it is needed then unless indicated otherwise, it should be assumed that ${\displaystyle X}$ denotes the universe set, which means that all sets that are used in the formula are subsets of ${\displaystyle X.}$ In particular, the complement of a set ${\displaystyle L}$ will be denoted by ${\displaystyle L^{\complement }}$ where unless indicated otherwise, it should be assumed that ${\displaystyle L^{\complement }}$ denotes the complement of ${\displaystyle L}$ in (the universe) ${\displaystyle X.}$

Assume ${\displaystyle L\subseteq X.}$

Identity:^[3]

Definition: ${\displaystyle e}$ is called a left identity element of a binary operator ${\displaystyle \,\ast \,}$ if ${\displaystyle e\,\ast \,R=R}$ for all ${\displaystyle R}$ and it is called a right identity element of ${\displaystyle \,\ast \,}$ if ${\displaystyle L\,\ast \,e=L}$ for all ${\displaystyle L.}$ A left identity element that is also a right identity element if called an identity element.

The empty set ${\displaystyle \varnothing }$ is an identity element of binary union ${\displaystyle \cup }$ and symmetric difference ${\displaystyle \triangle ,}$ and it is also a right identity element of set subtraction ${\displaystyle \,\setminus :}$

$${\displaystyle {\begin{alignedat}{10}L\cap X&\;=\;&&L&\;=\;&X\cap L~~~~{\text{ where }}L\subseteq X\\[1.4ex]L\cup \varnothing &\;=\;&&L&\;=\;&\varnothing \cup L\\[1.4ex]L\,\triangle \varnothing &\;=\;&&L&\;=\;&\varnothing \,\triangle L\\[1.4ex]L\setminus \varnothing &\;=\;&&L\\[1.4ex]\end{alignedat}}}$$ but ${\displaystyle \varnothing }$ is not a left identity element of ${\displaystyle \,\setminus \,}$ since $${\displaystyle \varnothing \setminus L=\varnothing }$$ so ${\textstyle \varnothing \setminus L=L}$ if and only if ${\displaystyle L=\varnothing .}$

Idempotence^[3] ${\displaystyle L\ast L=L}$ and Nilpotence ${\displaystyle L\ast L=\varnothing }$:

$${\displaystyle {\begin{alignedat}{10}L\cup L&\;=\;&&L&&\quad {\text{ (Idempotence)}}\\[1.4ex]L\cap L&\;=\;&&L&&\quad {\text{ (Idempotence)}}\\[1.4ex]L\,\triangle \,L&\;=\;&&\varnothing &&\quad {\text{ (Nilpotence of index 2)}}\\[1.4ex]L\setminus L&\;=\;&&\varnothing &&\quad {\text{ (Nilpotence of index 2)}}\\[1.4ex]\end{alignedat}}}$$

Domination^[3]/Absorbing element:

Definition: ${\displaystyle z}$ is called a left absorbing element of a binary operator ${\displaystyle \,\ast \,}$ if ${\displaystyle z\,\ast \,R=z}$ for all ${\displaystyle R}$ and it is called a right absorbing element of ${\displaystyle \,\ast \,}$ if ${\displaystyle L\,\ast \,z=z}$ for all ${\displaystyle L.}$ A left absorbing element that is also a right absorbing element if called an absorbing element. Absorbing elements are also sometime called annihilating elements or zero elements.

A universe set is an absorbing element of binary union ${\displaystyle \cup .}$ The empty set ${\displaystyle \varnothing }$ is an absorbing element of binary intersection ${\displaystyle \cap }$ and binary Cartesian product ${\displaystyle \times ,}$ and it is also a left absorbing element of set subtraction ${\displaystyle \,\setminus :}$

$${\displaystyle {\begin{alignedat}{10}X\cup L&\;=\;&&X&\;=\;&L\cup X~~~~{\text{ where }}L\subseteq X\\[1.4ex]\varnothing \cap L&\;=\;&&\varnothing &\;=\;&L\cap \varnothing \\[1.4ex]\varnothing \times L&\;=\;&&\varnothing &\;=\;&L\times \varnothing \\[1.4ex]\varnothing \setminus L&\;=\;&&\varnothing &\;\;&\\[1.4ex]\end{alignedat}}}$$ but ${\displaystyle \varnothing }$ is not a right absorbing element of set subtraction since $${\displaystyle L\setminus \varnothing =L}$$ where ${\textstyle L\setminus \varnothing =\varnothing }$ if and only if ${\textstyle L=\varnothing .}$

Double complement or involution law:

$${\displaystyle {\begin{alignedat}{10}X\setminus (X\setminus L)&=L&&\qquad {\text{ Also written }}\quad &&\left(L^{\complement }\right)^{\complement }=L&&\quad &&{\text{ where }}L\subseteq X\quad {\text{ (Double complement/Involution law)}}\\[1.4ex]\end{alignedat}}}$$

$${\displaystyle L\setminus \varnothing =L}$$ $${\displaystyle {\begin{alignedat}{4}\varnothing &=L&&\setminus L\\&=\varnothing &&\setminus L\\&=L&&\setminus X~~~~{\text{ where }}L\subseteq X\\\end{alignedat}}}$$^[3]

$${\displaystyle L^{\complement }=X\setminus L\quad {\text{ (definition of notation)}}}$$

$${\displaystyle {\begin{alignedat}{10}L\,\cup (X\setminus L)&=X&&\qquad {\text{ Also written }}\quad &&L\cup L^{\complement }=X&&\quad &&{\text{ where }}L\subseteq X\\[1.4ex]L\,\triangle (X\setminus L)&=X&&\qquad {\text{ Also written }}\quad &&L\,\triangle L^{\complement }=X&&\quad &&{\text{ where }}L\subseteq X\\[1.4ex]L\,\cap (X\setminus L)&=\varnothing &&\qquad {\text{ Also written }}\quad &&L\cap L^{\complement }=\varnothing &&\quad &&\\[1.4ex]\end{alignedat}}}$$^[3]

$${\displaystyle {\begin{alignedat}{10}X\setminus \varnothing &=X&&\qquad {\text{ Also written }}\quad &&\varnothing ^{\complement }=X&&\quad &&{\text{ (Complement laws for the empty set))}}\\[1.4ex]X\setminus X&=\varnothing &&\qquad {\text{ Also written }}\quad &&X^{\complement }=\varnothing &&\quad &&{\text{ (Complement laws for the universe set)}}\\[1.4ex]\end{alignedat}}}$$

In the left hand sides of the following identities, ${\displaystyle L}$ is the L eft most set and ${\displaystyle R}$ is the R ight most set. Assume both ${\displaystyle L{\text{ and }}R}$ are subsets of some universe set ${\displaystyle X.}$

In the left hand sides of the following identities, L is the L eft most set and R is the R ight most set. Whenever necessary, both L and R should be assumed to be subsets of some universe set X, so that ${\displaystyle L^{\complement }:=X\setminus L{\text{ and }}R^{\complement }:=X\setminus R.}$

$${\displaystyle {\begin{alignedat}{9}L\cap R&=L&&\,\,\setminus \,&&(L&&\,\,\setminus &&R)\\&=R&&\,\,\setminus \,&&(R&&\,\,\setminus &&L)\\&=L&&\,\,\setminus \,&&(L&&\,\triangle \,&&R)\\&=L&&\,\triangle \,&&(L&&\,\,\setminus &&R)\\\end{alignedat}}}$$

$${\displaystyle {\begin{alignedat}{9}L\cup R&=(&&L\,\triangle \,R)&&\,\,\cup &&&&L&&&&\\&=(&&L\,\triangle \,R)&&\,\triangle \,&&(&&L&&\cap \,&&R)\\&=(&&R\,\setminus \,L)&&\,\,\cup &&&&L&&&&~~~~~{\text{ (union is disjoint)}}\\\end{alignedat}}}$$

$${\displaystyle {\begin{alignedat}{9}L\,\triangle \,R&=&&R\,\triangle \,L&&&&&&&&\\&=(&&L\,\cup \,R)&&\,\setminus \,&&(&&L\,\,\cap \,R)&&\\&=(&&L\,\setminus \,R)&&\cup \,&&(&&R\,\,\setminus \,L)&&~~~~~{\text{ (union is disjoint)}}\\&=(&&L\,\triangle \,M)&&\,\triangle \,&&(&&M\,\triangle \,R)&&~~~~~{\text{ where }}M{\text{ is an arbitrary set. }}\\&=(&&L^{\complement })&&\,\triangle \,&&(&&R^{\complement })&&\\\end{alignedat}}}$$

$${\displaystyle {\begin{alignedat}{9}L\setminus R&=&&L&&\,\,\setminus &&(L&&\,\,\cap &&R)\\&=&&L&&\,\,\cap &&(L&&\,\triangle \,&&R)\\&=&&L&&\,\triangle \,&&(L&&\,\,\cap &&R)\\&=&&R&&\,\triangle \,&&(L&&\,\,\cup &&R)\\\end{alignedat}}}$$

De Morgan's laws state that for ${\displaystyle L,R\subseteq X:}$

$${\displaystyle {\begin{alignedat}{10}X\setminus (L\cap R)&=(X\setminus L)\cup (X\setminus R)&&\qquad {\text{ Also written }}\quad &&(L\cap R)^{\complement }=L^{\complement }\cup R^{\complement }&&\quad &&{\text{ (De Morgan's law)}}\\[1.4ex]X\setminus (L\cup R)&=(X\setminus L)\cap (X\setminus R)&&\qquad {\text{ Also written }}\quad &&(L\cup R)^{\complement }=L^{\complement }\cap R^{\complement }&&\quad &&{\text{ (De Morgan's law)}}\\[1.4ex]\end{alignedat}}}$$

Unions, intersection, and symmetric difference are commutative operations:^[3]

$${\displaystyle {\begin{alignedat}{10}L\cup R&\;=\;&&R\cup L&&\quad {\text{ (Commutativity)}}\\[1.4ex]L\cap R&\;=\;&&R\cap L&&\quad {\text{ (Commutativity)}}\\[1.4ex]L\,\triangle R&\;=\;&&R\,\triangle L&&\quad {\text{ (Commutativity)}}\\[1.4ex]\end{alignedat}}}$$

Set subtraction is not commutative. However, the commutativity of set subtraction can be characterized: from ${\displaystyle (L\,\setminus \,R)\cap (R\,\setminus \,L)=\varnothing }$ it follows that: $${\displaystyle L\,\setminus \,R=R\,\setminus \,L\quad {\text{ if and only if }}\quad L=R.}$$ Said differently, if distinct symbols always represented distinct sets, then the only true formulas of the form ${\displaystyle \,\cdot \,\,\setminus \,\,\cdot \,=\,\cdot \,\,\setminus \,\,\cdot \,}$ that could be written would be those involving a single symbol; that is, those of the form: ${\displaystyle S\,\setminus \,S=S\,\setminus \,S.}$ But such formulas are necessarily true for every binary operation ${\displaystyle \,\ast \,}$ (because ${\displaystyle x\,\ast \,x=x\,\ast \,x}$ must hold by definition of equality), and so in this sense, set subtraction is as diametrically opposite to being commutative as is possible for a binary operation. Set subtraction is also neither left alternative nor right alternative; instead, ${\displaystyle (L\setminus L)\setminus R=L\setminus (L\setminus R)}$ if and only if ${\displaystyle L\cap R=\varnothing }$ if and only if ${\displaystyle (R\setminus L)\setminus L=R\setminus (L\setminus L).}$ Set subtraction is quasi-commutative and satisfies the Jordan identity.

Absorption laws:

$${\displaystyle {\begin{alignedat}{4}L\cup (L\cap R)&\;=\;&&L&&\quad {\text{ (Absorption)}}\\[1.4ex]L\cap (L\cup R)&\;=\;&&L&&\quad {\text{ (Absorption)}}\\[1.4ex]\end{alignedat}}}$$

Other properties

$${\displaystyle {\begin{alignedat}{10}L\setminus R&=L\cap (X\setminus R)&&\qquad {\text{ Also written }}\quad &&L\setminus R=L\cap R^{\complement }&&\quad &&{\text{ where }}L,R\subseteq X\\[1.4ex]X\setminus (L\setminus R)&=(X\setminus L)\cup R&&\qquad {\text{ Also written }}\quad &&(L\setminus R)^{\complement }=L^{\complement }\cup R&&\quad &&{\text{ where }}R\subseteq X\\[1.4ex]L\setminus R&=(X\setminus R)\setminus (X\setminus L)&&\qquad {\text{ Also written }}\quad &&L\setminus R=R^{\complement }\setminus L^{\complement }&&\quad &&{\text{ where }}L,R\subseteq X\\[1.4ex]\end{alignedat}}}$$

Intervals:

$${\displaystyle (a,b)\cap (c,d)=(\max\{a,c\},\min\{b,d\})}$$ $${\displaystyle [a,b)\cap [c,d)=[\max\{a,c\},\min\{b,d\})}$$

The following statements are equivalent for any ${\displaystyle L,R\subseteq X:}$^[3]

1. ${\displaystyle L\subseteq R}$
   • Definition of subset: if ${\displaystyle l\in L}$ then ${\displaystyle l\in R}$
3. ${\displaystyle L\cap R=L}$
4. ${\displaystyle L\cup R=R}$
5. ${\displaystyle L\,\triangle \,R=R\setminus L}$
6. ${\displaystyle L\,\triangle \,R\subseteq R\setminus L}$
7. ${\displaystyle L\setminus R=\varnothing }$
8. ${\displaystyle L}$ and ${\displaystyle X\setminus R}$ are disjoint (that is, ${\displaystyle L\cap (X\setminus R)=\varnothing }$)
9. ${\displaystyle X\setminus R\subseteq X\setminus L\qquad }$ (that is, ${\displaystyle R^{\complement }\subseteq L^{\complement }}$)

The following statements are equivalent for any ${\displaystyle L,R\subseteq X:}$

1. ${\displaystyle L\not \subseteq R}$
2. There exists some ${\displaystyle l\in L\setminus R.}$

The following statements are equivalent:

1. ${\displaystyle L=R}$
2. ${\displaystyle L\,\triangle \,R=\varnothing }$
3. ${\displaystyle L\,\setminus \,R=R\,\setminus \,L}$

• If ${\displaystyle L\cap R=\varnothing }$ then ${\displaystyle L=R}$ if and only if ${\displaystyle L=\varnothing =R.}$
• Uniqueness of complements: If ${\textstyle L\cup R=X{\text{ and }}L\cap R=\varnothing }$ then ${\displaystyle R=X\setminus L}$

A set ${\displaystyle L}$ is empty if the sentence ${\displaystyle \forall x(x\not \in L)}$ is true, where the notation ${\displaystyle x\not \in L}$ is shorthand for ${\displaystyle \lnot (x\in L).}$

If ${\displaystyle L}$ is any set then the following are equivalent:

1. ${\displaystyle L}$ is not empty, meaning that the sentence ${\displaystyle \lnot [\forall x(x\not \in L)]}$ is true (literally, the logical negation of "${\displaystyle L}$ is empty" holds true).
2. (In classical mathematics) ${\displaystyle L}$ is inhabited, meaning: ${\displaystyle \exists x(x\in L)}$
   • In constructive mathematics, "not empty" and "inhabited" are not equivalent: every inhabited set is not empty but the converse is not always guaranteed; that is, in constructive mathematics, a set ${\displaystyle L}$ that is not empty (where by definition, "${\displaystyle L}$ is empty" means that the statement ${\displaystyle \forall x(x\not \in L)}$ is true) might not have an inhabitant (which is an ${\displaystyle x}$ such that ${\displaystyle x\in L}$).
3. ${\displaystyle L\not \subseteq R}$ for some set ${\displaystyle R}$

If ${\displaystyle L}$ is any set then the following are equivalent:

1. ${\displaystyle L}$ is empty (${\displaystyle L=\varnothing }$), meaning: ${\displaystyle \forall x(x\not \in L)}$
2. ${\displaystyle L\cup R\subseteq R}$ for every set ${\displaystyle R}$
3. ${\displaystyle L\subseteq R}$ for every set ${\displaystyle R}$
4. ${\displaystyle L\subseteq R\setminus L}$ for some/every set ${\displaystyle R}$
5. ${\displaystyle \varnothing /L=L}$

Given any ${\displaystyle x,}$ the following are equivalent:

1. ${\textstyle x\not \in L\setminus R}$
2. ${\textstyle x\in L\cap R\;{\text{ or }}\;x\not \in L.}$
3. ${\textstyle x\in R\;{\text{ or }}\;x\not \in L.}$

Moreover, $${\displaystyle (L\setminus R)\cap R=\varnothing \qquad {\text{ always holds}}.}$$

Inclusion is a partial order: Explicitly, this means that inclusion ${\displaystyle \,\subseteq ,\,}$ which is a binary operation, has the following three properties:^[3]

• Reflexivity: ${\textstyle L\subseteq L}$
• Antisymmetry: ${\textstyle (L\subseteq R{\text{ and }}R\subseteq L){\text{ if and only if }}L=R}$
• Transitivity: ${\textstyle {\text{If }}L\subseteq M{\text{ and }}M\subseteq R{\text{ then }}L\subseteq R}$

The following proposition says that for any set ${\displaystyle S,}$ the power set of ${\displaystyle S,}$ ordered by inclusion, is a bounded lattice, and hence together with the distributive and complement laws above, show that it is a Boolean algebra.

Existence of a least element and a greatest element: $${\displaystyle \varnothing \subseteq L\subseteq X}$$

Joins/supremums exist:^[3] $${\displaystyle L\subseteq L\cup R}$$

The union ${\displaystyle L\cup R}$ is the join/supremum of ${\displaystyle L}$ and ${\displaystyle R}$ with respect to ${\displaystyle \,\subseteq \,}$ because:

1. ${\displaystyle L\subseteq L\cup R}$ and ${\displaystyle R\subseteq L\cup R,}$ and
2. if ${\displaystyle Z}$ is a set such that ${\displaystyle L\subseteq Z}$ and ${\displaystyle R\subseteq Z}$ then ${\displaystyle L\cup R\subseteq Z.}$

The intersection ${\displaystyle L\cap R}$ is the join/supremum of ${\displaystyle L}$ and ${\displaystyle R}$ with respect to ${\displaystyle \,\supseteq .\,}$

Meets/infimums exist:^[3] $${\displaystyle L\cap R\subseteq L}$$

The intersection ${\displaystyle L\cap R}$ is the meet/infimum of ${\displaystyle L}$ and ${\displaystyle R}$ with respect to ${\displaystyle \,\subseteq \,}$ because:

1. if ${\displaystyle L\cap R\subseteq L}$ and ${\displaystyle L\cap R\subseteq R,}$ and
2. if ${\displaystyle Z}$ is a set such that ${\displaystyle Z\subseteq L}$ and ${\displaystyle Z\subseteq R}$ then ${\displaystyle Z\subseteq L\cap R.}$

The union ${\displaystyle L\cup R}$ is the meet/infimum of ${\displaystyle L}$ and ${\displaystyle R}$ with respect to ${\displaystyle \,\supseteq .\,}$

Other inclusion properties:

$${\displaystyle L\setminus R\subseteq L}$$ $${\displaystyle (L\setminus R)\cap L=L\setminus R}$$

• If ${\displaystyle L\subseteq R}$ then ${\displaystyle L\,\triangle \,R=R\setminus L.}$
• If ${\displaystyle L\subseteq X}$ and ${\displaystyle R\subseteq Y}$ then ${\displaystyle L\times R\subseteq X\times Y}$^[3]

In the left hand sides of the following identities, ${\displaystyle L}$ is the L eft most set, ${\displaystyle M}$ is the M iddle set, and ${\displaystyle R}$ is the R ight most set.

There is no universal agreement on the order of precedence of the basic set operators. Nevertheless, many authors use precedence rules for set operators, although these rules vary with the author.

One common convention is to associate intersection ${\displaystyle L\cap R=\{x:(x\in L)\land (x\in R)\}}$ with logical conjunction (and) ${\displaystyle L\land R}$ and associate union ${\displaystyle L\cup R=\{x:(x\in L)\lor (x\in R)\}}$ with logical disjunction (or) ${\displaystyle L\lor R,}$ and then transfer the precedence of these logical operators (where ${\displaystyle \,\land \,}$ has precedence over ${\displaystyle \,\lor \,}$) to these set operators, thereby giving ${\displaystyle \,\cap \,}$ precedence over ${\displaystyle \,\cup .\,}$ So for example, ${\displaystyle L\cup M\cap R}$ would mean ${\displaystyle L\cup (M\cap R)}$ since it would be associated with the logical statement ${\displaystyle L\lor M\land R~=~L\lor (M\land R)}$ and similarly, ${\displaystyle L\cup M\cap R\cup Z}$ would mean ${\displaystyle L\cup (M\cap R)\cup Z}$ since it would be associated with ${\displaystyle L\lor M\land R\lor Z~=~L\lor (M\land R)\lor Z.}$

Sometimes, set complement (subtraction) ${\displaystyle \,\setminus \,}$ is also associated with logical complement (not) ${\displaystyle \,\lnot ,\,}$ in which case it will have the highest precedence. More specifically, ${\displaystyle L\setminus R=\{x:(x\in L)\land \lnot (x\in R)\}}$ is rewritten ${\displaystyle L\land \lnot R}$ so that for example, ${\displaystyle L\cup M\setminus R}$ would mean ${\displaystyle L\cup (M\setminus R)}$ since it would be rewritten as the logical statement ${\displaystyle L\lor M\land \lnot R}$ which is equal to ${\displaystyle L\lor (M\land \lnot R).}$ For another example, because ${\displaystyle L\land \lnot M\land R}$ means ${\displaystyle L\land (\lnot M)\land R,}$ which is equal to both ${\displaystyle (L\land (\lnot M))\land R}$ and ${\displaystyle L\land ((\lnot M)\land R)~=~L\land (R\land (\lnot M))}$ (where ${\displaystyle (\lnot M)\land R}$ was rewritten as ${\displaystyle R\land (\lnot M)}$), the formula ${\displaystyle L\setminus M\cap R}$ would refer to the set ${\displaystyle (L\setminus M)\cap R=L\cap (R\setminus M);}$ moreover, since ${\displaystyle L\land (\lnot M)\land R=(L\land R)\land \lnot M,}$ this set is also equal to ${\displaystyle (L\cap R)\setminus M}$ (other set identities can similarly be deduced from propositional calculus identities in this way). However, because set subtraction is not associative ${\displaystyle (L\setminus M)\setminus R\neq L\setminus (M\setminus R),}$ a formula such as ${\displaystyle L\setminus M\setminus R}$ would be ambiguous; for this reason, among others, set subtraction is often not assigned any precedence at all.

Symmetric difference ${\displaystyle L\triangle R=\{x:(x\in L)\oplus (x\in R)\}}$ is sometimes associated with exclusive or (xor) ${\displaystyle L\oplus R}$ (also sometimes denoted by ${\displaystyle \,\veebar }$), in which case if the order of precedence from highest to lowest is ${\displaystyle \,\lnot ,\,\oplus ,\,\land ,\,\lor \,}$ then the order of precedence (from highest to lowest) for the set operators would be ${\displaystyle \,\setminus ,\,\triangle ,\,\cap ,\,\cup .}$ There is no universal agreement on the precedence of exclusive disjunction ${\displaystyle \,\oplus \,}$ with respect to the other logical connectives, which is why symmetric difference ${\displaystyle \,\triangle \,}$ is not often assigned a precedence.

Definition: A binary operator ${\displaystyle \,\ast \,}$ is called associative if ${\displaystyle (L\,\ast \,M)\,\ast \,R=L\,\ast \,(M\,\ast \,R)}$ always holds.

The following set operators are associative:^[3]

$${\displaystyle {\begin{alignedat}{5}(L\cup M)\cup R&\;=\;\;&&L\cup (M\cup R)\\[1.4ex](L\cap M)\cap R&\;=\;\;&&L\cap (M\cap R)\\[1.4ex](L\,\triangle M)\,\triangle R&\;=\;\;&&L\,\triangle (M\,\triangle R)\\[1.4ex]\end{alignedat}}}$$

For set subtraction, instead of associativity, only the following is always guaranteed: $${\displaystyle (L\,\setminus \,M)\,\setminus \,R\;~~{\color {red}{\subseteq }}~~\;L\,\setminus \,(M\,\setminus \,R)}$$ where equality holds if and only if ${\displaystyle L\cap R=\varnothing }$ (this condition does not depend on ${\displaystyle M}$). Thus ${\textstyle \;(L\setminus M)\setminus R=L\setminus (M\setminus R)\;}$ if and only if ${\displaystyle \;(R\setminus M)\setminus L=R\setminus (M\setminus L),\;}$ where the only difference between the left and right hand side set equalities is that the locations of ${\displaystyle L{\text{ and }}R}$ have been swapped.

Definition: If ${\displaystyle \ast {\text{ and }}\bullet }$ are binary operators then ${\displaystyle \,\ast \,}$ left distributes over ${\displaystyle \,\bullet \,}$ if $${\displaystyle L\,\ast \,(M\,\bullet \,R)~=~(L\,\ast \,M)\,\bullet \,(L\,\ast \,R)\qquad \qquad {\text{ for all }}L,M,R}$$ while ${\displaystyle \,\ast \,}$ right distributes over ${\displaystyle \,\bullet \,}$ if $${\displaystyle (L\,\bullet \,M)\,\ast \,R~=~(L\,\ast \,R)\,\bullet \,(M\,\ast \,R)\qquad \qquad {\text{ for all }}L,M,R.}$$ The operator ${\displaystyle \,\ast \,}$ distributes over ${\displaystyle \,\bullet \,}$ if it both left distributes and right distributes over ${\displaystyle \,\bullet \,.\,}$ In the definitions above, to transform one side to the other, the innermost operator (the operator inside the parentheses) becomes the outermost operator and the outermost operator becomes the innermost operator.

Right distributivity:^[3]

$${\displaystyle {\begin{alignedat}{9}(L\,\cap \,M)\,\cup \,R~&~~=~~&&(L\,\cup \,R)\,&&\cap \,&&(M\,\cup \,R)\qquad &&{\text{ (Right-distributivity of }}\,\cup \,{\text{ over }}\,\cap \,{\text{)}}\\[1.4ex](L\,\cup \,M)\,\cup \,R~&~~=~~&&(L\,\cup \,R)\,&&\cup \,&&(M\,\cup \,R)\qquad &&{\text{ (Right-distributivity of }}\,\cup \,{\text{ over }}\,\cup \,{\text{)}}\\[1.4ex](L\,\cup \,M)\,\cap \,R~&~~=~~&&(L\,\cap \,R)\,&&\cup \,&&(M\,\cap \,R)\qquad &&{\text{ (Right-distributivity of }}\,\cap \,{\text{ over }}\,\cup \,{\text{)}}\\[1.4ex](L\,\cap \,M)\,\cap \,R~&~~=~~&&(L\,\cap \,R)\,&&\cap \,&&(M\,\cap \,R)\qquad &&{\text{ (Right-distributivity of }}\,\cap \,{\text{ over }}\,\cap \,{\text{)}}\\[1.4ex](L\,\triangle \,M)\,\cap \,R~&~~=~~&&(L\,\cap \,R)\,&&\triangle \,&&(M\,\cap \,R)\qquad &&{\text{ (Right-distributivity of }}\,\cap \,{\text{ over }}\,\triangle \,{\text{)}}\\[1.4ex](L\,\cap \,M)\,\times \,R~&~~=~~&&(L\,\times \,R)\,&&\cap \,&&(M\,\times \,R)\qquad &&{\text{ (Right-distributivity of }}\,\times \,{\text{ over }}\,\cap \,{\text{)}}\\[1.4ex](L\,\cup \,M)\,\times \,R~&~~=~~&&(L\,\times \,R)\,&&\cup \,&&(M\,\times \,R)\qquad &&{\text{ (Right-distributivity of }}\,\times \,{\text{ over }}\,\cup \,{\text{)}}\\[1.4ex](L\,\setminus \,M)\,\times \,R~&~~=~~&&(L\,\times \,R)\,&&\setminus \,&&(M\,\times \,R)\qquad &&{\text{ (Right-distributivity of }}\,\times \,{\text{ over }}\,\setminus \,{\text{)}}\\[1.4ex](L\,\triangle \,M)\,\times \,R~&~~=~~&&(L\,\times \,R)\,&&\triangle \,&&(M\,\times \,R)\qquad &&{\text{ (Right-distributivity of }}\,\times \,{\text{ over }}\,\triangle \,{\text{)}}\\[1.4ex](L\,\cup \,M)\,\setminus \,R~&~~=~~&&(L\,\setminus \,R)\,&&\cup \,&&(M\,\setminus \,R)\qquad &&{\text{ (Right-distributivity of }}\,\setminus \,{\text{ over }}\,\cup \,{\text{)}}\\[1.4ex](L\,\cap \,M)\,\setminus \,R~&~~=~~&&(L\,\setminus \,R)\,&&\cap \,&&(M\,\setminus \,R)\qquad &&{\text{ (Right-distributivity of }}\,\setminus \,{\text{ over }}\,\cap \,{\text{)}}\\[1.4ex](L\,\triangle \,M)\,\setminus \,R~&~~=~~&&(L\,\setminus \,R)&&\,\triangle \,&&(M\,\setminus \,R)\qquad &&{\text{ (Right-distributivity of }}\,\setminus \,{\text{ over }}\,\triangle \,{\text{)}}\\[1.4ex](L\,\setminus \,M)\,\setminus \,R~&~~=~~&&(L\,\setminus \,R)&&\,\setminus \,&&(M\,\setminus \,R)\qquad &&{\text{ (Right-distributivity of }}\,\setminus \,{\text{ over }}\,\setminus \,{\text{)}}\\[1.4ex]~&~~=~~&&~~\;~~\;~~\;~L&&\,\setminus \,&&(M\cup R)\\[1.4ex]\end{alignedat}}}$$

Left distributivity:^[3]

$${\displaystyle {\begin{alignedat}{5}L\cup (M\cap R)&\;=\;\;&&(L\cup M)\cap (L\cup R)\qquad &&{\text{ (Left-distributivity of }}\,\cup \,{\text{ over }}\,\cap \,{\text{)}}\\[1.4ex]L\cup (M\cup R)&\;=\;\;&&(L\cup M)\cup (L\cup R)&&{\text{ (Left-distributivity of }}\,\cup \,{\text{ over }}\,\cup \,{\text{)}}\\[1.4ex]L\cap (M\cup R)&\;=\;\;&&(L\cap M)\cup (L\cap R)&&{\text{ (Left-distributivity of }}\,\cap \,{\text{ over }}\,\cup \,{\text{)}}\\[1.4ex]L\cap (M\cap R)&\;=\;\;&&(L\cap M)\cap (L\cap R)&&{\text{ (Left-distributivity of }}\,\cap \,{\text{ over }}\,\cap \,{\text{)}}\\[1.4ex]L\cap (M\,\triangle \,R)&\;=\;\;&&(L\cap M)\,\triangle \,(L\cap R)&&{\text{ (Left-distributivity of }}\,\cap \,{\text{ over }}\,\triangle \,{\text{)}}\\[1.4ex]L\times (M\cap R)&\;=\;\;&&(L\times M)\cap (L\times R)&&{\text{ (Left-distributivity of }}\,\times \,{\text{ over }}\,\cap \,{\text{)}}\\[1.4ex]L\times (M\cup R)&\;=\;\;&&(L\times M)\cup (L\times R)&&{\text{ (Left-distributivity of }}\,\times \,{\text{ over }}\,\cup \,{\text{)}}\\[1.4ex]L\times (M\,\setminus R)&\;=\;\;&&(L\times M)\,\setminus (L\times R)&&{\text{ (Left-distributivity of }}\,\times \,{\text{ over }}\,\setminus \,{\text{)}}\\[1.4ex]L\times (M\,\triangle R)&\;=\;\;&&(L\times M)\,\triangle (L\times R)&&{\text{ (Left-distributivity of }}\,\times \,{\text{ over }}\,\triangle \,{\text{)}}\\[1.4ex]\end{alignedat}}}$$

Intersection distributes over symmetric difference: $${\displaystyle {\begin{alignedat}{5}L\,\cap \,(M\,\triangle \,R)~&~~=~~&&(L\,\cap \,M)\,\triangle \,(L\,\cap \,R)~&&~\\[1.4ex]\end{alignedat}}}$$ $${\displaystyle {\begin{alignedat}{5}(L\,\triangle \,M)\,\cap \,R~&~~=~~&&(L\,\cap \,R)\,\triangle \,(M\,\cap \,R)~&&~\\[1.4ex]\end{alignedat}}}$$

Union does not distribute over symmetric difference because only the following is guaranteed in general: $${\displaystyle {\begin{alignedat}{5}L\cup (M\,\triangle \,R)~~{\color {red}{\supseteq }}~~\color {black}{\,}(L\cup M)\,\triangle \,(L\cup R)~&~=~&&(M\,\triangle \,R)\,\setminus \,L&~=~&&(M\,\setminus \,L)\,\triangle \,(R\,\setminus \,L)\\[1.4ex]\end{alignedat}}}$$

Symmetric difference does not distribute over itself: $${\displaystyle L\,\triangle \,(M\,\triangle \,R)~~{\color {red}{\neq }}~~\color {black}{\,}(L\,\triangle \,M)\,\triangle \,(L\,\triangle \,R)~=~M\,\triangle \,R}$$ and in general, for any sets ${\displaystyle L{\text{ and }}A}$ (where ${\displaystyle A}$ represents ${\displaystyle M\,\triangle \,R}$), ${\displaystyle L\,\triangle \,A}$ might not be a subset, nor a superset, of ${\displaystyle L}$ (and the same is true for ${\displaystyle A}$).

Failure of set subtraction to left distribute:

Set subtraction is right distributive over itself. However, set subtraction is not left distributive over itself because only the following is guaranteed in general: $${\displaystyle {\begin{alignedat}{5}L\,\setminus \,(M\,\setminus \,R)&~~{\color {red}{\supseteq }}~~&&\color {black}{\,}(L\,\setminus \,M)\,\setminus \,(L\,\setminus \,R)~~=~~L\cap R\,\setminus \,M\\[1.4ex]\end{alignedat}}}$$ where equality holds if and only if ${\displaystyle L\,\setminus \,M=L\,\cap \,R,}$ which happens if and only if ${\displaystyle L\cap M\cap R=\varnothing {\text{ and }}L\setminus M\subseteq R.}$

For symmetric difference, the sets ${\displaystyle L\,\setminus \,(M\,\triangle \,R)}$ and ${\displaystyle (L\,\setminus \,M)\,\triangle \,(L\,\setminus \,R)=L\,\cap \,(M\,\triangle \,R)}$ are always disjoint. So these two sets are equal if and only if they are both equal to ${\displaystyle \varnothing .}$ Moreover, ${\displaystyle L\,\setminus \,(M\,\triangle \,R)=\varnothing }$ if and only if ${\displaystyle L\cap M\cap R=\varnothing {\text{ and }}L\subseteq M\cup R.}$

To investigate the left distributivity of set subtraction over unions or intersections, consider how the sets involved in (both of) De Morgan's laws are all related: $${\displaystyle {\begin{alignedat}{5}(L\,\setminus \,M)\,\cap \,(L\,\setminus \,R)~~=~~L\,\setminus \,(M\,\cup \,R)~&~~{\color {red}{\subseteq }}~~&&\color {black}{\,}L\,\setminus \,(M\,\cap \,R)~~=~~(L\,\setminus \,M)\,\cup \,(L\,\setminus \,R)\\[1.4ex]\end{alignedat}}}$$ always holds (the equalities on the left and right are De Morgan's laws) but equality is not guaranteed in general (that is, the containment ${\displaystyle {\color {red}{\subseteq }}}$ might be strict). Equality holds if and only if ${\displaystyle L\,\setminus \,(M\,\cap \,R)\;\subseteq \;L\,\setminus \,(M\,\cup \,R),}$ which happens if and only if ${\displaystyle L\,\cap \,M=L\,\cap \,R.}$

This observation about De Morgan's laws shows that ${\displaystyle \,\setminus \,}$ is not left distributive over ${\displaystyle \,\cup \,}$ or ${\displaystyle \,\cap \,}$ because only the following are guaranteed in general: $${\displaystyle {\begin{alignedat}{5}L\,\setminus \,(M\,\cup \,R)~&~~{\color {red}{\subseteq }}~~&&\color {black}{\,}(L\,\setminus \,M)\,\cup \,(L\,\setminus \,R)~~=~~L\,\setminus \,(M\,\cap \,R)\\[1.4ex]\end{alignedat}}}$$ $${\displaystyle {\begin{alignedat}{5}L\,\setminus \,(M\,\cap \,R)~&~~{\color {red}{\supseteq }}~~&&\color {black}{\,}(L\,\setminus \,M)\,\cap \,(L\,\setminus \,R)~~=~~L\,\setminus \,(M\,\cup \,R)\\[1.4ex]\end{alignedat}}}$$ where equality holds for one (or equivalently, for both) of the above two inclusion formulas if and only if ${\displaystyle L\,\cap \,M=L\,\cap \,R.}$

The following statements are equivalent:

1. ${\displaystyle L\cap M\,=\,L\cap R}$
2. ${\displaystyle L\,\setminus \,M\,=\,L\,\setminus \,R}$
3. ${\displaystyle L\,\setminus \,(M\,\cap \,R)=(L\,\setminus \,M)\,\cap \,(L\,\setminus \,R);}$ that is, ${\displaystyle \,\setminus \,}$ left distributes over ${\displaystyle \,\cap \,}$ for these three particular sets
4. ${\displaystyle L\,\setminus \,(M\,\cup \,R)=(L\,\setminus \,M)\,\cup \,(L\,\setminus \,R);}$ that is, ${\displaystyle \,\setminus \,}$ left distributes over ${\displaystyle \,\cup \,}$ for these three particular sets
5.