Membership Is the Test
Set identities are equalities between sets, so a reliable way to prove one is to ask what it means for an arbitrary point to belong to each side. The two distributive laws have similar-looking forms, but each has a specific matching pattern. The absorption laws simplify expressions in which one set already contains the information in a union or intersection.
Throughout, let \(X\) be a set and let \(A,B,C\subseteq X\). We use the definitions \(x\in A\cup B\) if and only if \(x\in A\) or \(x\in B\), and \(x\in A\cap B\) if and only if \(x\in A\) and \(x\in B\). To prove two sets are equal, we show that an arbitrary \(x\in X\) belongs to the first set if and only if it belongs to the second. This proves both inclusions.
These identities hold for every choice of \(A,B,C\subseteq X\). They do not require that the sets be nonempty, distinct, or disjoint. In particular, the same proofs still apply if one or more of the sets is \(\varnothing\) or \(X\).
The Distributive Laws
Proof. Fix \(x\in X\). For the first identity, the definitions of intersection and union give \(x\in A\cap(B\cup C)\) if and only if \(x\in A\) and \((x\in B\text{ or }x\in C)\). By the distributive rule for logical “and” and “or,” this holds if and only if \((x\in A\text{ and }x\in B)\) or \((x\in A\text{ and }x\in C)\). Applying the definitions again, this is equivalent to \(x\in(A\cap B)\cup(A\cap C)\). Since this equivalence holds for every \(x\in X\), the sets are equal.
For the second identity, \(x\in A\cup(B\cap C)\) if and only if \(x\in A\) or \((x\in B\text{ and }x\in C)\). The distributive rule for logical “or” and “and” makes this equivalent to \((x\in A\text{ or }x\in B)\) and \((x\in A\text{ or }x\in C)\). By the definitions, the latter condition is equivalent to \(x\in(A\cup B)\cap(A\cup C)\). Thus the second equality also holds for every \(x\), proving both identities. \(\square\)
The membership argument works because membership in a union corresponds to “or,” while membership in an intersection corresponds to “and.” The first identity distributes intersection across union; the second distributes union across intersection. The expressions on the right are not formed by simply copying the inner operation: the same set \(A\) appears in both parentheses.
Worked Example: Check Both Distributive Laws
Work in \(X=\{1,2,3,4,5,6,7,8\}\) with \(A=\{1,2,4,7\}\), \(B=\{2,3,4,8\}\), and \(C=\{1,3,5,7\}\). Compute both sides of each distributive identity.
Intersection across union. First, \(B\cup C=\{1,2,3,4,5,7,8\}\). Therefore \(A\cap(B\cup C)=\{1,2,4,7\}\). Separately, \(A\cap B=\{2,4\}\) and \(A\cap C=\{1,7\}\), so \((A\cap B)\cup(A\cap C)=\{1,2,4,7\}\). The two sides agree.
Union across intersection. Here \(B\cap C=\{3\}\), so \(A\cup(B\cap C)=\{1,2,3,4,7\}\). On the other side, \(A\cup B=\{1,2,3,4,7,8\}\) and \(A\cup C=\{1,2,3,4,5,7\}\). Their intersection is \(\{1,2,3,4,7\}\), as required.
Check: the first calculation uses the fact that the only elements of \(B\cup C\) that also belong to \(A\) are \(1,2,4,7\). The second uses the fact that \(3\) is the only element shared by \(B\) and \(C\), and \(3\notin A\). These membership checks confirm the computed sets.
The Absorption Laws
Proof. Fix \(x\in X\). For the first identity, \(x\in A\cup(A\cap B)\) if and only if \(x\in A\) or \((x\in A\text{ and }x\in B)\). If this condition holds, then \(x\in A\): either it holds directly, or the second part includes \(x\in A\). Conversely, if \(x\in A\), then \(x\in A\cup(A\cap B)\) by the union condition. Thus \(x\in A\cup(A\cap B)\) if and only if \(x\in A\).
For the second identity, \(x\in A\cap(A\cup B)\) if and only if \(x\in A\) and \((x\in A\text{ or }x\in B)\). This implies \(x\in A\) because that is one of the required conditions. Conversely, if \(x\in A\), then \(x\in A\cup B\) as well, so \(x\in A\cap(A\cup B)\). The membership conditions agree in both cases, proving the equalities. \(\square\)
Absorption says that adjoining \(A\cap B\) to \(A\) by a union adds nothing: every element of \(A\cap B\) already belongs to \(A\). Likewise, intersecting \(A\) with \(A\cup B\) removes nothing: every element of \(A\) already belongs to \(A\cup B\).
Worked Example: Simplify a Nested Expression
For any \(A,B,C\subseteq X\), simplify \(E=(A\cup B)\cap\bigl(A\cup(B\cap C)\bigr)\).
Use the second distributive law with the common set \(A\), and with \(B\) and \(B\cap C\) as the other two sets: \[ E=A\cup\bigl(B\cap(B\cap C)\bigr). \] By associativity and idempotence of intersection, \(B\cap(B\cap C)=(B\cap B)\cap C=B\cap C\). Hence \[ E=A\cup(B\cap C). \] The basic associative and idempotent laws used here follow directly from the membership definitions: membership in an intersection of three sets requires membership in all three, regardless of grouping, and \(x\in B\cap B\) is equivalent to \(x\in B\).
Concrete check. Take \(X=\{1,2,3,4,5,6\}\), \(A=\{1,2\}\), \(B=\{2,3,5\}\), and \(C=\{3,4\}\). Then \(B\cap C=\{3\}\), so the simplified expression is \(A\cup(B\cap C)=\{1,2,3\}\). Directly, \(A\cup B=\{1,2,3,5\}\) and \(A\cup(B\cap C)=\{1,2,3\}\); intersecting those two sets gives \(\{1,2,3\}\). The direct computation agrees with the simplification.
Using Absorption in a Longer Simplification
When an expression has several operations, it helps to identify a matching pattern before changing it. A valid simplification can be applied to a part of the expression while leaving the rest unchanged. The next example uses distribution to expose an absorption pattern.
Worked Example: Reduce an Expression to One Set
For \(A,B,C\subseteq X\), simplify \(F=(A\cap B)\cup\bigl(A\cap(B\cup C)\bigr)\cup A\).
First apply the first distributive law: \(A\cap(B\cup C)=(A\cap B)\cup(A\cap C)\). Thus \[ F=(A\cap B)\cup(A\cap B)\cup(A\cap C)\cup A. \] Every element of \(A\cap B\) and every element of \(A\cap C\) belongs to \(A\). Therefore the union of those sets with \(A\) is just \(A\), giving \(F=A\).
Concrete check. Let \(X=\{1,2,3,4,5\}\), \(A=\{1,3\}\), \(B=\{3,4\}\), and \(C=\{1,4\}\). Then \(A\cap B=\{3\}\), \(B\cup C=\{1,3,4\}\), and \(A\cap(B\cup C)=\{1,3\}\). Consequently \(F=\{3\}\cup\{1,3\}\cup\{1,3\}=\{1,3\}=A\). No element outside \(A\) enters the union, and both elements of \(A\) occur in it.
A Common Distribution Error
Common pitfall. A frequent error is to write \(A\cup(B\cap C)=(A\cup B)\cap C\). This is not the union-across-intersection distributive law: the required right-hand side has \(A\cup C\) as its second factor.
For a concrete failure, take \(X=\{1,2\}\), \(A=\{1\}\), and \(B=C=\{2\}\). The left side of the incorrect equation is \(A\cup(B\cap C)=\{1\}\cup\{2\}=\{1,2\}\). The proposed right side is \((A\cup B)\cap C=\{1,2\}\cap\{2\}=\{2\}\). The two sets differ: \(1\) belongs to the left side but not the right side. The correct calculation is \[ A\cup(B\cap C)=(A\cup B)\cap(A\cup C)=\{1,2\}\cap\{1,2\}=\{1,2\}. \] When applying a distributive law, check that the set outside the parentheses is repeated in both resulting parentheses.
These laws matter because they let us rewrite set expressions without changing which elements they contain. Element chasing provides a dependable check: translate each operation into its membership condition, follow the logical “and” and “or,” and compare the resulting conditions. A pattern that does not preserve those conditions is not a valid set identity.
Check Your Understanding
For each question, state the membership condition or set identity that justifies your answer.
- Write the two distributive laws for sets and identify which operation is distributed in each one.
- Use an element-chasing argument to prove \(A\cap(B\cup C)=(A\cap B)\cup(A\cap C)\).
- Simplify \(A\cup(A\cap C)\), and explain why the result holds for every \(A,C\subseteq X\).
- For \(A=\{1,4\}\), \(B=\{2,4\}\), and \(C=\{1,2\}\), compute both sides of \(A\cup(B\cap C)=(A\cup B)\cap(A\cup C)\).
- Give a choice of \(A,B,C\) for which \(A\cup(B\cap C)\ne(A\cup B)\cap C\), and compute both sides.