Is (A ∩ B) ∪ C = A ∩ (B ∪ C)?

The identity $(A\cap B)\cup C=A\cap(B\cup C)$ does not hold for all sets. The issue is that the left-hand side includes every element of $C$, whereas the right-hand side includes only elements of $C$ that also belong to $A$.

The standard distributive law is $A\cap(B\cup C)=(A\cap B)\cup(A\cap C)$. Comparing that law with the proposed expression gives both a quick way to see the mismatch and a precise condition under which the proposed equality does hold.

The correct distributive law

Intersection distributes over union as follows:

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

The term involving $C$ is $A\cap C$, not all of $C$. An element must be in $A$ as well as in $C$ to be included in this part of the right-hand side.

By contrast, the proposed left-hand side is $(A\cap B)\cup C$. It includes $A\cap B$ and then adds every element of $C$, including elements outside $A$.

A counterexample

Take

$$A=\{1,2\},\qquad B=\{2\},\qquad C=\{3\}.$$

First calculate the left-hand side:

$$(A\cap B)\cup C=\{2\}\cup\{3\}=\{2,3\}.$$

Now calculate the right-hand side:

$$A\cap(B\cup C)=\{1,2\}\cap\{2,3\}=\{2\}.$$

Since $\{2,3\}\ne\{2\}$, the identity is false in general. The element $3$ makes the difference: it belongs to $C$, so it belongs to the left-hand side, but it does not belong to $A$, so it is excluded from the right-hand side.

Compare the two sides element by element

For any element $x$,

$$\begin{aligned} x\in (A\cap B)\cup C &\iff (x\in A\text{ and }x\in B)\text{ or }x\in C,\\ x\in A\cap(B\cup C) &\iff x\in A\text{ and }(x\in B\text{ or }x\in C)\\ &\iff (x\in A\text{ and }x\in B)\text{ or }(x\in A\text{ and }x\in C). \end{aligned}$$

The first condition allows any $x\in C$. The second allows an element from $C$ only if it is also in $A$. Therefore, every element on the right is on the left, but the left can have extra elements.

In set notation, this inclusion is

$$A\cap(B\cup C)\subseteq (A\cap B)\cup C.$$

Exactly when are the sets equal?

For fixed sets $A$, $B$, and $C$, the proposed equality holds exactly when $C\subseteq A$.

If $C\subseteq A$, then $A\cap C=C$. The correct distributive law gives

$$A\cap(B\cup C)=(A\cap B)\cup(A\cap C)=(A\cap B)\cup C,$$

which is the proposed equality.

Conversely, suppose the two sides are equal. Any $c\in C$ belongs to $(A\cap B)\cup C$, so it must belong to the equal set $A\cap(B\cup C)$. In particular, it belongs to $A$. Since this holds for every $c\in C$, we have $C\subseteq A$.

This condition includes the special case $C=\varnothing$, but it is not necessary for $B$ to have any particular relation to $A$ or $C$.

Common mistakes

• Distributing over only one term. The expression $A\cap(B\cup C)$ distributes to $(A\cap B)\cup(A\cap C)$; the intersection with $A$ applies to both parts of the union.
• Treating union as if it adds a condition. In $(A\cap B)\cup C$, membership in $C$ is enough to put an element in the result. It does not also have to be in $A$.
• Checking only one choice of sets. Some choices, such as $C\subseteq A$, make the proposed equality true. A single successful example does not establish an identity for all sets; a counterexample disproves it.

When checking a proposed set identity, translate membership into logical “and” and “or” statements. This makes it easier to see which conditions apply to each part and to construct a counterexample if one side admits an element the other excludes.