Why Is (A\C)\(B\C) Equal to (A\B)\C?

The identity is true for all sets $A$, $B$, and $C$. The key is to read set difference as a membership condition: $x\in X\setminus Y$ means that $x$ is in $X$ and is not in $Y$. Applying that meaning to both sides shows that they contain exactly the same elements.

Prove the identity element by element

Take any object $x$. It belongs to the left-hand side precisely when it belongs to $A\setminus C$ and does not belong to $B\setminus C$:

$$\begin{aligned} x\in (A\setminus C)\setminus(B\setminus C) &\iff x\in A\setminus C\text{ and }x\notin B\setminus C\\ &\iff x\in A,\ x\notin C,\text{ and not }(x\in B\text{ and }x\notin C)\\ &\iff x\in A,\ x\notin C,\text{ and }x\notin B\\ &\iff x\in A\setminus B\text{ and }x\notin C\\ &\iff x\in (A\setminus B)\setminus C. \end{aligned}$$

In the third line, the condition $x\notin C$ is already known. Therefore, for $x$ not to belong to $B\setminus C$, it must not belong to $B$. This is the step that makes the identity work.

Because the equivalence holds for every $x$, the two sets have the same elements, so

$(A\setminus C)\setminus(B\setminus C)=(A\setminus B)\setminus C.$

A shorter way to see it

Both sides can be described as “the elements of $A$ that are in neither $B$ nor $C$.” In set notation, that description is

$A\setminus(B\cup C).$

For the left side, first removing $C$ and then removing $B\setminus C$ removes all elements of $C$ and all elements of $B$ not already removed. Together, those are exactly the elements of $B\cup C$. For the right side, removing $B$ and then removing $C$ has the same effect. Hence both sides equal $A\setminus(B\cup C)$.

This argument also shows that the order in which you remove $B$ and $C$ does not matter:

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

Worked example

Let the universe for this example be the positive integers up to $8$, and choose

$A=\{1,2,3,4,5\},\quad B=\{2,4,6\},\quad C=\{3,4,7\}.$

Evaluate the left side in stages:

• $A\setminus C=\{1,2,5\}$.
• $B\setminus C=\{2,6\}$.
• Removing $\{2,6\}$ from $\{1,2,5\}$ leaves $\{1,5\}$.

Now evaluate the right side:

• $A\setminus B=\{1,3,5\}$.
• Removing $C=\{3,4,7\}$ leaves $\{1,5\}$.

The two sides agree. The common description $A\setminus(B\cup C)$ gives the same result, since $B\cup C=\{2,3,4,6,7\}$ and $A\setminus(B\cup C)=\{1,5\}$.

Common mistakes

Treating set difference as symmetric. In general, $X\setminus Y$ is not the same as $Y\setminus X$. The first keeps elements of $X$ that are not in $Y$; the second starts with $Y$ instead.

Dropping a condition too early. From $x\notin B\setminus C$ alone, you cannot conclude $x\notin B$: an element of $B$ might also be in $C$, and then it is not in $B\setminus C$. The conclusion follows here only because membership in the left side already gives $x\notin C$.

Ignoring parentheses. The expression $(A\setminus C)\setminus(B\setminus C)$ specifies exactly which sets are being subtracted. Parentheses matter when translating a nested expression into membership conditions.

Checking only one example. A finite example can help you understand the operations, but it does not prove a statement for all sets. The arbitrary-element argument establishes the identity in full generality.