Let U be the universal set and A, B, C subsets of U. Which of the following is correct? (1) A − (B ∪ C) = (A − B) ∪ (A − C); (2) (A ∩ B) − C = (A − C) ∩ (B − C); (3) A − (B − C) = A ∩ (B' ∩ C'); (4) (A ∪ B) − C = A ∪ (B − C).

Step-by-step solution

1. 1

   Convert every difference into an intersection with a complement. The single rule $X-Y=X\cap Y'$ turns all four statements into pure intersection/union/complement algebra, where De Morgan's laws can be applied mechanically. Guessing from Venn pictures is error-prone; this rewriting is not.

2. 2

   Test statement 1. Left side: $A-(B\cup C)=A\cap(B\cup C)'=A\cap B'\cap C'$. Right side: $(A\cap B')\cup(A\cap C')=A\cap(B'\cup C')$. These differ — an element of $A$ lying in $B$ but not in $C$ belongs to the right side but not the left. False. (The correct version is $A-(B\cap C)=(A-B)\cup(A-C)$.)

3. 3

   Test statement 2. Left side: $(A\cap B)-C=(A\cap B)\cap C'=A\cap B\cap C'$. Right side: $(A\cap C')\cap(B\cap C')=A\cap B\cap C'\cap C'=A\cap B\cap C'$, using idempotence $C'\cap C'=C'$. The two sides are identical. True.

4. 4

   Test statement 3. Left side: $A-(B-C)=A\cap(B\cap C')'=A\cap(B'\cup C)$. The claimed right side is $A\cap B'\cap C'$. Take $x\in A\cap B\cap C$: then $x$ is in the left side (since $x\in C$) but not the right (since $x\in B$ and $x\in C$). False.

5. 5

   Test statement 4. Left side: $(A\cup B)\cap C'=(A\cap C')\cup(B\cap C')$. Right side: $A\cup(B\cap C')$. Any $x\in A\cap C$ is in the right side but not the left, because the left removes all of $C$ including the part inside $A$. False.

6. 6

   Conclude.

   $\boxed{\text{Only statement 2 is correct}}$

   A concrete counterexample settles all three false ones at once: take $U=\{1,2,3\}$, $A=\{1,2,3\}$, $B=\{1\}$, $C=\{2\}$. Then statement 1 reads $\{3\}=\{1,2,3\}$, statement 3 reads $\{2,3\}=\{3\}$ and statement 4 reads $\{1,3\}=\{1,2,3\}$ — all false — while statement 2 reads $\{1\}=\{1\}$.