Suppose $U$ is the power set of the set $S = \{1,2,3,4,5,6\}$. For any $T \in U$, let $|T|$ denote the number of elements in $T$ and $T^{\prime}$ denote the complement of $T$. For any $T, R \in U$, let $T \setminus R$ be the set of all elements in $T$ which are not in $R$. Which one of the following is true?

Let us analyze the given options:

• Option (A): $\forall X \in U (|X| = |X^{\prime}|)$ is False. If $X = \emptyset$, then $|X| = 0$ while $|X^{\prime}| = |S| = 6 \neq 0$.

• Option (B): $\exists X \in U, \exists Y \in U (|X| = 5, |Y| = 5 \text{ and } X \cap Y = \emptyset)$ is False. By the principle of inclusion-exclusion:

  $$|X \cap Y| = |X| + |Y| - |X \cup Y| \ge 5 + 5 - 6 = 4 > 0$$

  Hence, $X$ and $Y$ cannot be disjoint.

• Option (C): $\forall X \in U, \forall Y \in U (|X| = 2, |Y| = 3 \text{ and } X \setminus Y = \emptyset)$ is False because not all subsets in $U$ have cardinality $2$ and $3$.

• Option (D): $\forall X \in U, \forall Y \in U (X \setminus Y = Y^{\prime} \setminus X^{\prime})$ is True: Using standard set identities:

  $$X \setminus Y = X \cap Y^{\prime}$$

  $$Y^{\prime} \setminus X^{\prime} = Y^{\prime} \cap (X^{\prime})' = Y^{\prime} \cap X = X \cap Y^{\prime}$$

  Since both sides simplify to $X \cap Y^{\prime}$, the equality holds for all $X, Y \in U$.

Therefore, (D) is the correct option.