Question: II. Let
$U = \{\alpha \mid \alpha \text{ is a letter in the English Alphabet}\} = \{a, b, c, \ldots, x, y, z\}$
$A = \{t, r, i, a, n, g, l, e, s\}$
$B = \{s, q, u, a, r, e\}$,
$C = \{h, e, x, a, g, o, n, s\}$
Determine the following.
$A \cap (C \cap U)^c$
1. **State the problem:** We need to find $A \cap (C \cap U)^c$ where $U$ is the universal set of English alphabet letters, $A$, $B$, and $C$ are given subsets.
2. **Understand the sets:**
- $U$ is all letters $\{a,b,c,\ldots,z\}$.
- $A = \{t, r, i, a, n, g, l, e, s\}$.
- $C = \{h, e, x, a, g, o, n, s\}$.
3. **Calculate $C \cap U$:** Since $C$ is a subset of $U$, $C \cap U = C = \{h, e, x, a, g, o, n, s\}$.
4. **Find the complement $(C \cap U)^c$ relative to $U$:**
$$
(C \cap U)^c = U \setminus C = \{\text{all letters in } U \text{ but not in } C\} = \{b, c, d, f, i, j, k, l, m, p, q, r, t, u, v, w, y, z\}
$$
5. **Find the intersection $A \cap (C \cap U)^c$:**
$$
A \cap (C \cap U)^c = \{t, r, i, a, n, g, l, e, s\} \cap \{b, c, d, f, i, j, k, l, m, p, q, r, t, u, v, w, y, z\}
$$
6. **Identify common elements:**
- From $A$: $t, r, i, a, n, g, l, e, s$
- From complement: $b, c, d, f, i, j, k, l, m, p, q, r, t, u, v, w, y, z$
Common elements are $\{i, l, r, t\}$.
7. **Final answer:** $A \cap (C \cap U)^c = \{i, l, r, t\}$.
This corresponds to option (d).