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To verify De Morgan's laws for complementation, we need to analyze the sets provided in the image.

### Given:

- Universal set \( U = \{x : x \in \mathbb{Z}, -2 \leq x \leq 10\} \)
- Set \( A = \{x : x = 2p + 1, \, p \in \mathbb{Z}, \, -1 \leq p \leq 4\} \)
- Set \( B = \{x : x = 3q + 1, \, q \in \mathbb{Z}, \, -1 \leq q < 4\} \)

### Step 1: Determine Elements in \( A \) and \( B \)

**Set \( A \):**
Given \( x = 2p + 1 \) and \( -1 \leq p \leq 4 \):

- If \( p = -1 \), \( x = 2(-1) + 1 = -1 \)
- If \( p = 0 \), \( x = 2(0) + 1 = 1 \)
- If \( p = 1 \), \( x = 2(1) + 1 = 3 \)
- If \( p = 2 \), \( x = 2(2) + 1 = 5 \)
- If \( p = 3 \), \( x = 2(3) + 1 = 7 \)
- If \( p = 4 \), \( x = 2(4) + 1 = 9 \)

Thus, \( A = \{-1, 1, 3, 5, 7, 9\} \).

**Set \( B \):**
Given \( x = 3q + 1 \) and \( -1 \leq q < 4 \):

- If \( q = -1 \), \( x = 3(-1) + 1 = -2 \)
- If \( q = 0 \), \( x = 3(0) + 1 = 1 \)
- If \( q = 1 \), \( x = 3(1) + 1 = 4 \)
- If \( q = 2 \), \( x = 3(2) + 1 = 7 \)
- If \( q = 3 \), \( x = 3(3) + 1 = 10 \)

Thus, \( B = \{-2, 1, 4, 7, 10\} \).

### Step 2: Determine the Complements of \( A \) and \( B \)

The complements of sets \( A \) and \( B \), denoted as \( A' \) and \( B' \), are the sets of elements in the universal set \( U \) that are not in \( A \) or \( B \), respectively.

- \( U = \{-2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \)

- \( A' = U \setminus A = \{-2, 0, 2, 4, 6, 8, 10\} \)

- \( B' = U \setminus B = \{-1, 0, 2, 3, 5, 6, 8, 9\} \)

### Step 3: Verify De Morgan's Laws

1. **First Law:** \( (A \cup B)' = A' \cap B' \)

   - \( A \cup B = \{-2, -1, 1, 3, 4, 5, 7, 9, 10\} \)
   
   - \( (A \cup B)' = U \setminus (A \cup B) = \{0, 2, 6, 8\} \)

   - \( A' \cap B' = \{-2, 0, 2, 4, 6, 8, 10\} \cap \{-1, 0, 2, 3, 5, 6, 8, 9\} = \{0, 2, 6, 8\} \)

   - Since \( (A \cup B)' = A' \cap B' \), the first law is verified.

2. **Second Law:** \( (A \cap B)' = A' \cup B' \)

   - \( A \cap B = \{1, 7\} \)
   
   - \( (A \cap B)' = U \setminus (A \cap B) = \{-2, -1, 0, 2, 3, 4, 5, 6, 8, 9, 10\} \)

   - \( A' \cup B' = \{-2, 0, 2, 4, 6, 8, 10\} \cup \{-1, 0, 2, 3, 5, 6, 8, 9\} = \{-2, -1, 0, 2, 3, 4, 5, 6, 8, 9, 10\} \)

   - Since \( (A \cap B)' = A' \cup B' \), the second law is verified.

### Conclusion

Both of De Morgan's laws for complementation are verified for the given sets \( A \) and \( B \) with the universal set \( U \).

Would you like more details or have any questions?

### Related Questions:

1. What are De Morgan's laws, and how are they used in set theory?
2. How do you find the complement of a set in a universal set?
3. Can De Morgan's laws be applied to more than two sets?
4. How do intersections and unions of sets differ?
5. What is the importance of universal sets in set theory?

### Tip: 
When working with sets, always start by clearly defining the universal set to correctly determine complements.

Verify De Morgan's laws for complementation, given the universal set U = {x : x ∈ Z, -2 ≤ x ≤ 10}, and sets A = {x : x = 2p + 1, p ∈ Z, -1 ≤ p ≤ 4}, B = {x : x = 3q + 1, q ∈ Z, -1 ≤ q < 4}.

Learn how to verify De Morgan's laws using the sets A and B, with elements derived from specific equations within a given universal set. Understand the laws of set complementation through step-by-step calculations.

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