Let A, B, C be distinct subsets of a universal set U. For a subset X of U, let X' denotes the complement of X in U. Consider the following sets: `1. (A'nnB')nn(AuuBuuC')=(Auu(BuuC))'`

Options

A. `1` only
B. `2` only
C. Both `1` and `2`
D. Neither `1` nor `2`

Answer

To prove the equation \( (A' \cap B') \cap (A \cup B \cup C') = (A \cup B \cup C)' \), we will analyze both sides step by step.

### Step 1: Understand the Left-Hand Side (LHS)

The left-hand side is given as:
\[ (A' \cap B') \cap (A \cup B \cup C') \]

**Hint:** Remember that \( A' \) and \( B' \) represent the complements of sets \( A \) and \( B \) respectively, meaning all elements in the universal set \( U \) that are not in \( A \) and \( B \).

### Step 2: Simplify \( A' \cap B' \)

The intersection \( A' \cap B' \) consists of all elements that are not in \( A \) and not in \( B \). Thus, it can be represented as:
\[ A' \cap B' = U - (A \cup B) \]

**Hint:** The intersection of complements gives us the complement of the union.

### Step 3: Substitute into LHS

Now substitute the result from Step 2 into the LHS:
\[ (A' \cap B') \cap (A \cup B \cup C') = (U - (A \cup B)) \cap (A \cup B \cup C') \]

**Hint:** You are now looking for the intersection of two sets: one that excludes \( A \) and \( B \) and another that includes \( A \), \( B \), and excludes \( C \).

### Step 4: Analyze the Intersection

The intersection \( (U - (A \cup B)) \cap (A \cup B \cup C') \) will include all elements that are in \( A \cup B \cup C' \) but not in \( A \) or \( B \). Therefore, we can simplify this to:
\[ C' \cap (U - (A \cup B)) = C' - (A \cup B) \]

**Hint:** The intersection will only retain elements from \( C' \) that are not in \( A \) or \( B \).

### Step 5: Understand the Right-Hand Side (RHS)

The right-hand side is given as:
\[ (A \cup B \cup C)' \]

This represents all elements in the universal set \( U \) that are not in the union of \( A \), \( B \), and \( C \).

**Hint:** The complement of a union means excluding all elements that are in any of those sets.

### Step 6: Rewrite the RHS

The RHS can be expressed as:
\[ (A \cup B \cup C)' = U - (A \cup B \cup C) \]

**Hint:** This means we are excluding everything that is in \( A \), \( B \), or \( C \).

### Step 7: Compare LHS and RHS

Now, we need to compare the simplified LHS and RHS:
- LHS: \( C' - (A \cup B) \)
- RHS: \( U - (A \cup B \cup C) \)

To show they are equal, notice that:
\[ C' - (A \cup B) = U - (A \cup B \cup C) \]

This is true because if an element is in \( C' \) and not in \( A \) or \( B \), it must also not be in \( C \).

### Conclusion

Thus, we have shown that:
\[ (A' \cap B') \cap (A \cup B \cup C') = (A \cup B \cup C)' \]
Hence, the statement is proved.