If `A ` and `B` are any sets, prove that `Ann(B-A)=phi`

Answer

To prove that \( A \cap (B - A) = \emptyset \), we will follow these steps:

### Step 1: Understand the Definitions
- **Set Difference**: \( B - A \) is the set of all elements that are in set \( B \) but not in set \( A \).
- **Intersection**: \( A \cap (B - A) \) is the set of all elements that are common to both set \( A \) and the set \( B - A \).

### Step 2: Express \( B - A \)
Let’s define \( B - A \):
\[ B - A = \{ x \in B \mid x \notin A \} \]
This means \( B - A \) contains all elements of \( B \) that are not in \( A \).

### Step 3: Find the Intersection
Now, we need to find \( A \cap (B - A) \):
\[ A \cap (B - A) = \{ x \in A \mid x \in (B - A) \} \]
Substituting the definition of \( B - A \):
\[ A \cap (B - A) = \{ x \in A \mid x \in B \text{ and } x \notin A \} \]

### Step 4: Analyze the Intersection
The condition \( x \in A \) and \( x \notin A \) cannot be true at the same time. Therefore, there are no elements that can satisfy both conditions.

### Step 5: Conclude the Proof
Since there are no elements that can be in both \( A \) and \( B - A \), we conclude that:
\[ A \cap (B - A) = \emptyset \]

Thus, we have proved that \( A \cap (B - A) = \emptyset \).

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