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

Answer

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

### Step 1: Understand the Sets
First, let's clarify what \( B - A \) means. The expression \( B - A \) represents the set of all elements that are in set \( B \) but not in set \( A \).

### Step 2: Assume an Element in the Intersection
We want to prove that the intersection \( A \cap (B - A) \) is empty. To do this, we will assume that there exists an element \( x \) such that \( x \in A \cap (B - A) \).

### Step 3: Analyze the Assumption
If \( x \in A \cap (B - A) \), then by the definition of intersection, \( x \) must satisfy both conditions:
1. \( x \in A \)
2. \( x \in (B - A) \)

### Step 4: Understand the Implication of \( x \in (B - A) \)
From the second condition, \( x \in (B - A) \) means that \( x \) is in \( B \) but not in \( A \). This can be expressed as:
- \( x \in B \)
- \( x \notin A \)

### Step 5: Reach a Contradiction
Now we have:
- \( x \in A \) (from our assumption)
- \( x \notin A \) (from the definition of \( B - A \))

This creates a contradiction because \( x \) cannot be both in \( A \) and not in \( A \) at the same time.

### Step 6: Conclude the Proof
Since our assumption that there exists an element \( x \) in \( A \cap (B - A) \) leads to a contradiction, we conclude that no such element exists. Therefore, we have:
\[ A \cap (B - A) = \emptyset \]

### Final Statement
Thus, we have proven that \( A \cap (B - A) = \emptyset \).

---