`bar(A.barB+barA.B)` is equivalent to

Options

A. `A.bar(B)+bar(A).B`
B. `(A+barB)(bar(A)+B)`
C. `bar(A.bar(B)+bar(AB)`
D. `(A+B).(bar(A)+B)`

Answer

To solve the expression \( \overline{A \cdot \overline{B} + \overline{A} \cdot B} \), we can use De Morgan's Theorems. Let's break it down step by step.

### Step 1: Identify the expression
The expression we need to simplify is:
\[ \overline{A \cdot \overline{B} + \overline{A} \cdot B} \]

### Step 2: Apply De Morgan's Theorem
According to De Morgan's Theorem, the complement of a sum is the product of the complements. Therefore, we can rewrite the expression as:
\[ \overline{A \cdot \overline{B}} \cdot \overline{\overline{A} \cdot B} \]

### Step 3: Simplify each term
Now, we simplify each term:
1. For \( \overline{A \cdot \overline{B}} \):
- Using De Morgan's Theorem again, this becomes:
\[ \overline{A} + B \]

2. For \( \overline{\overline{A} \cdot B} \):
- Again using De Morgan's Theorem, this becomes:
\[ A + \overline{B} \]

### Step 4: Combine the results
Now, we combine the results from Step 3:
\[ (\overline{A} + B) \cdot (A + \overline{B}) \]

### Step 5: Expand the expression
Now we can expand this expression:
\[ \overline{A} \cdot A + \overline{A} \cdot \overline{B} + B \cdot A + B \cdot \overline{B} \]

### Step 6: Simplify the expression
Using the property that \( X \cdot \overline{X} = 0 \):
1. \( \overline{A} \cdot A = 0 \)
2. \( B \cdot \overline{B} = 0 \)

Thus, the expression simplifies to:
\[ \overline{A} \cdot \overline{B} + B \cdot A \]

### Final Result
The final simplified expression is:
\[ \overline{A} \cdot \overline{B} + A \cdot B \]