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

Options

A. `A.bar(B)+bar(A).B`
B. `(A+bar(B))(bar(A)+B)`
C. `bar(A.bar(B)+bar(A).B)`
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 laws. Let's break it down step by step:

### Step 1: Apply De Morgan's Theorem
According to De Morgan's theorem, the complement of a sum is equal to the product of the complements. Therefore, we can rewrite the expression as follows:

\[ \overline{A \cdot \overline{B} + \overline{A} \cdot B} = \overline{A \cdot \overline{B}} \cdot \overline{\overline{A} \cdot B} \]

### Step 2: Apply De Morgan's Theorem to Each Term
Now we will apply De Morgan's theorem to each term separately.

1. For the first term \( \overline{A \cdot \overline{B}} \):
\[ \overline{A \cdot \overline{B}} = \overline{A} + \overline{\overline{B}} = \overline{A} + B \]

2. For the second term \( \overline{\overline{A} \cdot B} \):
\[ \overline{\overline{A} \cdot B} = \overline{\overline{A}} + \overline{B} = A + \overline{B} \]

### Step 3: Combine the Results
Now we can combine the results from Step 2:

\[ \overline{A \cdot \overline{B} + \overline{A} \cdot B} = (\overline{A} + B) \cdot (A + \overline{B}) \]

### Step 4: Expand the Expression
Now we can expand the expression:

\[ (\overline{A} + B)(A + \overline{B}) = \overline{A}A + \overline{A}\overline{B} + BA + B\overline{B} \]

Since \( \overline{A}A = 0 \) and \( B\overline{B} = 0 \), we can simplify:

\[ = \overline{A}\overline{B} + AB \]

### Final Result
Thus, the expression \( \overline{A \cdot \overline{B} + \overline{A} \cdot B} \) simplifies to:

\[ \overline{A} \cdot \overline{B} + A \cdot B \]