In Boolean algebra `bar((A+barB))*C` will be equal to

Options

A. `(barA*B)+barC`
B. `(A*barB)+C`
C. `(A*B)*C`
D. `(A+barB)+C`

Explanation

To solve the Boolean algebra expression `bar((A + barB) * C)`, we will apply De Morgan's Theorem step by step.

### Step-by-Step Solution:

1. **Identify the Expression**:
We start with the expression:
\[ \overline{(A + \overline{B}) \cdot C} \]

2. **Apply De Morgan's Theorem**:
According to De Morgan's Theorem, the negation of a product is equal to the sum of the negations:
\[ \overline{(X \cdot Y)} = \overline{X} + \overline{Y} \]
Here, let \( X = (A + \overline{B}) \) and \( Y = C \). Thus, we can write:
\[ \overline{(A + \overline{B}) \cdot C} = \overline{(A + \overline{B})} + \overline{C} \]

3. **Negate the Sum**:
Now we need to negate \( (A + \overline{B}) \). Again, we apply De Morgan's Theorem:
\[ \overline{(A + \overline{B})} = \overline{A} \cdot \overline{\overline{B}} = \overline{A} \cdot B \]
Therefore, substituting this back into our equation gives:
\[ \overline{(A + \overline{B}) \cdot C} = (\overline{A} \cdot B) + \overline{C} \]

4. **Final Expression**:
The final simplified expression is:
\[ \overline{A} \cdot B + \overline{C} \]

### Conclusion:
Thus, the expression `bar((A + barB) * C)` simplifies to:
\[ \overline{A} \cdot B + \overline{C} \]