In the Boolean algebra `bar((bar(A).bar(B))). A` equal to

Options

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

Explanation

To solve the Boolean algebra expression `bar((bar(A).bar(B))). A`, we will follow these steps:

### Step 1: Understand the Expression
The expression given is `bar((bar(A).bar(B))). A`. Here, `bar` denotes the NOT operation, and the dot (.) denotes the AND operation.

### Step 2: Apply De Morgan's Theorem
According to De Morgan's Theorem, we have:
\[ \bar(X \cdot Y) = \bar(X) + \bar(Y) \]
In our case, we can apply this to `bar(A) . bar(B)`.

So, we will rewrite the expression:
\[ \bar((\bar(A) \cdot \bar(B))) = \bar(\bar(A)) + \bar(\bar(B)) \]

### Step 3: Simplify the Expression
Using the property that `bar(bar(X)) = X`, we can simplify:
\[ \bar(\bar(A)) = A \]
\[ \bar(\bar(B)) = B \]

Thus, we have:
\[ \bar((\bar(A) \cdot \bar(B))) = A + B \]

### Step 4: Substitute Back into the Expression
Now, substituting back into the original expression:
\[ (A + B) \cdot A \]

### Step 5: Apply Distribution
Using the distribution property of Boolean algebra:
\[ (A + B) \cdot A = A \cdot A + B \cdot A \]

### Step 6: Simplify Further
Using the property that \( A \cdot A = A \):
\[ A + (B \cdot A) \]

### Step 7: Factor Out A
Now we can factor out A:
\[ A(1 + B) \]

### Step 8: Simplify Using Boolean Properties
Using the property that \( 1 + B = 1 \) (since anything ORed with 1 is 1):
\[ A \cdot 1 = A \]

### Final Result
Thus, the final result of the expression `bar((bar(A).bar(B))). A` is:
\[ A \]