In the Boolean algebra : `overline(A).overline(B) = ….`

Options

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

Explanation

To solve the Boolean algebra expression `overline(A) . overline(B)`, we will use De Morgan's Theorem, which states that the negation of a conjunction is the disjunction of the negations. The expression can be simplified step by step as follows:

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression `overline(A) . overline(B)`, which means the logical AND of the negations of A and B.

2. **Applying De Morgan's Theorem**:
According to De Morgan's Theorem, the expression `overline(A) . overline(B)` can be rewritten as `overline(A + B)`. This means that the negation of the conjunction (AND operation) is equivalent to the disjunction (OR operation) of the negations.

3. **Final Expression**:
Therefore, we can conclude that:
\[ \overline(A) . \overline(B) = \overline(A + B) \]

### Summary:
The expression `overline(A) . overline(B)` simplifies to `overline(A + B)` according to De Morgan's Theorem.