Boolean Algebra and De Morgan's Theorem

The question asks us to find the value of the Boolean expression \(\rm \overline{A+B}\). This expression represents the NOR operation between variables A and B. To simplify or find an equivalent expression for \(\rm \overline{A+B}\), we use one of the fundamental theorems in Boolean algebra known as De Morgan's Theorem.

De Morgan's Theorems Explained

De Morgan's Theorems are a pair of transformation rules in Boolean algebra that are essential for simplifying expressions and designing digital circuits. They allow us to convert sums into products and products into sums, along with inversion.

There are two main theorems:

• First Theorem: The complement of a sum is equal to the product of the complements. Mathematically, this is expressed as:

  \(\rm \overline{A+B} = \overline{A} \cdot \overline{B}\)

  This means the NOT-OR (NOR) operation is equivalent to the NOT-A AND NOT-B operation.

• Second Theorem: The complement of a product is equal to the sum of the complements. Mathematically, this is expressed as:

  \(\rm \overline{A \cdot B} = \overline{A} + \overline{B}\)

  This means the NOT-AND (NAND) operation is equivalent to the NOT-A OR NOT-B operation.

Applying De Morgan's Theorem to \(\rm \overline{A+B}\)

Our expression is \(\rm \overline{A+B}\). To simplify this, we apply De Morgan's First Theorem directly.

According to De Morgan's First Theorem:

\(\rm \overline{A+B} = \overline{A} \cdot \overline{B}\)

Here, \(\rm \overline{A}\) represents NOT A, and \(\rm \overline{B}\) represents NOT B. The dot \(\rm (\cdot)\) represents the logical AND operation. So, \(\rm \overline{A} \cdot \overline{B}\) can also be written as \(\rm A̅ B̅ \).

Comparing with Options

Let's examine each given option to see which one matches our result \(\rm A̅ B̅ \):

• Option 1: \(\rm A \cdot B\)

  This represents the logical AND operation between A and B, which is not equivalent to \(\rm \overline{A+B}\).

• Option 2: \(\rm A̅ B̅ \)

  This notation implies \(\rm A̅ \cdot B̅ \), which is exactly what we derived from De Morgan's First Theorem (\(\rm \overline{A} \cdot \overline{B}\)). Therefore, this option is correct.

• Option 3: \(\rm A B̅ \)

  This implies \(\rm A \cdot B̅ \), representing A AND NOT B. This is not equivalent to \(\rm \overline{A+B}\).

• Option 4: \(\rm A̅ B\)

  This implies \(\rm A̅ \cdot B\), representing NOT A AND B. This is not equivalent to \(\rm \overline{A+B}\).

Based on De Morgan's Theorem, the expression \(\rm \overline{A+B}\) is equivalent to \(\rm \overline{A} \cdot \overline{B}\) or \(\rm A̅ B̅ \).