CBSE Class 12 Computer Science Chapter 13: Boolean Algebra NCERT Solutions

The logic gates known as universal gates are the NAND (Not-AND) gate and the NOR (Not-OR) gate. These gates are termed 'universal' because any other basic logic gate, such as an AND gate, an OR gate, or a NOT gate, can be constructed using only NAND gates or only NOR gates. Consequently, any complex Boolean function or digital circuit can be implemented using just one type of universal gate, making them highly versatile in digital circuit design.

The given Boolean expression is $(A.B)+C$. This expression represents a logic circuit where:

1. First, inputs A and B are ANDed together ($A.B$).
2. The output of the AND gate is then ORed with input C ($(A.B)+C$).

The logic circuit would consist of one AND gate and one OR gate. The AND gate takes inputs A and B, and its output is fed as one of the inputs to the OR gate. The other input to the OR gate is C.

(Note: A visual representation of the circuit diagram would typically be included here. Since this is text-based, the description above outlines the structure.)

The OR operator ($A\; +\; B$) can be expressed in terms of AND ($.$) and NOT ($\text{'}$) operators using De Morgan's theorems. One way to achieve this is by using the second form of De Morgan's theorem, which states that $(X\; +\; Y)\text{'}\; =\; X\text{'}\; .\; Y\text{'}$. If we apply the complement operation twice to the left side, we get $X\; +\; Y\; =\; ((X\; +\; Y)\text{'})\text{'}$. Substituting De Morgan's theorem into this, we get $X\; +\; Y\; =\; (X\text{'}\; .\; Y\text{'})\text{'}$.

Therefore, the OR operation $A\; +\; B$ can be expressed as $(A\text{'}\; .\; B\text{'})\text{'}$.

The steps are:

1. Complement both inputs: $A\text{'}$ and $B\text{'}$.
2. AND the complemented inputs: $A\text{'}\; .\; B\text{'}$.
3. Complement the result of the AND operation: $(A\text{'}\; .\; B\text{'})\text{'}$.

1. The Boolean reduction $(be)\text{'}\; +\; be\; =\; 1$ uses the Complementary Law. This law states that for any Boolean variable $x$, $x\; +\; x\text{'}\; =\; 1$. In this case, $x$ is represented by the term $be$ and $x\text{'}$ by $(be)\text{'}$.
2. The Boolean reduction $xyz\; +\; zx\; =\; xz$ uses the Absorption Law. The Absorption Law states that for any Boolean variables $x$ and $y$, $x\; +\; xy\; =\; x$ and $x(x+y)\; =\; x$. In the given expression, we can factor out $xz$ from both terms: $xz(y\; +\; 1)$. Since $y\; +\; 1\; =\; 1$ (Identity Law), the expression becomes $xz(1)$, which simplifies to $xz$ by the Identity Law. Alternatively, recognizing that $zx$ is a factor of $xyz$ (since $xyz\; =\; xz\; \backslash cdot\; y$), the term $zx$ is absorbed by $xyz$ according to the Absorption Law.

Statement of Associative Law: The Associative Law in Boolean Algebra states that the order in which operations are performed does not affect the final result when grouping operands. It applies to both the OR (+) and AND (.) operations:

• For OR operation: $(A\; +\; B)\; +\; C\; =\; A\; +\; (B\; +\; C)$
• For AND operation: $(A\; .\; B)\; .\; C\; =\; A\; .\; (B\; .\; C)$

Verification using Truth Table for OR operation:

We need to show that the columns for $(A\; +\; B)\; +\; C$ and $A\; +\; (B\; +\; C)$ are identical.

As observed from the truth table, the column for $(A\; +\; B)\; +\; C$ is identical to the column for $A\; +\; (B\; +\; C)$. Thus, the Associative Law for OR operation is verified.

Verification using Truth Table for AND operation:

Similarly, we can verify $(A\; .\; B)\; .\; C\; =\; A\; .\; (B\; .\; C)$.

The truth table shows that the column for $(A\; .\; B)\; .\; C$ is identical to the column for $A\; .\; (B\; .\; C)$, thus verifying the Associative Law for the AND operation.

Here are the corrected Boolean statements:

1. The statement $X+1\; =\; X$ is incorrect. According to the Identity Law (or Dominance Law), any variable ORed with 1 results in 1. The correct statement is $X+1\; =\; 1$. Alternatively, if the intention was to show that ORing with 0 leaves the variable unchanged, the correct statement would be $X+0\; =\; X$.
2. The statement $(A\text{'})\text{'}\; =\; A\text{'}$ is incorrect. The Double Complement Law states that complementing a variable twice returns the original variable, i.e., $(A\text{'})\text{'}\; =\; A$. Therefore, the correct statement is $(A\text{'})\text{'}\; =\; A$.
3. The statement $A+A\text{'}=0$ is incorrect. According to the Complementary Law, a variable ORed with its complement always results in 1. The correct statement is $A+A\text{'}=1$. The expression $A.A\text{'}=0$ is true based on the Complementary Law.
4. The statement $(A+B)\text{'}\; =\; A.B$ is incorrect. This is a misapplication of De Morgan's Theorem. De Morgan's theorem states that the complement of a sum is the product of the complements, i.e., $(A+B)\text{'}\; =\; A\text{'}.B\text{'}$. The expression $A.B$ is the complement of $(A\text{'}+B\text{'})$.