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

The gates known as universal gates are the NAND gate and the NOR gate.

They are called universal gates 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. This means that any complex digital circuit or Boolean function can be implemented using just one type of universal gate, making them highly versatile in digital circuit design.

The Boolean expression is $(A.B)+C$. This expression involves two operations: an AND operation between A and B, and an OR operation between the result of (A.B) and C.

To draw the logic circuit:

1. First, implement the AND operation for inputs A and B. This requires an AND gate.
2. Next, take the output of the AND gate (which represents A.B) and use it as one input to an OR gate.
3. The other input to the OR gate will be C.
4. The output of the OR gate will be the final result, $(A.B)+C$.

The circuit diagram would consist of an AND gate with inputs A and B, and an OR gate with inputs from the AND gate's output and C.

The OR operator ($A\; +\; B$) can be expressed in terms of AND ($.$) and NOT ($\text{'}$ or overbar) operators using De Morgan's theorems. One of De Morgan's theorems states that the complement of a sum is equal to the product of the complements: $(A\; +\; B)\text{'}\; =\; A\text{'}\; \backslash cdot\; B\text{'}$.

To express the OR operation itself ($A\; +\; B$), we can apply the complement operation twice to the above equation:

$(A\; +\; B)\; =\; ((A\; +\; B)\text{'})\text{'}$

Now, substituting the result from De Morgan's theorem ($(A\; +\; B)\text{'}\; =\; A\text{'}\; \backslash cdot\; B\text{'}$):

$A\; +\; B\; =\; (A\text{'}\; \backslash cdot\; B\text{'})\text{'}$

Therefore, the OR operator can be expressed as the complement of the AND of the complements of the operands.

1. The Boolean reduction $(be)\text{'}\; +\; be\; =\; 1$ uses the Complementary Law. The Complementary Law states that for any Boolean variable x, $x\; +\; x\text{'}\; =\; 1$. In this case, $be$ can be considered as a variable, and its complement is $(be)\text{'}$.
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, $zx$ can be factored out, resulting in $zx(y\; +\; 1)$. Since $y\; +\; 1\; =\; 1$ (Identity Law), this simplifies to $zx\; \backslash cdot\; 1\; =\; zx$. Alternatively, viewing $zx$ as 'x' and $xyz$ as 'xy' in the form $x\; +\; xy\; =\; x$, the expression simplifies to $zx$.

Associative Law: This law states that the grouping of operands does not affect the result of the operation when the same operation is performed multiple times. It applies to both OR and AND operations.

For the OR operation:

$(A\; +\; B)\; +\; C\; =\; A\; +\; (B\; +\; C)$

For the 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 columns for $(A\; +\; B)\; +\; C$ and $A\; +\; (B\; +\; C)$ have identical values for all possible input combinations. Thus, the Associative Law for the 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 columns for $(A\; .\; B)\; .\; C$ and $A\; .\; (B\; .\; C)$ are identical, 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 for OR operation, $X+1\; =\; 1$. The correct statement for $X$ would be $X+0\; =\; X$.
2. The statement $(A\text{'})\text{'}\; =\; A\text{'}$ is incorrect. The Double Complement Law states that $(A\text{'})\text{'}\; =\; A$. Therefore, the correct statement is $(A\text{'})\text{'}\; =\; A$.
3. The statement $A+A\text{'}=0$ is incorrect. According to the Complementary Law, the sum of a variable and its complement is always 1, i.e., $A+A\text{'}=1$. The correct statement for 0 would be $A.A\text{'}=0$.
4. The statement $(A+B)\text{'}\; =\; A.B$ is incorrect. This is a direct application of De Morgan's Theorem, which states that $(A+B)\text{'}\; =\; A\text{'}\; .\; B\text{'}$. The expression $A.B$ is the complement of the AND operation, not the OR operation.