Which of the following is not a valid expression of Boolean Algebra ?

Question

Accepted Answer

The correct option is (3):
The expression x'.y' + x.y is an identity for the Exclusive NOR (XNOR) operation, also known as the equivalence function. This XNOR expression, represented as $x \odot y$, is a two-variable function that is only equal to $x$ when the variable $y$ is logically equal to 1. Since a law in Boolean Algebra must hold true for all possible combinations of values for $x$ and $y$, the statement $\text{x'}.\text{y'} + \text{x.y} = \text{x}$ is not a universally valid law. For instance, if we test with $x=1$ and $y=0$, the Left Hand Side (LHS) evaluates to $\text{1'}.\text{0'} + \text{1.0} = \text{0.1} + \text{1.0} = 0 + 0 = 0$, while the Right Hand Side (RHS) is $x=1$. As $0 \neq 1$, this equation is proven false for that specific case. Therefore, this expression is the one that is NOT a valid law of Boolean Algebra.

The incorrect option is (1):
The expression $\text{x} + \text{x.y} = \text{x}$ is a fundamental identity known as the **Absorption Law** in Boolean Algebra. This law is valid because the term $\text{x.y}$ is redundant in the expression $\text{x} + \text{x.y}$. We can formally prove this using other laws: $\text{x} + \text{x.y} = \text{x}.(1) + \text{x.y}$. By factoring out the common term $x$, the expression becomes $\text{x}.(1 + \text{y})$. Because the Identity Law states that $1 + \text{y}$ is always equal to $1$, the entire expression simplifies to $\text{x}.(1)$, which is simply $\text{x}$. Thus, the statement $\text{x} + \text{x.y} = \text{x}$ is always true.

The incorrect option is (2):
The expression $\text{x'}.\text{y} + \text{x.y} = \text{y}$ is a valid and commonly used simplification in Boolean Algebra. This identity is proven by applying the **Distributive Law** followed by the **Inverse Law**. By factoring out the common term $\text{y}$, the expression transforms into $\text{y}.(\text{x'} + \text{x})$. The Inverse Law dictates that any variable ORed with its complement, such as $\text{x'} + \text{x}$, always results in a logic '1'. Consequently, the entire expression simplifies to $\text{y}.(1)$, which is simply equal to $\text{y}$. This demonstrates that if $y$ is true, the entire statement is true, regardless of $x$'s value, confirming its validity.

The incorrect option is (4):
The expression $(\text{x} + \text{y} + \text{z})' = \text{x'}.\text{y'}.\text{z'}$ is a correct and direct application of **De Morgan's Theorem**. De Morgan's theorem provides a rule for negating compound propositions or Boolean expressions. Specifically, it states that the complement of a sum (OR operation) is equivalent to the product (AND operation) of the complements of the individual variables. This law is essential for manipulating and simplifying complex Boolean functions and is universally true for any number of variables, including three as shown in this option.