The Boolean expression $F=(x.y'+x'.y+x'.y')$ is equivalent to :

Question

Accepted Answer

The correct option is (2):
The correct answer is derived by simplifying the given Boolean expression $F = x.y' + x'.y + x'.y'$ using the fundamental laws of Boolean Algebra. Notice that the last two terms, $x'.y$ and $x'.y'$, share the common term $x'$. We can group these using the Distributive Law ($A.B + A.C = A(B+C)$) to get $F = x.y' + x'(y + y')$. By applying the Complement Law ($y + y' = 1$), the expression simplifies to $F = x.y' + x'.(1)$, which is further simplified to $F = x.y' + x'$ using the Identity Law ($A.1 = A$). Finally, we use the Absorption Law variation, $A + A'.B = A + B$. Here, with $A=x'$ and $B=y'$, the expression $x' + x.y'$ simplifies directly to $\boldsymbol{x' + y'}$. This process demonstrates the logical equivalence of the complex sum-of-products form to a simpler sum term.

The incorrect option is (1):
The expression $x + y'$ is incorrect because it is missing the logic covered by the $x'.y$ term in the original function. The simplified form $x' + y'$ implies that the output $F$ is HIGH (1) if $x$ is LOW (0) OR if $y$ is LOW (0). The simplified option (1) $x + y'$ would only be correct if the original expression was $x.y' + x.y$, which simplifies to $x.(y' + y) = x.1 = x$. However, adding the $x'.y'$ term into $x$ simplifies it to $x + x'.y' = x + y'$, which is a different function than the correct answer.

The incorrect option is (3):
The expression $x' + y$ is incorrect. This expression is the logical dual of the correct answer $x' + y'$. The expression $x' + y$ would be the correct simplification for $x'.y + x'.y' + x.y$, where the $x.y'$ term in the original expression is replaced by $x.y$. In the expression $x' + y$, $F$ is HIGH (1) if $x$ is LOW (0) OR $y$ is HIGH (1). The presence of the $y'$ term in the original expression dictates that the simplified form must also contain $y'$, not $y$.

The incorrect option is (4):
The expression $x + y$ represents the OR operation of the two input variables, meaning the output is HIGH unless both $x$ and $y$ are LOW. This result is too simple and omits critical parts of the original logic. For instance, if $x=0$ and $y=0$, the original function $F$ gives $0.1 + 1.0 + 1.1 = 0+0+1 = 1$, but $x+y$ gives $0+0=0$. Since the results differ for a valid input combination, $F$ is not equivalent to $x+y$.