Question

Match List - I with List - II. 

List - I		List - II
A.	\(x(x' + y)\)	I.	\(xy + x'z\)
B.	\(x'y'z + x'yz + xy'\)	II.	\((x + y)(x' + z)\)
C.	\(xy + x'z + yz\)	III.	\(xy\)
D.	\((x + y)(x' + z)(y + z)\)	IV.	\(x'z + xy'\)

Choose the correct answer from the options given below:

• A. A-III, B-IV, C-I, D-II
• B. A-I, B-III, C-II, D-IV
• C. A-III, B-IV, C-II, D-I
• D. A-I, B-III, C-IV, D-II

Hint:

The Consensus Theorem is a lifesaver for matching questions: $AB + A'C + BC = AB + A'C$. Look for the "bridge" term ($BC$) where the variables from the other two terms ($A$ and $A'$) cancel out!

Answer 1

The Correct Option is A

Concept:
• Boolean algebra simplification relies on basic laws like Distributive Law (\(A(B+C) = AB + AC\)), Complement Law (\(AA' = 0\)), and the Consensus Theorem (\(AB + A'C + BC = AB + A'C\)).

Step 1: Simplify expression A
Using Distributive Law and Complement Law: \[ \begin{aligned} x(x' + y) &= x \cdot x' + x \cdot y &= 0 + xy &= xy \end{aligned} \] This matches with III. So, A-III.

Step 2: Simplify expression B
Factoring out \(x'z\) from the first two terms: \[ \begin{aligned} x'y'z + x'yz + xy' &= x'z(y' + y) + xy' &= x'z(1) + xy' &= x'z + xy' \end{aligned} \] This matches with IV. So, B-IV.

Step 3: Simplify expression C using Consensus Theorem
The Consensus Theorem states: \(AB + A'C + BC = AB + A'C\). Here, let \(A=x, B=y, C=z\): \[ xy + x'z + yz = xy + x'z \] The term \(yz\) is redundant. This matches with I. So, C-I.

Step 4: Simplify expression D using Dual Consensus Theorem
The Dual Consensus Theorem states: \((A + B)(A' + C)(B + C) = (A + B)(A' + C)\). Here, let \(A=x, B=y, C=z\): \[ (x + y)(x' + z)(y + z) = (x + y)(x' + z) \] This matches with II. So, D-II.