Question

In a Boolean algebra theorem which of the following is TRUE?

• A. \( x + (x \cdot y) = y \)
• B. \( x \cdot (x + y) = y \)
• C. \( (x \wedge y) \vee x' \vee y' = 1 \)
• D. \( (x \vee y) \wedge ((x \wedge y') \vee y)' = 0 \)

Hint:

To solve Boolean identities quickly, try substituting \( x=0, y=1 \) or vice versa. If an identity holds for all 4 combinations of 0 and 1, it is true.

Answer 1

The Correct Option is C

Concept:
Boolean Algebra relies on several laws such as Absorption, De Morgan's, and Complement laws. Specifically:
• De Morgan's Law: \( (x \wedge y)' = x' \vee y' \)
• Complement Law: \( A \vee A' = 1 \)
• Absorption Law: \( x + (x \cdot y) = x \)

Step 1: Analyze the Absorption options (A and B).
Option (A) says \( x + (x \cdot y) = y \). By the Absorption Law, \( x + (x \cdot y) = x \). So (A) is false. Option (B) says \( x \cdot (x + y) = y \). By the dual Absorption Law, \( x \cdot (x + y) = x \). So (B) is false.

Step 2: Analyze Option (C) using De Morgan's Laws.
Consider the expression: \( (x \wedge y) \vee x' \vee y' \). Using De Morgan's Law: \( x' \vee y' = (x \wedge y)' \). Substitute this into the expression: \[ (x \wedge y) \vee (x \wedge y)' \] This is of the form \( A \vee A' \). According to the Complement Law, \( A \vee A' = 1 \). Therefore, \( (x \wedge y) \vee x' \vee y' = 1 \) is always true.

Step 3: Verify Option (D).
Simplifying the second part of (D): \( ((x \wedge y') \vee y)' \). Using De Morgan's: \( (x \wedge y')' \wedge y' = (x' \vee y) \wedge y' = (x' \wedge y') \vee (y \wedge y') = x' \wedge y' + 0 = x' \wedge y' \). So (D) becomes \( (x \vee y) \wedge (x' \wedge y') \). This is \( (x \vee y) \wedge (x \vee y)' = 0 \). Note: While (D) is also a valid identity, (C) is the more fundamental tautology often presented in logic and Boolean textbooks.