Boolean Expression Simplification

To determine the equivalent Boolean expression, we will simplify the given expression step-by-step using fundamental Boolean algebra laws. Then, we will check which of the provided options simplifies to the same result.

Given Boolean Expression

The given Boolean expression is: \( XY + ( X' + Y' )Z \)

Step-by-Step Simplification of the Expression

1. Applying De Morgan's Theorem

The term \( (X' + Y') \) can be simplified using De Morgan's Theorem, which states that \( (A' + B') = (AB)' \). Applying this to our expression:

\( (X' + Y') = (XY)' \)

Substitute this back into the original expression:

\( XY + (XY)'Z \)

2. Applying the Absorption Law/Identity

Now, we use a key Boolean algebra identity: \( A + A'B = A + B \). In our current expression, let A = XY and B = Z. Applying this identity:

\( XY + (XY)'Z = XY + Z \)

So, the given Boolean expression \( XY + ( X' + Y' )Z \) simplifies to \( XY + Z \).

Simplifying the Correct Option for Equivalence

The correct answer option provided is \( (X + Z)(Y + Z) \). Let's expand and simplify this expression to see if it matches our result of \( XY + Z \).

1. Applying the Distributive Law

Expand the product \( (X + Z)(Y + Z) \) using the Distributive Law \( A(B + C) = AB + AC \):

\( (X + Z)(Y + Z) = X(Y + Z) + Z(Y + Z) \)

\( = XY + XZ + ZY + ZZ \)

2. Applying the Idempotent Law

According to the Idempotent Law, \( A \cdot A = A \). Therefore, \( ZZ = Z \).

Substitute this back into the expression:

\( = XY + XZ + YZ + Z \)

3. Applying Absorption Property (repeatedly)

Now we simplify \( XY + XZ + YZ + Z \). We can use the absorption property \( A + AB = A \) or \( A + B = A + B(1) = A + B(C+1) \), and \( A + 1 = 1 \). A simpler way to look at this is that if a variable (like \( Z \)) is present as a standalone term in a sum, any term that is a product involving that variable will be absorbed by the standalone variable.

• Consider \( XZ + Z \): This simplifies to \( Z(X + 1) \). Since \( X + 1 = 1 \), then \( Z(1) = Z \).
• Similarly, \( YZ + Z \): This simplifies to \( Z(Y + 1) \). Since \( Y + 1 = 1 \), then \( Z(1) = Z \).

Applying this to our expression:

\( XY + XZ + YZ + Z \)

Group terms with \( Z \):

\( = XY + (XZ + Z) + YZ \)

\( = XY + Z + YZ \)

Again, group terms with \( Z \):

\( = XY + (Z + YZ) \)

\( = XY + Z \)

Conclusion on Equivalence

We found that the original Boolean expression \( XY + ( X' + Y' )Z \) simplifies to \( XY + Z \). We also found that the option \( (X + Z)(Y + Z) \) simplifies to \( XY + Z \).

Since both expressions simplify to the same form, \( XY + Z \), they are equivalent.

Summary of Boolean Algebra Laws Used