Gate-count after simplification Simplify the Boolean expression XY + X(X + Z) + Y(X + Z) algebraically and determine the minimum number of logic gates required to implement the simplified result (assume 2-input gates).

Introduction / Context:
Algebraic simplification directly reduces hardware. By applying Boolean identities, we can shrink gate count, improve speed, and reduce power. This problem illustrates how eliminating redundancy can cut the implementation from several gates to just a couple.

Given Data / Assumptions:

• Expression: XY + X(X + Z) + Y(X + Z).
• Use standard Boolean identities.
• Assume final implementation with 2-input gates.

Concept / Approach:
Use absorption and distribution: X(X + Z) = X. Also expand Y(X + Z) = YX + YZ = XY + YZ, then combine like terms and apply absorption X + XY = X to collapse redundancy.

Step-by-Step Solution:

Verification / Alternative check:
A quick truth-table or Karnaugh map confirms F = X + YZ matches the original on all input combinations. This cross-check ensures no functional change occurred during simplification.

Why Other Options Are Wrong:

• 1: Impossible with both an AND product (YZ) and an OR sum with X required.
• 4 or 5: Reflect pre-simplification or non-shared implementations; simplification reduces to 2.

Common Pitfalls:
Missing the key identity X(X + Z) = X; failing to notice absorption X + XY = X; or forgetting that the expanded XY term is redundant once X is present.

Final Answer:
2