Simplify the following Boolean expressions and draw their logic circuits: i) X = (A + B)(B + C) + (B + C) ii) Y = A . (A + B)(B + C) iii) Z = A . B + A . B . (B + C)

Let $P = (B+C)$. The expression becomes $X = (A+B)P + P$.
Using the Boolean algebra absorption law $Q + QP = Q$, where $Q = P$, we have:
$X = P$
Substitute $P = (B+C)$ back:
$X = B + C$

The circuit for $X = B + C$ consists of one OR gate.

• Inputs: B, C
• Gate: OR gate
• Output: X

Circuit for i):

   B ---|
        |--- OR --- X
   C ---|

First, simplify $A \cdot (A+B)$. Using the Boolean algebra absorption law $A \cdot (A+B) = A$.
So, the expression becomes:
$Y = A \cdot (B + C)$

The circuit for $Y = A \cdot (B + C)$ consists of one OR gate and one AND gate.

• Inputs for OR gate: B, C
• Output of OR gate: $(B+C)$
• Inputs for AND gate: A, output of OR gate
• Output of AND gate: Y

Circuit for ii):

   B ---|
        |--- OR ---|
   C ---|          |--- AND --- Y
                   |
   A --------------|

Let $Q = A \cdot B$. The expression becomes $Z = Q + Q \cdot (B+C)$.
Using the Boolean algebra absorption law $Q + QP = Q$, where $Q = A \cdot B$, we have:
$Z = Q$
Substitute $Q = A \cdot B$ back:
$Z = A \cdot B$

The circuit for $Z = A \cdot B$ consists of one AND gate.

• Inputs: A, B
• Gate: AND gate
• Output: Z

Circuit for iii):

   A ---|
        |--- AND --- Z
   B ---|