To reduce the given Boolean expression, we will apply fundamental Boolean algebra laws step-by-step. The expression to be reduced is:

\((B + BC) (B + \bar{B}C) (B + D)\)

Boolean Expression Simplification

We will simplify each term of the Boolean expression individually first, and then combine them to find the final reduced result.

First Term: Simplifying \((B + BC)\)

The first part of the Boolean expression is \((B + BC)\). This can be simplified using the Absorption Law. The Absorption Law states that \(A + AB = A\).

• Let \(A = B\).
• Then, \((B + BC)\) becomes \(B\).

So, the first term reduces to \(\mathbf{B}\).

Second Term: Simplifying \((B + \bar{B}C)\)

The second part of the Boolean expression is \((B + \bar{B}C)\). This can be simplified using the Consensus Theorem (also known as the Simplification Theorem or Postulate 1.8 in some texts). The Consensus Theorem states that \(A + \bar{A}B = A + B\).

• Let \(A = B\).
• Let \(B = C\).
• Then, \((B + \bar{B}C)\) becomes \(B + C\).

Alternatively, we can use the Distributive Law \(A(B+C) = AB + AC\) and then the Complement Law \(A + \bar{A} = 1\):

• \((B + \bar{B}C) = (B + \bar{B})(B + C)\) (Distributive Law)
• We know that \((B + \bar{B}) = 1\) (Complement Law)
• So, \((1)(B + C) = B + C\)

Thus, the second term reduces to \(\mathbf{B + C}\).

Combining and Further Reducing the Boolean Expression

Now, substitute the simplified terms back into the original Boolean expression:

\((B) (B + C) (B + D)\)

Reducing \((B) (B + C)\)

We now consider the first two terms: \((B) (B + C)\). This can again be simplified using the Absorption Law. The Absorption Law also states that \(A(A + B) = A\).

• Let \(A = B\).
• Then, \((B) (B + C)\) becomes \(B\).

So, the expression simplifies to \((B) (B + D)\).

Reducing the Final Expression \((B) (B + D)\)

Finally, we have \((B) (B + D)\). Applying the Absorption Law \(A(A + B) = A\) one more time:

• Let \(A = B\).
• Then, \((B) (B + D)\) becomes \(B\).

Therefore, the completely reduced result of the Boolean expression is \(\mathbf{B}\).

Summary of Boolean Laws Used

The following Boolean algebra laws were crucial in reducing the expression: