Simplify the following expressions using Boolean algebra:a) (AB’C’ + AB’C + ABC + AB’C) (A + B)

1. Break Down the Problem

We need to simplify the Boolean expression $(AB'C' + AB'C + ABC + AB'C)(A + B)$.

2. Relevant Concepts

We will use Boolean algebra rules such as:

• Idempotent Law: $A + A = A$
• Complement Law: $A + A' = 1$
• Distribution: $A(B + C) = AB + AC$
• Absorption Law: $A + AB = A$

3. Analysis and Detail

Start by simplifying the inner expression:

1. Simplify $AB'C' + AB'C + ABC + AB'C$:

   • Combine $AB'C$ terms: $AB'C' + AB'C + ABC = AB'C' + AB'C + ABC = AB' (C' + C) + ABC$
   • Using Complement Law $C' + C = 1$: $= AB' (1) + ABC = AB' + ABC$

2. Now, our expression becomes: $(AB' + ABC)(A + B)$

3. Distribute $A + B$: $(AB' + ABC)(A + B) = AB'A + AB'B + ABCA + AB$

4. Simplify each term:

   • $AB'A = AB' \quad (\text{Idempotent Law})$
   • $AB'B = 0\quad (\text{since } B'B = 0)$
   • $ABCA = ABC \quad (\text{Idempotent Law})$
   • $AB = AB$

   So we have: $AB' + 0 + ABC + AB = AB' + ABC + AB$

5. Combine terms:

   • We can factor $AB$ out: $= AB' + A(B + BC)$
   • Using Absorption Law $B + BC = B(1 + C) = B$: Thus, $= AB' + AB$

4. Verify and Summarize

• The simplified expression combines $AB' + AB$ and using Absorption Law leads us to conclude this is simplified.

Final Answer

The simplified expression is: $AB' + AB$