P5: Simplify the following Boolean expression.
X=A’B’C’D’E’+A’B’CD’E’+A’BCD’E’+A’BC’D’E’+A’B’C’D’E+A’BCD’E+A’BCDE+AB’C’D’E+AB’C’D’E+ABCD’E+ABCDE+AB’CDE

To simplify the given Boolean expression, we will follow these steps:

### Expression:
\[ 
X = A'B'C'D'E' + A'B'CD'E' + A'BCD'E' + A'BC'D'E' + A'B'C'D'E + A'BCD'E + A'BCDE + AB'C'D'E + AB'C'D'E + ABCD'E + ABCDE + AB'CDE
\]

#### Step 1: Group terms for simplification
Group terms based on their common literals. Let’s rewrite:

1. \( A'B'C'D'E' \) and \( A'B'C'D'E \)
2. \( A'B'CD'E' \)
3. \( A'BCD'E' \) and \( A'BCD'E \)
4. \( A'BC'D'E' \)
5. \( A'BCDE \)
6. \( AB'C'D'E \)
7. \( ABCD'E \)
8. \( ABCDE \)
9. \( AB'CDE \)

#### Step 2: Apply simplification rules
1. \( A'B'C'D'E' + A'B'C'D'E \):
   \[
   A'B'C'D'(E' + E) = A'B'C'D'
   \]

2. \( A'B'CD'E' \) remains as is.

3. \( A'BCD'E' + A'BCD'E \):
   \[
   A'BCD'(E' + E) = A'BCD'
   \]

4. \( A'BC'D'E' \) remains as is.

5. \( A'BCDE \) remains as is.

6. \( AB'C'D'E \) remains as is.

7. \( ABCD'E \) remains as is.

8. \( ABCDE \) remains as is.

9. \( AB'CDE \) remains as is.

#### Step 3: Combine simplified terms
The simplified Boolean expression is:

\[
X = A'B'C'D' + A'B'CD'E' + A'BCD' + A'BC'D'E' + A'BCDE + AB'C'D'E + ABCD'E + ABCDE + AB'CDE
\]

#### Step 4: Combine further if possible
Inspect for more common terms:

- The terms \( A'B'C'D' \), \( A'B'CD'E' \), \( A'BCD' \), \( A'BC'D'E' \), and \( A'BCDE \) cannot combine further as they differ in too many variables.
- Similarly, \( AB'C'D'E \), \( ABCD'E \), \( ABCDE \), and \( AB'CDE \) remain distinct.

#### Final Simplified Expression:
\[
X = A'B'C'D' + A'B'CD'E' + A'BCD' + A'BC'D'E' + A'BCDE + AB'C'D'E + ABCD'E + ABCDE + AB'CDE
\]

This is the most simplified version of the given Boolean expression.

Would you like me to explain any part of the process or check for further optimization?

P5: Simplify the following Boolean expression. X=A’B’C’D’E’+A’B’CD’E’+A’BCD’E’+A’BC’D’E’+A’B’C’D’E+A’BCD’E+A’BCDE+AB’C’D’E+AB’C’D’E+ABCD’E+ABCDE+AB’CDE

Learn how to simplify the Boolean expression X = A’B’C’D’E’ + A’B’CD’E’ + A’BCD’E’ + A’BC’D’E’ + A’B’C’D’E + A’BCD’E + A’BCDE + AB’C’D’E + AB’C’D’E + ABCD’E + ABCDE + AB’CDE. The solution involves Boolean algebra concepts like the Consensus Theorem, Distributive Law, and Complement Law, ideal for undergraduate engineering or computer science students.

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