$$ \overline{(A\prime B\prime C\prime D+A\prime B\prime CD\prime+A\prime B\prime C\prime D\prime+A\prime B\prime CD+A\prime BC\prime D\prime+AB\prime C\prime D\prime+AB\prime C\prime D+AB\prime CD\prime+AB\prime CD+ABC\prime D\prime)} $$

The given expression is a Boolean algebra expression. To simplify it, we'll follow the steps of applying Boolean algebra rules.

### Expression:
\[
\overline{(A'B'C'D + A'B'CD' + A'B'C'D' + A'B'CD + A'BC'D' + AB'C'D' + AB'C'D + AB'CD' + AB'CD + ABC'D')}
\]

### Step 1: Simplify the expression inside the complement.

Notice that we can group terms where only one variable changes:

1. $ A'B'C'D + A'B'C'D' = A'B'C'(D + D') = A'B'C' $  
2. $ A'B'CD' + A'B'CD = A'B'C(D' + D) = A'B'C $  
3. $ AB'C'D' + AB'C'D = AB'C'(D' + D) = AB'C' $  
4. $ AB'CD' + AB'CD = AB'C(D' + D) = AB'C $

Now the expression becomes:
\[
\overline{(A'B'C' + A'B'C + A'BC'D' + AB'C' + AB'C)}
\]

### Step 2: Combine like terms

- $ A'B'C' + A'B'C = A'B' $  (since $ C' + C = 1 $)
- $ AB'C' + AB'C = AB' $ (same reasoning as above)

So, the expression reduces to:
\[
\overline{(A'B' + A'BC'D' + AB')}
\]

### Step 3: Further simplification

Combine $ A'B' $ and $ AB' $:
- $ A'B' + AB' = B' $  (factoring $ B' $)

So, the expression simplifies to:
\[
\overline{(B' + A'BC'D')}
\]

Since $ A'BC'D' $ can't simplify further, the final expression inside the complement is:
\[
\overline{(B' + A'BC'D')}
\]

### Step 4: Apply De Morgan's law

Now apply De Morgan's law:
\[
\overline{(B' + A'BC'D')} = B \cdot \overline{A'BC'D'}
\]

### Final simplified expression:
The expression remains:
\[
B \cdot \overline{A'BC'D'}
\]

Since $\overline{A'BC'D'}$ is not further simplifiable without additional information about the variables, this is the simplified expression.

Would you like more details or have any questions?

### Related Questions:
1. How does De Morgan's law apply to Boolean algebra?
2. What are the basic Boolean algebra rules used in simplifications?
3. How can truth tables help in verifying Boolean expressions?
4. What are Karnaugh maps, and how do they assist in simplifications?
5. How does the distributive property work in Boolean algebra?

### Tip:
When simplifying Boolean expressions, always look for opportunities to factor common terms and apply Boolean identities.

Learn how to simplify a complex Boolean algebra expression using De Morgan's Law. Follow a detailed step-by-step guide with examples and explanations. Suitable for undergraduate students studying Boolean algebra and digital logic.

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