To simplify the expression \( Y = ((A + B)'[C(A + B)']')' \), we will follow the rules of Boolean algebra step by step:

### Step 1: Distribute the negation (De Morgan's Law)
The negation of a sum \( (A + B)' \) is:
\[
(A + B)' = A'B'
\]
So, substitute this into the original expression:
\[
Y = ((A'B')[C(A'B')']')'
\]

### Step 2: Simplify \( (A'B')' \) (De Morgan's Law again)
The negation of a product \( (A'B')' \) is:
\[
(A'B')' = A + B
\]
Now, the expression becomes:
\[
Y = ((A'B')[C(A + B)'])'
\]

### Step 3: Simplify the inner negation
The negation of a sum \( (A + B)' \) is:
\[
(A + B)' = A'B'
\]
So the expression becomes:
\[
Y = ((A'B')[C A'B'])'
\]

### Step 4: Apply distribution of terms inside
We can simplify the product \( (A'B')[C A'B'] \):
\[
(A'B')[C A'B'] = A'B'C A'B' = A'B'C
\]
This leaves us with:
\[
Y = (A'B'C)'
\]

### Step 5: Apply De Morgan's Law once again
The negation of a product \( A'B'C \) is:
\[
(A'B'C)' = A + B + C'
\]

### Final Simplified Expression:
\[
Y = A + B + C'
\]

Let me know if you want any further details, or if you have more questions!

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**Related Questions:**
1. How does De Morgan’s Law work in Boolean algebra?
2. Can this Boolean expression be simplified further if constraints are added?
3. What are other simplification techniques in Boolean algebra?
4. How do Karnaugh maps help in simplifying Boolean expressions?
5. What is the difference between SOP (Sum of Products) and POS (Product of Sums) forms?

**Tip:** De Morgan's Laws are essential for simplifying complex Boolean expressions by transforming sums into products and vice versa!

Learn how to simplify the Boolean expression Y = ((A + B)'[C(A + B)']')' using De Morgan's Laws and other Boolean algebra simplification techniques. This step-by-step guide is perfect for students in Grades 10-12 who want to master Boolean algebra.

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