Simplification of 𝑌

To simplify the Boolean expression:
$Y = A + B + A'B'C + A'B'C'$

We will simplify step-by-step:

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1. Combine the terms $A'B'C$ and $A'B'C'$:

Observe that $A'B'C + A'B'C'$ shares the common factor $A'B'$.
So, factor $A'B'$:
$A'B'C + A'B'C' = A'B'(C + C')$
From the Complement Law: $C + C' = 1$.
Thus:
$A'B'(C + C') = A'B' \cdot 1 = A'B'$

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2. Update the expression:

Substitute $A'B'C + A'B'C'$ with $A'B'$ into the original expression:
$Y = A + B + A'B'$

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3. Simplify further:

Now, examine the expression $A + B + A'B'$:

• $A$ and $B$ already cover many possibilities.
• $A'B'$ represents the case when both $A$ and $B$ are 0.

Thus, the expression $A + B + A'B'$ cannot be simplified further because $A'B'$ ensures that the case when $A = 0$ and $B = 0$ is included.

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Final Simplified Expression:

The simplified expression is:
$Y = A + B + A'B'$

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Verification:

The truth table confirms this result:

A	B	C	A'B'C	A'B'C'	A'B'	A + B + A'B'
0	0	0	0	1	1	1
0	0	1	1	0	1	1
0	1	X	0	0	0	1
1	X	X	0	0	0	1

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Do you want more explanation or a truth table construction?

Here are 5 related questions:

1. How do you simplify a Boolean expression using Karnaugh maps?
2. What is the difference between SOP (Sum of Products) and POS (Product of Sums)?
3. How do you implement this simplified expression in digital circuits?
4. What are the Boolean laws used in simplification?
5. How can we derive the minimal expression using Quine-McCluskey?

Tip: Factor common terms in Boolean expressions to reduce complexity efficiently.