Simplify the Boolean function `(x* y) + [(x+ y') * y]'`.

Options

A. 0
B. 1
C. x + y
D. xy

Explanation

To simplify the Boolean function \( (xy) + [(x + y') \cdot y]' \), we will follow these steps:

### Step 1: Rewrite the expression
The original expression is:
\[ xy + [(x + y') \cdot y]' \]

### Step 2: Apply De Morgan's Theorem
According to De Morgan's Theorem, the negation of a conjunction is the disjunction of the negations:
\[ [(x + y') \cdot y]' = (x + y')' + y' \]

### Step 3: Simplify \( (x + y')' \)
Using De Morgan's Theorem again, we can simplify \( (x + y')' \):
\[ (x + y')' = x' \cdot (y')' = x' \cdot y \]
Since \( (y')' = y \).

### Step 4: Substitute back into the expression
Now substitute back into the expression:
\[ xy + (x' \cdot y + y') \]

### Step 5: Distribute and combine terms
Now we can combine the terms:
\[ xy + x'y + y' \]

### Step 6: Factor out common terms
Notice that \( xy + x'y \) can be factored:
\[ y(x + x') + y' = y(1) + y' = y + y' \]

### Step 7: Apply the Complement Law
According to the Complement Law, \( y + y' = 1 \):
\[ 1 \]

### Final Result
Thus, the simplified form of the Boolean function is:
\[ 1 \]