The set `(A cap B')'cup (B cap C)` equals a).`A' cup B cup C ` b).`A' cup B` c).`A'cup C` d).`A' cap B`

Options

A. `A' cup B cup C `
B. `A' cup B `
C. `A'cup C`
D. `A' cap B`

Explanation

To solve the problem, we need to simplify the expression \((A \cap B')' \cup (B \cap C)\) step by step.

### Step 1: Apply De Morgan's Law
We start with the expression:
\[ (A \cap B')' \]
According to De Morgan's Law, the complement of an intersection is the union of the complements. Therefore, we have:
\[ (A \cap B')' = A' \cup B \]

### Step 2: Rewrite the Expression
Now we can rewrite the original expression using the result from Step 1:
\[ (A \cap B')' \cup (B \cap C) = (A' \cup B) \cup (B \cap C) \]

### Step 3: Apply the Distributive Law
Next, we can apply the distributive property of set operations:
\[ (A' \cup B) \cup (B \cap C) = A' \cup (B \cup (B \cap C)) \]
Using the property that \(B \cup (B \cap C) = B\), we simplify this to:
\[ A' \cup B \]

### Step 4: Final Expression
Thus, the expression simplifies to:
\[ A' \cup B \]

### Conclusion
The final expression we derived is \(A' \cup B\), which corresponds to option (b).

### Answer
The answer is:
**b). \(A' \cup B\)**