We are tasked with simplifying the following expression using set identities:

$[B \cup A^c \ (B \cap A^c)]^c$

Step-by-Step Simplification:

1. Break down the initial expression: The expression inside the brackets can be written as: $B \cup (A^c \setminus (B \cap A^c))$ Here, $A^c \setminus (B \cap A^c)$ represents the difference of sets, i.e., elements in $A^c$ but not in $B \cap A^c$.

2. Simplify the difference of sets: Using the identity $A \setminus B = A \cap B^c$, we get: $A^c \setminus (B \cap A^c) = A^c \cap (B \cap A^c)^c$ Simplify $(B \cap A^c)^c$ using De Morgan's Law: $(B \cap A^c)^c = B^c \cup A$ Therefore: $A^c \setminus (B \cap A^c) = A^c \cap (B^c \cup A)$ Now, distribute the intersection: $A^c \cap (B^c \cup A) = (A^c \cap B^c) \cup (A^c \cap A)$ Since $A^c \cap A = \emptyset$, we are left with: $A^c \setminus (B \cap A^c) = A^c \cap B^c$

3. Substitute back into the main expression: Now we substitute $A^c \cap B^c$ back into the original expression: $B \cup (A^c \cap B^c)$ Using the distributive property: $B \cup (A^c \cap B^c) = (B \cup A^c) \cap (B \cup B^c)$ Since $B \cup B^c = U$ (where $U$ is the universal set), we get: $B \cup (A^c \cap B^c) = B \cup A^c$

4. Apply the complement: Now apply the complement to the simplified expression: $(B \cup A^c)^c$ Using De Morgan's Law again: $(B \cup A^c)^c = B^c \cap A$

Final Answer:

The simplified expression is:

$B^c \cap A$

Would you like further details on any step, or do you have any questions?

Here are 5 related questions to deepen understanding:

1. How does De Morgan's Law help in simplifying set expressions?
2. What is the significance of the complement in set theory?
3. Can you prove the identity $A \setminus B = A \cap B^c$ with an example?
4. How does the distributive property apply in set operations?
5. What is the result of $(A^c)^c$, and how does it simplify?

Tip: When simplifying set expressions, always try applying De Morgan's Laws and distributive properties early—they often lead to major simplifications!