The new image contains the same set theory problem as before:

$A - B = A \cap B^c = B^c - A^c$

Let’s go through the proof step by step once again.

1. Proving $A - B = A \cap B^c$

• Definition of $A - B$: The set difference $A - B$ is defined as the set of elements that are in $A$ but not in $B$. Symbolically: $A - B = \{x \mid x \in A \text{ and } x \notin B\}.$

• Definition of $A \cap B^c$: The complement of $B$, denoted $B^c$, consists of all elements not in $B$. The intersection $A \cap B^c$ represents the set of elements that are both in $A$ and not in $B$: $A \cap B^c = \{x \mid x \in A \text{ and } x \notin B\}.$

Both expressions define the same set of elements, meaning: $A - B = A \cap B^c.$

2. Proving $A \cap B^c = B^c - A^c$

• Definition of $B^c - A^c$: The set $B^c - A^c$ consists of elements in $B^c$ that are not in $A^c$. Since $A^c$ is the complement of $A$, this translates to elements that are in $B^c$ and in $A$ (because $x \notin A^c$ implies $x \in A$). Symbolically: $B^c - A^c = \{x \mid x \in B^c \text{ and } x \notin A^c\} = \{x \mid x \in B^c \text{ and } x \in A\}.$

This is precisely the definition of $A \cap B^c$. Hence: $A \cap B^c = B^c - A^c.$

Conclusion:

The full equality is proven as follows: $A - B = A \cap B^c = B^c - A^c.$

This completes the proof.

Would you like any further clarification on any step?

Here are five related questions to explore:

1. How does the proof change if we consider the union of sets instead of intersections and differences?
2. Can you prove the identity $(A \cup B)^c = A^c \cap B^c$?
3. What would happen if the sets $A$ and $B$ were disjoint? Does the equality still hold?
4. Can you represent the equality $A - B = A \cap B^c = B^c - A^c$ using a Venn diagram?
5. How do De Morgan’s laws assist in simplifying complex set expressions?

Tip: When working with set identities, try applying specific examples (like finite sets of numbers) to see the behavior and verify results step by step!