We are tasked with proving the following statement:

Statement: If $A \subseteq B$, then $A \cap B^c = \emptyset$.

Here, $A \subseteq B$ means that every element of set $A$ is also in set $B$, and $B^c$ represents the complement of set $B$, i.e., the set of elements that are not in $B$.

Proof:

1. Assumption:
   $A \subseteq B$.
   This means that if $x \in A$, then $x \in B$. In other words, every element of $A$ is contained in $B$.

2. Complement of $B$:
   The complement of set $B$, denoted by $B^c$, is defined as:
   $B^c = \{x \mid x \notin B\}$ So, any element of $B^c$ is not in $B$.

3. Intersection of $A$ and $B^c$:
   We are tasked with proving that the intersection $A \cap B^c = \emptyset$.
   That is, we need to show that there is no element $x$ that belongs to both $A$ and $B^c$.

4. Suppose there is an element in the intersection:
   Let $x \in A \cap B^c$.
   By the definition of intersection, $x \in A$ and $x \in B^c$.

   • Since $x \in A$ and $A \subseteq B$, this implies that $x \in B$.
   • But $x \in B^c$ means $x \notin B$.

   This is a contradiction because $x$ cannot be both in $B$ and not in $B$ at the same time.

5. Conclusion:
   Therefore, our assumption that such an $x$ exists is false.
   Hence, $A \cap B^c = \emptyset$, which means that there are no elements that are in both $A$ and the complement of $B$.

Thus, we have proven that if $A \subseteq B$, then $A \cap B^c = \emptyset$.

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Would you like any further clarification or details on this proof? Here are some related questions you might want to consider:

1. What is the complement of a set in a universal set?
2. Can a set have a non-empty intersection with its own complement?
3. How would the proof change if $A$ is not a subset of $B$?
4. What is the difference between a strict subset and a subset?
5. How does set theory relate to logic in terms of proving properties?

Tip: Remember that a set and its complement are always disjoint!