Let's go through each of these proofs step by step:

1. $A - A = \emptyset$

• Proof: The difference $A - A$ consists of all elements that are in $A$ but not in $A$. Since there are no such elements (every element in $A$ is also in $A$), $A - A = \emptyset$.

2. $A - B = A \cap B'$

• Proof: The difference $A - B$ consists of all elements that are in $A$ but not in $B$. The complement $B'$ contains all elements not in $B$. Thus, $A - B$ is exactly the set of elements that are in $A$ and not in $B$, which is the definition of $A \cap B'$. Therefore, $A - B = A \cap B'$.

3. $A = U - A'$

• Proof: $A'$ represents the complement of $A$ in the universal set $U$, which means all elements not in $A$. The difference $U - A'$ consists of all elements in $U$ that are not in $A'$, which is exactly the set $A$. Thus, $A = U - A'$.

4. $A - \emptyset = A$

• Proof: The difference $A - \emptyset$ consists of all elements in $A$ but not in $\emptyset$. Since $\emptyset$ has no elements, the set $A$ remains unchanged. Therefore, $A - \emptyset = A$.

5. $A - B = B - A$ if and only if $A = B$

• Proof:
  • If $A = B$: If $A$ and $B$ are identical sets, then $A - B$ and $B - A$ would both result in $\emptyset$. Thus, $A - B = B - A = \emptyset$.
  • If $A - B = B - A$: Assume that $A - B = B - A$. Consider any element $x$. If $x \in A$ but $x \notin B$, then $x$ should be in $A - B$, but not in $B - A$, which leads to a contradiction. Similarly, if $x \in B$ but $x \notin A$, then $x$ should be in $B - A$, but not in $A - B$, which also leads to a contradiction. Thus, no such $x$ exists, implying $A = B$.

6. $A - B = \emptyset$ if and only if $A \subseteq B$

• Proof:
  • If $A \subseteq B$: If every element of $A$ is also in $B$, then there are no elements left in $A - B$. Therefore, $A - B = \emptyset$.
  • If $A - B = \emptyset$: If $A - B = \emptyset$, then there are no elements in $A$ that are not in $B$. This implies that every element of $A$ must be in $B$, meaning $A \subseteq B$.

7. $A - B = A$ if and only if $A \cap B = \emptyset$

• Proof:
  • If $A \cap B = \emptyset$: If $A$ and $B$ have no elements in common, then $A - B$ consists of all elements in $A$, since none of the elements are in $B$. Hence, $A - B = A$.
  • If $A - B = A$: If $A - B = A$, this means no elements in $A$ are in $B$, otherwise they would be removed by the difference operation. Therefore, $A \cap B = \emptyset$.

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Would you like any more details or explanations on these proofs?

Here are 5 related questions to consider:

1. How do we prove that $A \cap A = A$?
2. Can you prove that $A \cup A = A$?
3. How do we show that $(A - B) \cup B = A \cup B$?
4. How does the distributive property work with sets, such as $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$?
5. Can you prove that $A - (B \cap C) = (A - B) \cup (A - C)$?

Tip: When working with set operations, drawing Venn diagrams can help visualize and understand the relationships between different sets.