Let's go through each of the statements and prove them rigorously. We'll use the basic definitions of set operations: complement, intersection, and difference, as well as some basic properties of set theory.

(i) $A - A = \emptyset$

Proof:

The set difference $A - A$ is defined as: $A - A = \{ x \in A \mid x \notin A \}$ However, if $x$ is in $A$, then $x$ cannot be in $A - A$ because $x$ would have to simultaneously not be in $A$. Therefore: $A - A = \emptyset$

(ii) $A - B = A \cap B'$

Proof:

The set difference $A - B$ is defined as: $A - B = \{ x \in A \mid x \notin B \}$ This is exactly the definition of the intersection of $A$ with the complement of $B$ (denoted $B'$): $A - B = A \cap B'$

(iii) $A' = U - A$

Proof:

By definition, the complement of $A$ in a universal set $U$ is: $A' = \{ x \in U \mid x \notin A \}$ This is the same as the set difference between the universal set $U$ and $A$: $A' = U - A$

(iv) $A - \emptyset = A$

Proof:

The set difference $A - \emptyset$ is defined as: $A - \emptyset = \{ x \in A \mid x \notin \emptyset \}$ Since the empty set has no elements, every element in $A$ satisfies $x \notin \emptyset$. Thus: $A - \emptyset = A$

(v) $A - B = B - A$ if and only if $A = B$

Proof:

(⇒) Assume $A - B = B - A$. Consider any element $x$ in $A$. If $x$ is in $A$, then $x$ cannot be in $B$ (otherwise $x$ would not be in $A - B$). Similarly, if $x$ is in $B$, then $x$ cannot be in $A$. Therefore, $A$ and $B$ must contain exactly the same elements, meaning $A = B$.

(⇐) If $A = B$, then $A - B = \emptyset$ and $B - A = \emptyset$, so $A - B = B - A$.

(vi) $A - B = \emptyset$ if and only if $A \subseteq B$

Proof:

(⇒) Assume $A - B = \emptyset$. This means there is no element in $A$ that is not in $B$, so every element of $A$ must be in $B$. Therefore, $A \subseteq B$.

(⇐) If $A \subseteq B$, then there is no element in $A$ that is not in $B$, so $A - B = \emptyset$.

(vii) $A - B = A$ if and only if $A \cap B = \emptyset$

Proof:

(⇒) Assume $A - B = A$. This means that no element of $A$ is in $B$. Therefore, the intersection $A \cap B = \emptyset$.

(⇐) If $A \cap B = \emptyset$, then no element in $A$ is in $B$, so the set difference $A - B$ contains all elements of $A$, which means $A - B = A$.

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Do you have any questions or need further details on these proofs?

Here are some related questions to explore:

1. How does the symmetric difference relate to set differences?
2. Can the results above be extended to more than two sets?
3. How do these properties change in the context of infinite sets?
4. What happens if we replace set difference with intersection or union in these statements?
5. How do these proofs relate to Venn diagrams?

Tip: Venn diagrams can be a useful tool to visually confirm set identities and relationships.