Let $$P(S)$$ denote the power set of $$S = \{1, 2, 3, \ldots, 10\}$$. Define the relations $$R_1$$ and $$R_2$$ on $$P(S)$$ as $$AR_1B$$ if $$(A \cap B^c) \cup (B \cap A^c) = \phi$$ and $$AR_2 B$$ if $$A \cup B^c = B \cup A^c, \forall A, B \in P(S)$$. Then :

The set $$S = \{1, 2, 3, \ldots, 10\}$$, and $$P(S)$$ is its power set. The relations $$R_1$$ and $$R_2$$ are defined on $$P(S)$$.

**Analysis of $$R_1$$:**

The relation $$A R_1 B$$ holds if $$(A \cap B^c) \cup (B \cap A^c) = \emptyset$$.

Note that $$(A \cap B^c)$$ is the set of elements in $$A$$ but not in $$B$$, and $$(B \cap A^c)$$ is the set of elements in $$B$$ but not in $$A$$. Their union is the symmetric difference $$A \Delta B$$.

Thus, $$A R_1 B$$ iff $$A \Delta B = \emptyset$$.

$$A \Delta B = \emptyset$$ implies that $$A - B = \emptyset$$ and $$B - A = \emptyset$$, so $$A \subseteq B$$ and $$B \subseteq A$$, hence $$A = B$$.

Therefore, $$R_1$$ is the equality relation.

**Check for equivalence relation:**

- **Reflexive:** For any $$A \in P(S)$$, $$A = A$$, so $$(A \cap A^c) \cup (A \cap A^c) = \emptyset \cup \emptyset = \emptyset$$, thus $$A R_1 A$$.

- **Symmetric:** If $$A R_1 B$$, then $$A = B$$, so $$B = A$$, thus $$B R_1 A$$.

- **Transitive:** If $$A R_1 B$$ and $$B R_1 C$$, then $$A = B$$ and $$B = C$$, so $$A = C$$, thus $$A R_1 C$$.

Hence, $$R_1$$ is an equivalence relation.

**Analysis of $$R_2$$:**

The relation $$A R_2 B$$ holds if $$A \cup B^c = B \cup A^c$$.

Consider an element $$x \in S$$:

- If $$x \in A$$ and $$x \in B$$, then $$x \in A \cup B^c$$ and $$x \in B \cup A^c$$.

- If $$x \notin A$$ and $$x \notin B$$, then $$x \in B^c$$ (so $$x \in A \cup B^c$$) and $$x \in A^c$$ (so $$x \in B \cup A^c$$).

- If $$x \in A$$ but $$x \notin B$$, then $$x \in A \cup B^c$$, but $$x \notin B$$ and $$x \notin A^c$$ (since $$x \in A$$), so $$x \notin B \cup A^c$$. For equality, this case must not occur.

- Similarly, if $$x \in B$$ but $$x \notin A$$, then $$x \in B \cup A^c$$, but $$x \notin A \cup B^c$$. For equality, this case must not occur.

Thus, for all $$x$$, $$x$$ must be in both $$A$$ and $$B$$ or in neither, implying $$A = B$$.

Therefore, $$R_2$$ is also the equality relation.

**Check for equivalence relation:**

- **Reflexive:** For any $$A \in P(S)$$, $$A \cup A^c = S$$ and $$A \cup A^c = S$$, so $$A \cup A^c = A \cup A^c$$, thus $$A R_2 A$$.

- **Symmetric:** If $$A R_2 B$$, then $$A = B$$, so $$B = A$$, thus $$B R_2 A$$.

- **Transitive:** If $$A R_2 B$$ and $$B R_2 C$$, then $$A = B$$ and $$B = C$$, so $$A = C$$, thus $$A R_2 C$$.

Hence, $$R_2$$ is an equivalence relation.

Both $$R_1$$ and $$R_2$$ are equivalence relations. The correct option is A.

\boxed{$$\text{A}$$}