Let S = \ 1, 2, 3, 4, 5\ and P(S) be the power set of S . Let f(A)...

For any two subsets $A$ and $B$, the sum of elements satisfies the inclusion-exclusion principle:
$f(A \cup B) = f(A) + f(B) - f(A \cap B)$

The given relation $A \, R \, B$ requires $f(A \cup B) = f(A) + f(B)$.
Substituting this into the principle yields $f(A \cap B) = 0$.
Since all elements of $S$ are strictly positive integers, the sum of elements in $A \cap B$ can be zero if and only if $A \cap B = \phi$.
Thus, $A \, R \, B$ is equivalent to saying $A$ and $B$ are disjoint sets.

For reflexivity:
$A \cap A = A$. For $A \, R \, A$ to hold, we need $A = \phi$.
Since this is not true for any non-empty subset (e.g., $A = \{1\}$), $R$ is not reflexive.

For symmetry:
If $A \, R \, B$, then $A \cap B = \phi$.
This implies $B \cap A = \phi$, so $B \, R \, A$.
Thus, $R$ is symmetric.

For transitivity:
Let $A = \{1\}$, $B = \{2\}$, and $C = \{1\}$.
We have $A \cap B = \phi \Rightarrow A \, R \, B$.
And $B \cap C = \phi \Rightarrow B \, R \, C$.
But $A \cap C = \{1\} \neq \phi$, so $A$ is not related to $C$.
Thus, $R$ is not transitive.

Therefore, the relation $R$ is symmetric only.

Answer: symmetric only