Animated Solution for Mathematics - Sets and Relations: Let $P(S)$ denote the power set of $S=\{1,2,3,\dots ,10\}$. Define the relations $R_1$ and $R_2$ on $P(S)$ as $A{R}_{1}B$ if $(A\cap B^c)\cup (B\cap A^c)=\varnothing$ and $A{R}_{2}B$ if $A\cup B^c=B\cup A^c,\forall A,B\in P(S)$. Then:

We are exploring the power set $P(S)$ of a set $S=\{1,2,3,\dots ,10\}$. We are tasked with evaluating two relations, $R_1$ and $R_2$, to determine if they qualify as equivalence relations.

Let us begin with $R_1$, defined by the condition $A{R}_{1}B$ if $(A\cap B^c)\cup (B\cap A^c)=\varnothing$.

Visualize this expression: $A\cap B^c$ represents elements in $A$ but not in $B$, while $B\cap A^c$ represents elements in $B$ but not in $A$. Their union is the symmetric difference, denoted as $A\Delta B$, which contains all elements belonging to exactly one of the two sets.

If this union is the empty set, it implies there are no elements that belong to $A$ but not $B$, and no elements that belong to $B$ but not $A$. Consequently, we arrive at the conclusion that $A=B$.

Since $R_1$ is equivalent to the equality relation, it is reflexive ($A=A$), symmetric ($A=B⟹B=A$), and transitive ($A=B$ and $B=C⟹A=C$). Thus, $R_1$ is an equivalence relation.

A Deeper Dive

Next, we examine $R_2$, defined by $A\cup B^c=B\cup A^c$. We apply the 'element-chasing' technique to simplify this condition.

Suppose an arbitrary element $x$ belongs to $A$. By definition, $x\in A\cup B^c$. Given the relation $A\cup B^c=B\cup A^c$, it follows that $x\in B\cup A^c$.

Since $x\in A$, it cannot be in $A^c$. Therefore, the only remaining possibility is that $x\in B$. This proves that $A\subseteq B$.

Conversely, if we take an element $y\in B$, then $y\in B\cup A^c$. By the equality of the sets, $y$ must also be in $A\cup B^c$. Since $y$ is in $B$, it cannot be in $B^c$, which forces $y\in A$. This proves that $B\subseteq A$.

When we establish both $A\subseteq B$ and $B\subseteq A$, the fundamental laws of set theory dictate that $A=B$.

Just like $R_1$, the relation $R_2$ collapses into the identity relation. Because both $R_1$ and $R_2$ are equivalent to $A=B$, they both satisfy the criteria for equivalence relations.

Conclusion: Both $R_1$ and $R_2$ are equivalence relations.