Let P(S) denote the power set of S = {1, 2, 3 | vedclass

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(A) For relation $R_1$: The condition $(A \cap B^c) \cup (B \cap A^c) = \varnothing$ is the definition of the symmetric difference $A \Delta B = \varn

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both $R_1$ and $R_2$ are equivalence relations

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(A) For relation $R_1$: The condition $(A \cap B^c) \cup (B \cap A^c) = \varnothing$ is the definition of the symmetric difference $A \Delta B = \varnothing$,which implies $A = B$. Since $A = B$ is an equivalence relation (reflexive,symmetric,and transitive),$R_1$ is an equivalence relation.For relation $R_2$: The condition $A \cup B^c = B \cup A^c$ can be analyzed using set properties. $A \cup B^c = B \cup A^c \iff (A \cup B^c) \cap (A \cap B) = (B \cup A^c) \cap (A \cap B) \iff A = B$. Alternatively,using the Venn diagram regions where $a, b, c, d$ represent disjoint regions: $A = a \cup c$ and $B = b \cup c$. $A \cup B^c = (a \cup c) \cup (a \cup d) = a \cup c \cup d$. $B \cup A^c = (b \cup c) \cup (b \cup d) = b \cup c \cup d$. Equating these gives $a \cup c \cup d = b \cup c \cup d$,which implies $a = b$. Since $a$ and $b$ are the regions unique to $A$ and $B$ respectively,$a = b = \varnothing$ implies $A = B$. Thus,$R_2$ is also an equivalence relation.Therefore,both $R_1$ and $R_2$ are equivalence relations.

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 $A R_1 B$ 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:

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