Let P S denote the power set of S = 1 , 2 , 3 , … , 10 . Define the relations R 1 and R 2 on P S as A…

Given,

$S=\left\{1,2,3,\dots \dots 10\right\}$

$P\left(S\right)=$ power set of $S$

$A{R}_{1}B\Rightarrow \left(A\cap B^c\right)\cup \left(A^c\cap B\right)=\varphi$

Now for reflexive property, replacing $B$ with $A$ we get,

$\left(A\cap A^c\right)\cup \left(A^c\cap A\right)$ which $\varphi$ always,

Now checking symmetric we will interchange $A \& B$,

So, $\left(B\cap A^c\right)\cup \left(B^c\cap A\right)$ which is same as $\left(A\cap B^c\right)\cup \left(A^c\cap B\right)$, hence the relation is symmetric,

So, $R_1$ is reflexive, symmetric

Now checking for transitive
$\left(A\cap B^c\right)\cup \left(A^c\cap B\right)=\varphi ;$

Now from diagram the elements in $\left(A\cap B^c\right)\cup \left(A^c\cap B\right)$ will be,

$\left\{a\right\}\cup \left\{b\right\}$ which is given as empty set $\varphi$

Hence, we can say that, $\left\{a\right\}=\varphi =\left\{b\right\}\Rightarrow A=B$

Now taking, $(B\cap C^c)\cup (B^c\cap C)=\varphi \therefore B=C$

$\therefore A=C$ equivalence.

Now solving,

$R_2\equiv A\cup B^c=A^c\cup B$

Now for reflexive replacing $B\to A$ we get,

$A\cup A^c=A^c\cup A$ which is true,

And for symmetric interchanging $A\iff B$ we get,

$B\cup A^c=B^c\cup A$ which is again true,

Hence, $R_2\to$ Reflexive, symmetric

Now for transitive,

From diagram the elements in$A\cup B^c=A^c\cup B\Rightarrow \left\{a,c,d\right\}=\left\{b,c,d\right\}$

On comparing both side, we get $\left\{a\right\}=\left\{b\right\} \therefore A=B$

And, $B\cup C^c=B^c\cup C\Rightarrow B=C$

$\therefore A=C \therefore A\cup C^c=A^c\cup C$

$\therefore$ Equivalence

Hence, both given relation are equivalence.