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 AR_1B if (A∩ B^ )∪ (B∩ A^ )=φ and AR_2B if A∪ B^ =B∪ A^…

Given,

$$S={1,2,3,\ldots \ldots 10}$$S=1,2,3,……10

$P(S)=$P(S)= power set of $S$S

$$A{R}_{1}B\Rightarrow (A\cap {B}^{c})\cup ({A}^{c}\cap B)=\phi$$AR1​B⇒(A∩Bc)∪(Ac∩B)=ϕ

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

$(A\cap {A}^{c})\cup ({A}^{c}\cap A)$(A∩Ac)∪(Ac∩A) which $\phi$ϕ always,

Now checking symmetric we will interchange A&B,

So, $(B\cap {A}^{c})\cup ({B}^{c}\cap A)$(B∩Ac)∪(Bc∩A) which is same as $(A\cap {B}^{c})\cup ({A}^{c}\cap B)$(A∩Bc)∪(Ac∩B), hence the relation is symmetric,

So, ${R}_{1}$R1​ is reflexive, symmetric

Now checking for transitive

$$(A\cap {B}^{c})\cup ({A}^{c}\cap B)=\phi ;$$(A∩Bc)∪(Ac∩B)=ϕ;

Now from diagram the elements in $(A\cap {B}^{c})\cup ({A}^{c}\cap B)$(A∩Bc)∪(Ac∩B) will be,

${a}\cup {b}$a∪b which is given as empty set $\phi$ϕ

Hence, we can say that, ${a}=\phi ={b}\Rightarrow A=B$a=ϕ=b⇒A=B

Now taking, $(B\cap {C}^{c})\cup ({B}^{c}\cap C)=\phi \therefore B=C$(B∩Cc)∪(Bc∩C)=ϕ∴B=C

$\therefore A=C$∴A=C equivalence.

Now solving,

$${R}_{2}\equiv A\cup {B}^{c}={A}^{c}\cup B$$R2​≡A∪Bc=Ac∪B

Now for reflexive replacing $B\rightarrow A$B→A we get,

$A\cup {A}^{c}={A}^{c}\cup A$A∪Ac=Ac∪A which is true,

And for symmetric interchanging $A\Leftrightarrow B$A⇔B we get,

$B\cup {A}^{c}={B}^{c}\cup A$B∪Ac=Bc∪A which is again true,

Hence, ${R}_{2}\rightarrow$R2​→ Reflexive, symmetric

Now for transitive,

From diagram the elements in$A\cup {B}^{c}={A}^{c}\cup B\Rightarrow {a,c,d}={b,c,d}$A∪Bc=Ac∪B⇒a,c,d=b,c,d

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

And, $B\cup {C}^{c}={B}^{c}\cup C\Rightarrow B=C$B∪Cc=Bc∪C⇒B=C

$$\therefore A=C\therefore A\cup {C}^{c}={A}^{c}\cup C$$∴A=C∴A∪Cc=Ac∪C

$\therefore$∴ Equivalence

Hence, both given relation are equivalence.