Prove that the intersection of two equivalence relations is an equivalence relation. I am reading this chapter of the Book of Proof, and I'm stuck at the Exercise 10 of section 11.2. It is as follows.

Suppose $R$ and $S$ are two equivalence relations on a set $A$. Prove
that $R \cap S$ is also an equivalence relation.

Thanks for helps!
 A: Hint: Use the fact that $R$ and $S$ are EQUIVALENCE relations on THE SAME set, and hence both must be reflexive, symmetric, and transitive on that set.
Then use the definition of set intersection: $R\cap S$ is the set of all pairs of elements in the set such that $(x, y) \in R$ AND $(x, y) \in S$ or, put differently, $(x, y) \in R\cap S \iff (x, y)\in R$ and $(x, y) \in S$. 
Try to figure out what elements must necessarily be in $R\cap S$ and check to see that they must then be in both $R$ and $S$.

Another approach would be to use an indirect proof with the hints above:
"Given $R$ and $S$ are equivalence relations on a set $A$, suppose for the sake of contradiction, that $R\cap S$ is NOT an equivalence relation...". If not an equivalence relation, then $R\cap S$ fails to be reflexive and/or fails to be symmetric, and/or fails to be transitive. If you can work towards a contradiction (that this assumption must contradict the fact that both $R$ and $S$ are equivalence relations), then you are done.
A: For the solution of this exercise, you have to show that $R \cap S$ keeps the three properties of equivalence relations (reflexive, symmetric and transitive).
This means that for each $x\in R\cap S$ you have to show that $\langle x,x\rangle \in R \cap S$ and for each pair $\langle x,y\rangle \in R \cap S$, you have to show that $\langle y,x\rangle \in R \cap S$ and for each pairs $\langle x,y\rangle, \langle y,z\rangle \in R \cap S$ you have to show that $\langle x,z\rangle \in R \cap S$
A: Proof. Suppose $R$ and $S$ are both equivalence relations on a set $A$. In what follows we will show that $R$ and $S$ both being equivalence relations on the set $A$ implies that $R\cap S$ is also an equivalence relation.
It is immediately apparent that since $R$ and $S$ are equivalence relations then
$$\begin{align*} x\in A \Rightarrow (xRx)\land(xSx) \Rightarrow (\langle x,x\rangle\in R) \land (\langle x,x\rangle\in S)\Rightarrow\langle x, x\rangle\in R\cap S.\end{align*}$$
Thus $x\in A\Rightarrow x(R\cap S)x$ therefore $R\cap S$ is reflexive.
Now suppose $\langle x,y \rangle\in R\cap S$ then $(\langle x,y \rangle\in R) \land (\langle x,y \rangle\in S)$ by definition of the intersection of two sets. But both $R$ and $S$ are symmetric so it must be that  $(\langle y,x \rangle\in R) \land (\langle y,x \rangle\in S)$ which implies $\langle y,x \rangle\in R\cap S$. Thus we have shown $x(R\cap S)y\Rightarrow y(R\cap S)x$ therefore $R\cap S$ is symmetric.
Finally, consider $(\langle x,y \rangle\in R\cap S)\land(\langle y,z \rangle\in R\cap S)$. Then since $(\langle x,y \rangle\in R)\land\langle (y,z \rangle\in R)\Rightarrow\langle x,z \rangle\in R$, since $R$ is transitive, and since $(\langle x,y \rangle\in S)\land\langle (y,z \rangle\in S)\Rightarrow\langle x,z \rangle\in S$, since $S$ is transitive, therefore $\langle x,z\rangle\in R$ and $\langle x,z\rangle\in S$ so by definition of the intersection of two sets $\langle x,z\rangle\in R\cap S$. Thus we have shown $(\langle x,y \rangle\in R\cap S)\land(\langle y,z \rangle\in R\cap S)\Rightarrow\langle x,z\rangle\in R\cap S$.
It follows that $R\cap S$ is an equivalence relation since it is reflexive, symmetric and transitive. $\Box$
A: Hint (did this in school): 
Let's have 2 relations 
$R- antisymmetric$
$S-antisymmetric$
I had to prove that $R \cap S$ is also antisymmetric.
$P=R \cap S$
$(x,y) \in P$
$\implies (x,y) \in R \cap S $
$\implies (x,y) \in R \wedge (x,y) \in S $
$\implies ((y,x) \notin R \wedge (y,x) \notin S) \Rightarrow(x \ne y) $
$\implies (y,x) \notin R \cap S $
$\implies (y,x) \notin P $
