Composition of two symmetric relations Let $R1$, $R2$ be symmetric relations on a set. I want to prove that $R1\circ R2$ is symmetric if and only if $R1\circ R2=R2\circ R1$.
I have tried a few problems of this type by doing something like this:
$(a,b) \in R2 \implies (b,a) \in R2$.
 Also $(b,c) \in R1 \implies (c,b) \in R1$.
Let $R1 \circ R2(a)=R1(b)=c$. So $(a,c) \in R1\circ R2$ and $(c,a) \in R1\circ R2.$
But in this problem, I am unable to proceed.
Please help. Thanks in advance.
 A: Just write out the definitions: 
$R_1\circ R_2=R_2\circ R_1$ means that for every $x,y_1,z$ such that $(x,y_1)\in R_1$ and $(y_1,z)\in R_2$, there exists $y_2$ such that $(z,y_2)\in R_1$ and $(y_2,x)\in R_2$.  
$R_1\circ R_2$ is symmetric means that if $(x,z)\in R_1\circ R_2$ then $(z,x)\in R_1\circ R_2$.  This in turn means that for every $x,y_1,z$ such that $(x,y_1)\in R_1$ and $(y_1,z)\in R_2$, there exists $y_2$ such that $(z,y_2)\in R_1$ and $(y_2,x)\in R_2$.  
Can you see why these two properties are equivalent?  
[Note: this shows that $R_2\circ R_1$ is the converse or inverse relation to $R_1\circ R_2$; i.e., $(x,z)\in R_1\circ R_2$ if and only if $(z,x)\in R_2\circ R_1$.  If you can show this (which is fairly easy from the definition of relation composition) then you can solve the problem very quickly and neatly, since a relation is symmetric if and only if it equals its converse.]  
A: A possible follow up to the existing answer, but little more in detail:
$$\begin{aligned}\boxed{\Leftarrow:}\quad R_1\circ R_2&=R_2\circ R_1\\ (x,y)&\in R_1\circ R_2\\&\implies \exists z,(x,z)\in R_2\land (z,y)\in R_1\\&\overset{R_1,R_2\text{symmetric }}{\implies} (z,x)\in R_2
\land (y,z)\in R_1\\&\implies (y,x)\in R_2\ \circ R_1=R_1\circ R_2\\&\implies R_1\circ R_2\text{ is symmetric }\end{aligned}$$
$$\begin{aligned}\boxed{\Rightarrow: }&\color{white}\implies R_1\circ R_2\text{ is symmetric }\\
&\implies R_1\circ R_2=(R_1\circ R_2)^{-1}=R_2^{-1}\circ R_1^{-1}\\&\overset{R_1,R_2\text{ symmetric}}{=}R_2\circ 
R_1\end{aligned}$$
