If $S$ and $R$ are both symmetric relations with the same domain, will $S\circ R$ also be symmetric?
I think it should be symmetric. If $S$ and $R$ are both symmetric, then there should be a $x,y$ $\in$ $S$, such that $xSy$ and $ySx$ are true. Also, there should be a $x,z$ $\in$ $R$, such that $xSz$ and $zSx$, combined them together, we get $y$ $S\circ R$ $z$ and $z$ $S\circ R$ $y$ are also true, so $S\circ R$ should be symmetric. I didn't know if this is the right proof or not. Please help...
This is not true in general :
$R$ and $S$ have to fullfill condition $S \circ R = R \circ S$.
See here.
Besides, a powerful approach is to use the representation of relations by their matrix, using for composition a particular multiplication rule denoted by $\odot$ (ordinary multiplication followed by a thresholding operation in order that entries $>1$ are set equal to $1$). This is well explained here and in a even more didactic way there at the bottom of the text. ).
As the product of 2 symmetric matrices $R$ and $S$ is symmetric iff $RS=SR$ (I intentionnaly take the same name for relations and matrices), we are done.
For example, one obtains a counterexample with
$$R=\begin{pmatrix}1&0&0\\ 0&0&1\\ 0&1&1\end{pmatrix}; S=\begin{pmatrix}0&0&1\\ 0&1&0\\ 1&0&0\end{pmatrix}$$
giving
$$RS=R \odot S=\begin{pmatrix}0&0&1\\ 1&0&0\\ 1&1&0\end{pmatrix} \ \text{and} \ SR=S \odot R=\begin{pmatrix}0&1&1\\ 0&0&1\\ 1&0&0\end{pmatrix}$$
(none of them being symmetrical).