As the title suggests, I want to prove that $R = R^{-1} \circ R$ implies that R is symmetric and transitive. I found an example solution, but it is different from mine. I am interested if my solution is valid.
First $\forall x, y$ if $xRy$ then $x R^{-1}\circ R y$, so $\exists t$ such that $ xRt \land tR^{-1}y$. By definition of $R^{-1}$, $yRt \land tR^{-1}x$ so $y R^{-1}\circ R x$ and $x R y$.
For transitivity,
$\forall x, y, z$ if $xRy \land yRz$ then, using symmetry $zRy$ and $yR^{-1}z$. Combining the two, we get $z R^{-1} \circ R x$ which gives us $z R x$ and, using symmetry again, $x R z$ as desired.