I'm given two equivalence relations $E_1$ and $E_2$ over a set A and need to show whether the composition $E_1 \circ E_2$ is reflexive, symmetric and transitive.
I only managed to show that $E_1 \circ E_2$ is reflexive, but I'm struggling with symmetry and transitivity.
Let $(x,z)\in E_1 \circ E_2$ be arbitrary. Then it follows that $\exists y\in A : xE_1y \wedge yE_2z$.
Since $E_1$ and $E_2$ are reflexive: $xE_1x$ and $zE_2z$. Then $y=x$, so that $xE_1x$ and $y=z$, so that $zE_2z$, leads to $x=z$ and therefore $E_1 \circ E_2$ is reflexive.
For the symmetry property, I would need: $xE_1\circ E_2y => yE_1\circ E_2x$.
Let $(x,y) \in E_1\circ E_2$ be arbitrary. It follows $\exists z \in A : xE_1z \wedge zE_2y$. But now I was unable to find a way to show that $yE_1\circ E_2x$ by applying the properties of $E_1$ and $E_2$, but also couldn't show that it was not symmetric.
I had the same problem with the transitive property.