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.
$E_1\circ E_2$ is not necessarily an equivalence relation. Let $A = \{x,y,z\}$, $E_1$ be given by the equivalence classes $\{x,y\}$ and $\{z\}$ and $E_2$ be given by the equivalence classes $\{x\}$ and $\{y,z\}$.
Then $(x,z) \in E_1\circ E_2$, since $(x,y)\in E_1$ and $(y,z)\in E_2$. However, $z$ is connected in $E_1$ only to itself, and $x$ is connected in $E_2$ only to itself. Hence you won't be able to find some $y$ such that both $(z,y)\in E_1$ and $(y,x)\in E_2$.
Moreover, your writing of the reflexivity proof isn't correct. You shouldn't "pick $(x,z)\in E_1\circ E_2$ arbitrary", because you cannot deduce what is $y$ as you did once it is given. It is indeed reflexive, but the way of proving it is to state that taking $y:=x$ indeed statisfies the defining formula of $E_1\circ E_2$.