Proof Help: Let $R$ be an equivalence relation on $A$. And, let $x, y, z \in A$. If $x \in y/R$ and $z \not \in x/R$, then $z \not \in y/R$.

Question

Need help with the following proof:

Let $R$ be an equivalence relation on $A$. And, let $x, y, z \in A$. If $x \in y/R$ and $z \not \in x/R$, then $z \not \in y/R$.

Here is the outline so far and I don't think it is correct.

This is proven through contradiction. Suppose $x \in y/R$, $z \not \in x/R$, and $z \in y/R$. Since $x \in y/R, yRx$, and since $R$ is symmetric, $xRy$ too. Since $z \in y/R$, $yRz$, and since $R$ is transitive, $xRz$. That is, $z \in x/R$. This contradicts the hypothesis $z \not \in x/R$.

Answer 1

The proof stated in the question is a valid proof.