Prove that if $R$ is a symmetric, transitive relation on $A$ and the domain of $R$ is $A$, then $R$ is reflexive on $A$. Assume, $R$ is a symmetric, transitive relation on $A$ and the domain of $R$ is $A$. 
$Dom(R)=A$ implies $(\forall x \in A)(\exists y \in A)[xRy]$. 
Since, $xRy$ is true it follows that $yRx$ is true, by the symmetric property. 
By transitivity, $xRy$ and $yRx$ give $xRx$. Therefore, it follows that, $(\forall x \in A)[xRx]$, which means x is reflexive on A.
Please critique or give any advice, thank you!
 A: Apart from the fact that your proof could benefit from some better wording, it is perfectly correct. I would still suggest you try to reword it more clearly as that will help you later on. Make it more clear what you want to do. Something like:

We wish to prove that $R$ is reflexive i.e. that for every $x$, $xRx$ is true:
$$\forall x: xRx$$
Let $x_0\in A$ be an arbitrary element. Then, because we know that
$$\forall x\in A \exists y\in A: xRy,$$
we know there exists some $y_0$ such that....
...
Therefore, we know that $x_0Rx_0$.
Because $x_0$ was chosen arbitrarily, we know that this is true for every $x\in A$, so $R$ is reflexive.

Such a proof is clearer because:


*

*You state in the beginning what you want to prove

*You show clearly how each step follows from the previous.

*You do not mix up the quantified variables and the bound variables (in your proof, you take an arbitrary $x$, but then you also cite a relation as $\forall x$, which can be confusing and can lead to serious mistakes in harder proofs).

