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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!

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  • $\begingroup$ Your logic is absolutely fine. $\endgroup$ Commented Jun 27, 2016 at 11:10

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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:

  1. You state in the beginning what you want to prove
  2. You show clearly how each step follows from the previous.
  3. 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).
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  • $\begingroup$ Yes, point number 3 is something I must pay attention to from now on. $\endgroup$
    – user223868
    Commented Jun 27, 2016 at 11:20
  • $\begingroup$ @Ibrahim I'd say all three points are worth remembering. Also, note that point 3 is not necccesary (i.e. if you are careful enough, you can get away with some variable bending), but it's still better to be on the safe side. $\endgroup$
    – 5xum
    Commented Jun 27, 2016 at 11:22
  • $\begingroup$ Of course, I will try to take all of them to heart. Thank you! $\endgroup$
    – user223868
    Commented Jun 27, 2016 at 11:23

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