If $R$ is a binary relation over $A$ where $I_A \subseteq R $ and $R \circ R \subseteq R$ then $R$ is reflexive and transitive My proof:
Lets assume $A$ is an arbitrary set, where $R$ is a relation over the set $A$. Consider $I_A$ to be an arbitrary identity relation. We will prove if $I_A \subseteq R$ and $R \circ R \subseteq R$ then $R$ is a preorder relation. 
Consequently we know that $I_A \subseteq R$, therefore for every $x$, we know $(x,x) \in R$. Therefore we can conlude that $R$ is reflexive. We need to prove that $R$ is transitive.  
To prove $R$ is transitive, we know $R \circ R \subseteq R$. This tells us that when we combined R with itself, no new relations were created. Because we know no new relations were created,lets assume, a,b and c are arbitrary elements, we can conclude if aRb and bRc then aRc must be true. Therefore the we can conclude $R$ is transitive. This implies that $R$ is reflexive and transitive. $\blacksquare$
My problem:


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*I'm not sure if i've done the proof right, so if anyone can check it. That would be of great help

*When I tried writing the proof, I felt like i didn't go into too much mathematical detail regarding the composition. To check that if $R \circ R \subseteq$ is really transitive, I did some examples and checked if that was true. Then i did the same thing but with untransitive relations, which resulted in new relations. Which is what i sort of expected, but I couldn't really mathematically explain what $R \circ R$ was actually doing and the mathematical explanation behind why $R \circ R \subseteq R$ is transitive. And like i'm still really confused when I read my proof .

 A: *

*Your proof that $R$ is reflexive is correct, it's still missing some important details though. First of all, we need to show what we prove and how we are going to prove this:


First, we prove that $R$ is reflexive, by showing that for any $x \in A$, that $xRx$.

This helps a reader of your proof a lot, so you and he/she are on the same page. Then you need to pick up an arbitrary $x$ and conclude somehow that $(x, x) \in R$ or $xRx$.

So, consider any $x \in A$. Since $I_a \subseteq R$ and $x I_a x$, we know that $xRx$, thus $R$ is reflexive.

See how I write how I got to that conclusion that $xRx$.


*The same here: first of all state what you're going to prove and how. We can always reach out to the definition of the transitivity. This helps us to understand what we can assume and what exactly we need to prove.


Now we prove that $R$ is transitive, so consider any $x, y, z \in A$, where $x R y$ and $y R z$. We will show that $x R z$.

Now see carefully at the definition of composition of relations, which states that:

Given two binary relations $R_1$ and $R_2$ over some set $A$, the composition of $R_1$ and $R_2$, denoted $R_1 \circ R_2$, is the binary relation over $A$ defined as $$x(R \circ R)y \quad \text{if} \quad \exists z\in A.(x R_1 z \land z R_2 x )  $$

So go back to our proof. Let $R_1 = R$ and $R_2 = R$. Then we see that $x R y$ and $y R z$, or $x R_1 y$ and $y R_2 z$. So as it turned out we have such $z \in A$ (namely $y$), where $x R_1 z$ and $z R_2 y$. Therefore we can conclude that $x(R_1 \circ R_2)y$.

Since $x R y$ and $y R z$ and $y \in A$ we know that $x(R \circ R)z$ by definition of composition of relations. Since $x(R \circ R)z$ and $R \circ R \subseteq R$ we know that $xRz$, thus R is transitive.

