Prove that two different concatenations of relations are equivalent I've had this question on my exam today and I couldn't figure it out, I would like to know the answer.
The question:

Given relations $R$, $S$ on a set $U$.
$R$ is transitive, $S$ is reflexive.
Prove that $(R;S;R)^2$ is a subset of $(R;S)^3$.

Partial answer of mine:

By using associativity you get an equivalent question of:
Prove that $R;S;R;R;S;R \subseteq R;S;R;S;R;S$.
By monotonicity you have that $R;S;R \subseteq S;R;S$.
Then I took $(x, y) \in R;S;R$, so there exist $a, b \in U$ such that $xRa \land aSb \land bRy$.
Now I need to conclude that $xSa \land aRb \land bSy$.

Here I am stuck, nowhere I see how I should be using that $R$ is transitive and $S$ is reflexive, and I know they are obvious hints. I tried using them directly, proving by contradiction, but I am just stuck.
This is best carried out using notations from algebra of relations.

You're asked to prove $RSRRSR \leq RSRSRS$. That $R$ is transitive means $RR\leq R$, so  $RSRRSR \leq RSRSR$ by monotonicity of composition. That $S$ is reflexive means $I\leq S$, where $I$, the identity relation, is neutral element for composition: $I\leq S$ entails $(RSRSR)I\leq (RSRSR)S$ (monotonicity of composition again). Finally, you have

$(RSR)^2 = RSRRSR \leq RSRSR = RSRSRI \leq RSRSRS = (RS)^3$

as required.