Proposition. Suppose $R$ and $S$ are transitive relations on $A$. Prove that if $S \circ R \subseteq R \circ S$ then $R \circ S$ is transitive.
My attempt:
Suppose $S \circ R \subseteq R \circ S$.
Suppose $(x,y) \in R \circ S$ and $(y,z) \in R \circ S$
Then there exists $a$ such that $(x,a) \in S$ and $(a,y) \in R$
And there exists $b$ such that $(y,b) \in S$ and $(b,z) \in R$
Since $(a,y) \in R$ and $(y,b) \in S$, we conclude that $(a,b) \in S \circ R$. Since $S \circ R \subseteq R \circ S$, we have $(a,b) \in R \circ S$
Since $(a,b) \in R \circ S$, there exists some $c$ such that $(a,c) \in S$ and $(c,b) \in R$.
We know that $S$ is transitive. Since $(x,a) \in S$ and $(a,c) \in S$, we conclude that $(x,c) \in S.$
We know that $R$ is transitive. Since $(c,b) \in R$ and $(b,z) \in R$, we conclude that $(c,z) \in R.$
Since $(x,c) \in S$ and $(c,z) \in R$, we conclude that $(x,z) \in R \circ S$.
Arbitrary elements were considered, hence $R \circ S$ is transitive. $\Box$
Is it correct?
The proof is quite correct. Note that it basically writes itself : It's clear what the goals are and there is only one way to use the given inclusion of relations. Well done.