Alternative Proof of "Suppose R and S are transitive relations Prove that if $S\circ R \subseteq R\circ S$ then $R\circ S$ is transitive"

Question

This question has already been solved here. But I wonder whether there is an easier prove if one uses that R is transitive if $R\circ R \subseteq R$. I am not sure whether this can be used but this is what I tried: $R\circ S$ is transitive if $$(R\circ S) \circ (R \circ S) \subseteq R\circ S$$ Since relations are associative we write: $$(R\circ S) \circ (R \circ S) \subseteq R\circ (S \circ R) \circ S$$ Now comes the step I am not sure about since $S\circ R \subseteq R\circ S$: $$R\circ (S \circ R) \circ S\subseteq R\circ (R \circ S)\circ S$$

Now I can use associativity again and get: $$R\circ (R \circ S)\circ S\subseteq(R\circ R) \circ (S \circ S)$$ But since $R$ and $S$ are transitive $R\circ R \subseteq R$ and $S\circ S \subseteq S$ $$(R\circ R) \circ (S \circ S)\subseteq R\circ S$$ as required. Is this proof correct?

Answer 1

I think the proof is correct. The only doubt I had was when I used $$S\circ R \subseteq R\circ S$$. It is sufficient here to prove that if $A,B,C$ are relations then $$A\circ B \circ C \subseteq A\circ B' \circ C$$ if $$B\subseteq B'$$ For this purpose suppose $(x,z)\in A\circ B \circ C$ Then from the definition of composition: $$\exists y,w:(x,w)\in C \land(w,y)\in B\land(y,z)\in A$$ Since $B\subseteq B'$ $$(w,y)\in B'$$ And therefore we have Then from the definition of composition: $$\exists y,w:(x,w)\in C \land(w,y)\in B'\land(y,z)\in A$$ which means that $$(x,z)\in A\circ B' \circ C$$