composition of binary relation equivalent to R,S symmetric

Question

I have to prove this: $R \circ S$ is symmetric $\iff R \circ S = S \circ R$. I tried to prove it, and I think I got the proof but I am not sure it is correct: \begin{eqnarray*} R \circ S \iff (\exists u)(x S u \land (u R y) \\ \iff (\exists u)(u S x \land y R u) \\ \iff (\exists u)(y R u \land u S x) \\ \iff (\exists u)(y R u \land y R y \land u S x \land x S x) \\ \iff (\exists u)( u R^{-1} \circ R y \land x S^{-1} \circ S u) \\ \iff (\exists u)( u R y \land x S u)\\ \iff (\exists u)( y R u \land u S x) \\ \iff y S \circ R x \\ \iff S \circ R \end{eqnarray*} please tell me if this is okay, and should I also prove that $ S \circ R = R \circ S $ ?? thank you

Answer 1

In this answer I preassume that $R$ and $S$ are symmetric. You make use of that in your proof, but the extra condition is not mentioned in your question.

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For a relation $T$ on a set $X$ the following statements are equivalent:

• $T$ is symmetric.
• $T\subseteq T^{-1}$.
• $T=T^{-1}$.

Next to that it can be shown that for relation $R,S$ on $X$ we have: $$(R\circ S)^{-1}=S^{-1}\circ R^{-1}$$

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So if $R$ and $S$ are symmetric then $(R\circ S)^{-1}=S^{-1}\circ R^{-1}=S\circ R$

That means that the statement that $R\circ S$ is symmetric comes to the same as the statement that $S\circ R=R\circ S$.