For a homework assignment, we have receive the question: Let A and B be sets and let $R \subseteq A \times B$ be a relation. The inverse relation of R is the subset $R^{-1} = \{\langle b,a \rangle : \langle a,b\rangle \in R\}$. Prove or disprove each of the following statements.
e) If $R \subseteq A\times A$, R is symmetric if and only if $R^{-1}$ is symmetric ($\forall x: \forall y: R(x,y) \rightarrow R(y,x)$)
I believe that this statement is true, as far as I understand it symmetry means that all points in a graph have two-sided arrows between them, and the inverse $R^{-1}$ of $R$ simply changes the direction of these arrows. Because they're two-sided in this case then both the relation and its inverse are symmetric. However, I'm not entirely sure how to prove this.
My first step so far is to define an arbitrary relation $R$, as well as two items in the set $A$. The next is to assume $R$ is symmetric, but I don't know exactly how to go past this point.
One possibility I've thought of are to use the definition of symmetry, which is defined above, however it uses an implication and not an equivalence, so I wasn't sure how to use it. Any ideas of how to approach this?
Edit: This is what I've managed to come up with, I'm not sure if it's completely correct but in case others come to this page:
True: Let R be an arbitrary relation from A to A, where x and y are arbitrary items in the set A. assuming R is true only when symmetric, then $R(x,y) \leftrightarrow (R(x,y) \rightarrow R(y,x))$. Following this, we can also state that by symmetry $R(y,x) \leftrightarrow (R(y,x) \rightarrow R(x,y))$. Putting these two together gives us $R(x,y) \leftrightarrow R(y,x)$ through right separation, and this statement is equivalent to our original statement, R is only symmetric if and only if $R^{-1}$ is symmetric.