Given a relation $R$, is the symmetric closure of the transitive closure of $R$ equal to the transitive closure of the symmetric closure of $R$? If yes, prove it. If not, give a counterexample.
Well for this, the only reasoning I can come up with is if I have an arbitrary transitive closure set $R_t = (2,3),(1,3)$ then the symmetric closure of $R_t$ is $R_s= \{ (3,2),(3,1) \}$. Therefore:
$$R_s \cup R_t=\{ (2,3),(1,3),(3,2),(3,1) \}$$
Now, if we have the symmetric closure $R_s= \{ (3,2),(3,1) \}$ the the transitive closure $R_s$ would be $R_t= \{(2,1)\}$
Thus making: $$R_t \cup R_s =\{ (3,2),(3,1),(2,1) \}$$
This means that they're not equal. First off, is this counterexample valid in this case? If not, then how can I prove otherwise?
if a relation on $\mathbb{N}$ consists of the single element (1,2) then the symmetric closure adds (2,1) and then transitive closure adds the further elements (1,1) and (2,2).
the other way round we only get (2,1)