Regarding Symmetric transitive closures let $Q'$ be the transitive closure of some relation $R'$, and let $S'$ be the symmetric closure of $Q'$. Must it be the case that $S'$ is transitive as well.

Please point out any mistakes 
Sol. No. Consider the following counterexample.
Let $A = \{a,b,c,d,e\}$.
Let $R = \{(a,b),(a,e),(e,d),(c,d),(c,b)\}$
then $Q' = R\cup\{(a,d)\}$
then $S' = R\cup\{(a,d)\}\cup\{(d,a),(b,a),(e,a),(d,e),(d,c),(b,c)\}$
$S'$ is not transitive because $(b,c)\in S'$ and $(c,e)\in S'$ but $(b,e)\not\in S'$.
