Proof that $ 1_A \cup s \cdot s \cdot s $ is equivalence relation

Question

I have problem with this task:
Proof that $$ 1_A \cup s \cdot s \cdot s $$ is equivalence relation
where $r \subset A \times A$ and $r$ is a relation such that $$ \forall_{x,y,z} r(x,y) \wedge r(x,z) \rightarrow r(y,z) $$ and $$ s = r \cup r^{-1} $$

Reflexivity results directly from the definition and symmetry results from the symmetry of the relationship $s$. This part was not difficult for me, but I am trying to show that this relation is transitive and I have no idea how to show that...

Answer 1

Let R be a reflexive, symmetrical relation with six elements in a chain: xRa, aRb, bRz, zRc, cRd, dRz.
Thus R = S and $1_A$ $\cup$ S•S•S is not transitive
where A = { x,a,b,y,c,d,z }