Let $R$ and $S$ be equivalence relations on the same set.
I have to prove that when $R ∪ S$ is an equivalence relation, then $R∘S=R∪S$.
I know that I will have to use the definitions of composition and union to achieve this but I have no clue how I can show that those are the same.
And I know that $R ∪ S = R \lor S$ otherwise the union couldn't be transitive.
$R∪S=(x, y)\in R\lor(x, y)\in S$
$R$ ∘ $S$ $=$ $\{(x,z)\in X\times Z\mid \exists y\in Y:(x,y)\in S\land (y,z)\in R\}$.
Suppose $(x, z)\in R\circ S$. Then there is $y$ such that $(x, y)\in S$ and $(y, z)\in R$. So $(x, y), (y, z)\in R\cup S$ which implies $(x, z)\in R\cup S$ because $R\cup S$ is an equivalence relation.
If you assume that $R$ and $S$ are also equivalence relations, suppose $(x, z)\in R\cup S$. In particular let $(x, z)\in S$. Then $(z, z)\in R$ and which implies $(x, z)\in R\circ S$ as well. Similar case if $(x, z)\in R$.