I'm trying to prove the following statement:
Proof that if $R$ is a relation from a set $X$ to a set $Y$, $S$ is a relation from $Y$ to $Z$, and $A \subseteq X$, then $I_{SoR}(A) = I_s(I_r(A))$
I know the general outline of equality proof of sets: I have to prove that both sets are subsets of each other.
I first began trying to prove that $I_{sor}(A) \subseteq I_s(I_r(A))$:
Proof: Suppose $R$ is a relation from a set $X$ to a set $Y$, $S$ is a relation from a set $Y$ to a set $Z$ and $A \subset X$. Futher suppose $z \in I_{S o R}(A)$. By the definition of image of relations of sets, there exists $a \in A$ such that $(a, z) \in S o R$. By the definition of composition, there exists a $y$ such that $(a, y) \in R$ and $(y, z) \in S$. From there, I think the way to go is to say that $y \in I_R(A)$ and $(y, z) \in S$, hence $z \in I_S(I_R(A))$, but I don't know if I'm making any logical mistake here.