A inclusion property for relations Suppose that $R$ and $S$ are two well-defined relations,
then
$$
(R \cap S)^{-1} \circ R^{-1} \subset (R \circ S)^{-1} ?
$$
Here, $M \circ N$ denotes the composition of the relations $M$ and $N$ and $L^{-1}$ the inverse of a relation $L$.
I try many counter-examples, but none works. For instance $A = \{ a,b\}, B = \{ a,d\}, C = \{ a,c\}$ and $D = \{ b,d\}$ for $R \subset A \times B$ and $S \subset C \times D$. Also, the identity $(M \circ N)^{-1} = N^{-1} \circ M^{-1}$ does not seem to help.
 A: I'm guessing you write $R \circ S$ for $S$ first and then $R$.
Note that the inverse of an intersection is the intersection of the inverses:
$$(R \cap S)^{-1} = R^{-1} \cap S^{-1}.$$
So the inclusion you are interested in is
$$(R^{-1} \cap S^{-1}) \circ R^{-1} \subset S^{-1} \circ R^{-1}$$
so this is not really about inverses and we can just think about:
$$(R  \cap S ) \circ R  \subset S  \circ R.$$
This does hold because for any $(x,z) \in (R  \cap S ) \circ R $, there must be
$y$ where
$(x,y) \in R$ and $(y,z) \in (R  \cap S)$. But then we have $(x,y) \in R$ and $(y,z) \in  S$, so that $(x,z) \in   S  \circ R $.
In a more general setting, if relations $R$ and $S$ satisfy $R \subset S$ and $T$ is a relation then $R \circ T \subset S \circ T$ and $T \circ R \subset T \circ S$.
A: Let $A$ be a set, $R, S \subseteq A^2$ any two binary relations and $x,y \in A$. Then
\begin{align*}
x (R \cap S)^{-1} \circ R^{-1}y & \implies \exists z \in A \colon x R^{-1}z \wedge z (R \cap S)^{-1}y \\
& \implies \exists z \in A \colon zRx \wedge y R \cap S z \\
& \implies \exists z \in A \colon z R x \wedge y R z \wedge y S z \\
& \implies \exists z \in A \colon y S z \wedge z R x \\
& \implies y R \circ S x \\
& \implies x (R \circ S)^{-1}y.
\end{align*}
Therefore, $(R \cap S)^{-1} \circ R^{-1} \subseteq (R \circ S)^{-1}$. $\square$
