Relations $R$ and $S$ on a two-element set with $S \circ R \neq R \circ S$

Given: $u := \varnothing$, $v := \{\varnothing\}$, $A := \{u,v\}$, $R := \{(u,v)\}$ and $S := \{(v,u)\}$.

$(a,c) \in S \circ R$ holds if and only if $(a,b) \in R$ and $(b,c) \in S$ for some $b$ (The inverse relation $R^{-1}$, the composite $S \circ R$, and the restriction $R \restriction A$).

A relation is a set $R$ every element of which is an ordered pair (Relation, $\operatorname{dom} R$, $\operatorname{ran} R$, $\operatorname{fld} R$, and the specialisations "relation from $A$ to $B$" and "relation on $A$").

$(a,b) = (c,d)$ if and only if $a = c$ and $b = d$ ($(a,b) = (c,d)$ if and only if $a = c$ and $b = d$).

$(a,b) := \{\{a\},\{a,b\}\}$ (The Kuratowski ordered pair $(a,b) := \{\{a\},\{a,b\}\}$).

$\{x,y\}$ is the set whose elements are exactly $x$ and $y$, and $\{x\} := \{x,x\}$ (The unordered pair $\{x,y\}$ and the singleton $\{x\} = \{x,x\}$).

There is exactly one set with no elements, written $\varnothing$ (There is exactly one set with no elements, written $\varnothing$).

If every $z$ satisfies $z \in x$ if and only if $z \in y$, then $x = y$ (The Axiom of Extensionality: $\forall x\,\forall y\,(\forall z\,(z \in x \leftrightarrow z \in y) \to x = y)$).

$T \circ (S \circ R) = (T \circ S) \circ R$ ($T \circ (S \circ R) = (T \circ S) \circ R$, $(S \circ R)^{-1} = R^{-1} \circ S^{-1}$, $(R^{-1})^{-1} = R$, $\operatorname{dom}(R^{-1}) = \operatorname{ran} R$, and $\Delta_B \circ R = R = R \circ \Delta_A$ for a relation $R$ from $A$ to $B$).

$z \in A \times B$ holds if and only if $z = (a,b)$ for some $a \in A$ and some $b \in B$ (The Cartesian product $A \times B := \{\, z \in \mathcal{P}(\mathcal{P}(A \cup B)) : \exists a \in A\ \exists b \in B\ z = (a,b) \,\}$).