In the following: :$S_n$ denotes the set $S_n = \set {s_1, s_2, \ldots, s_n}$ of cardinality $n$, where $n \in \Z_{\ge 0}$ is a non-negative integer. :$\circledcirc$ denotes an arbitrary relation on $S_n$. Let: :$\map R \circledcirc$ denote that $\circledcirc$ is reflexive :$\map S \circledcirc$ denote that $\circledcirc$ is symmetric :$\map T \circledcirc$ denote that $\circledcirc$ is transitive === Set of Cardinality $0$ === When $S = S_0 = \O$, the result Relation on Empty Set is Equivalence applies. Hence $\circledcirc$ is always reflexive, symmetric and transitive: :$\map R \circledcirc$ :$\map S \circledcirc$ :$\map T \circledcirc$ {{qed|lemma}} === Set of Cardinality $1$ === Let $S = S_1 = \set {s_1}$. From Relation on Set of Cardinality 1 is Symmetric and Transitive, $\circledcirc$ symmetric and transitive. Thus when $n = 1$, it is not possible for a relation on $S_n$ to be either non-symmetric or non-transitive Hence either: :$\neg \map R \circledcirc$ :$\map S \circledcirc$ :$\map T \circledcirc$ when $\circledcirc$ is the null relation, or: :$\map R \circledcirc$ :$\map S \circledcirc$ :$\map T \circledcirc$ when $\circledcirc$ is the trivial relation. {{qed|lemma}} === Set of Cardinality $2$ === Let $S = S_2 = \set {s_1, s_2}$. From Reflexive Relation on Set of Cardinality 2 is Transitive, it is not possible for $\circledcirc$ to be both reflexive and non-transitive. From Relation on Set of Cardinality 2 cannot be Non-Symmetric and Non-Transitive, it is not possible for $\circledcirc$ to be neither symmetric nor transitive. Thus the following combinations of $R$, $S$ and $T$ are not possible with $S_2$: :$\neg \map R \circledcirc, \neg \map S \circledcirc, \neg \map T \circledcirc$ :$\map R \circledcirc, \neg \map S \circledcirc, \neg \map T \circledcirc$ :$\map R \circledcirc, \map S \circledcirc, \neg \map T \circledcirc$ However, we note the following examples of the possible combinations of properties for relations on $S_2$: $(1): \quad \circledcirc := \O$, which is the null relation on $S_2$. From Null Relation is Antireflexive, Symmetric and Transitive, $\circledcirc$ is always antireflexive, symmetric and transitive, and so: :$\neg \map R \circledcirc$ :$\map S \circledcirc$ :$\map T \circledcirc$ $(2):$ :$\circledcirc := \set {\tuple {s_1, s_1}, \tuple {s_1, s_2}, \tuple {s_2, s_1}, \tuple {s_2, s_2} } = S \times S$ that is, the trivial relation. From Trivial Relation is Equivalence:. :$\map R \circledcirc$ :$\map S \circledcirc$ :$\map T \circledcirc$ $(3): \quad \circledcirc := \set {\tuple {s_1, s_2} }$: By inspection: :$\neg \map R \circledcirc$ :$\map S \circledcirc$ :$\neg \map T \circledcirc$ $(4):$ :$\circledcirc := \set {\tuple {s_1, s_1}, \tuple {s_1, s_2} }$ By inspection: :$\neg \map R \circledcirc$ :$\neg \map S \circledcirc$ :$\map T \circledcirc$ $(5):$ :$\circledcirc := \set {\tuple {s_1, s_1}, \tuple {s_2, s_2}, \tuple {s_1, s_2} }$ By inspection: :$\map R \circledcirc$ :$\neg \map S \circledcirc$ :$\map T \circledcirc$ {{qed|lemma}} === Set of Cardinality $\ge 3$ === Let $S_n = \set {s_1, s_2, \ldots, s_n}$. In the following, the smallest examples of relations on $S_n$ will be given to exemplify each combination of $R$, $S$ and $T$, $(1): \quad \circledcirc = \set {\tuple {s_1, s_2}, \tuple {s_2, s_3} }$ We have: :$\neg \map R \circledcirc$ because for example $\tuple {s_1, s_1} \notin \circledcirc$ :$\neg \map S \circledcirc$ because for example $\tuple {s_1, s_2} \in \circledcirc$ but $\tuple {s_2, s_1} \notin \circledcirc$ :$\neg \map T \circledcirc$ because for example $\tuple {s_1, s_2} \in \circledcirc$ and $\tuple {s_2, s_3} \in \circledcirc$ but $\tuple {s_1, s_3} \notin \circledcirc$. $(2): \quad \circledcirc = \set {\tuple {s_1, s_2}, \tuple {s_2, s_3}, \tuple {s_1, s_3} }$ We have: :$\neg \map R \circledcirc$ because for example $\tuple {s_1, s_1} \notin \circledcirc$ :$\neg \map S \circledcirc$ because for example $\tuple {s_1, s_2} \in \circledcirc$ but $\tuple {s_2, s_1} \notin \circledcirc$ We note that there exists exactly one pair of elements of $\circledcirc$ of the form $\tuple {\tuple {x, y}, \tuple {y, z} }$: :$\tuple {\tuple {s_1, s_2}, \tuple {s_2, s_3} }$ and in this case we also have that: :$\tuple {\tuple {s_1, s_3} }$ Hence: :$\map T \circledcirc$ $(3): \quad \circledcirc = \set {\tuple {s_1, s_2}, \tuple {s_2, s_1} }$ We have: :$\neg \map R \circledcirc$ because for example $\tuple {s_1, s_1} \notin \circledcirc$ :$\neg \map T \circledcirc$ because for example $\tuple {s_1, s_2} \in \circledcirc$ and $\tuple {s_2, s_1} \in \circledcirc$ but $\tuple {s_1, s_1} \notin \circledcirc$. We have: :$\tuple {s_1, s_2} \in \circledcirc$ and $\tuple {s_2, s_1} \in \circledcirc$ and so in all cases: :$\tuple {x, y} \in \circledcirc \implies \tuple {y, x} \in \circledcirc$ Hence: :$\map S \circledcirc$ $(4): \quad \circledcirc = \O$, that is, the null relation. From Null Relation is Antireflexive, Symmetric and Transitive: :$\neg \map R \circledcirc$ :$\map S \circledcirc$ :$\map T \circledcirc$ It remains to provide examples of $\circledcirc$ for which $\circledcirc$ is reflexive. By definition, $\circledcirc$ is reflexive {{iff}} it is a superset of the diagonal relation:  :$\Delta_S \subseteq \circledcirc$ So in the remaining examples it is taken as read that $\circledcirc$ is reflexive $(5): \quad \circledcirc = \Delta_{S_n} \cup \set {\tuple {s_1, s_2}, \tuple {s_2, s_3} }$ :$\map R \circledcirc$ as discussed :$\neg \map S \circledcirc$ because $\tuple {s_1, s_2} \in \circledcirc$ but $\tuple {s_2, s_1} \notin \circledcirc$ :$\neg \map T \circledcirc$ because for example $\tuple {s_1, s_2} \in \circledcirc$ and $\tuple {s_2, s_3} \in \circledcirc$ but $\tuple {s_1, s_3} \notin \circledcirc$. $(6): \quad \circledcirc = \Delta_{S_n} \cup \set {\tuple {s_1, s_2} }$ :$\map R \circledcirc$ as discussed :$\neg \map S \circledcirc$ because $\tuple {s_1, s_2} \in \circledcirc$ but $\tuple {s_2, s_1} \notin \circledcirc$ Apart from the trivial $\tuple {\tuple {x, x}, \tuple {x, x} }$ pairs there exist exactly two pair of elements of $\circledcirc$ of the form $\tuple {\tuple {x, y}, \tuple {y, z} }$: :$\tuple {\tuple {s_1, s_1}, \tuple {s_1, s_2} } \implies \tuple {s_1, s_2}$ :$\tuple {\tuple {s_1, s_2}, \tuple {s_2, s_2} } \implies \tuple {s_1, s_2}$ and so: :$\map T \circledcirc$ $(7): \quad \circledcirc = \Delta_{S_n} \cup \set {\tuple {s_1, s_2}, \tuple {s_2, s_3}, \tuple {s_3, s_2}, \tuple {s_2, s_1} }$ :$\map R \circledcirc$ as discussed :$\map S \circledcirc$ by inspection :$\neg \map T \circledcirc$ because for example $\tuple {s_1, s_2} \in \circledcirc$ and $\tuple {s_2, s_3} \in \circledcirc$ but $\tuple {s_1, s_3} \notin \circledcirc$. $(8): \quad \circledcirc = \Delta_{S_n}$ From Diagonal Relation is Equivalence: :$\map R \circledcirc$ :$\map S \circledcirc$ :$\map T \circledcirc$ {{qed}}