Task: Let $A$ be a set. Describe all the relations on $A$ that are at the same time:
i) symmetric and antisymmetric ii) equivalence and partial order
i) Theorem. Let $R$ be a relation on $A$, $R$ is symmetric and anti-symmetric if and only $R\subseteq i_{A}$. Where $i_{A}=\left\{ \left(a,b\right)\in A\times A\mid a=b\right\}$.
$\left(\rightarrow\right)$
Let $a,b$ be arbitrary elements of $A$. Suppose $aRb$. Since $aRb$ and $R$ is symmetric, then $bRa$. Since $aRb$, $bRa$ and $R$ is anti-symmetric, then $a=b$. Therefore since $a$ and $b$ are arbitrary we can conclude that $\forall a,b\in A\left(aRb\rightarrow ai_{A}b\right)$.
$\left(\leftarrow\right)$
Let $p,q,r,s$ be arbitrary elements of $A$. Suppose $pRq$. Suppose $rRs$ and $sRr$. Since $pRq$ and $R\subseteq i_{A}$, then $p=q$ and therefore $qRp$. Since $rRs$ and $R\subseteq i_{A}$, then $r=s$. Therefore since $p,q,r,s$ are arbitrary we can conclude that $\forall p,q\in A\left(pRq\rightarrow qRp\right)$ and $\forall r,s\in A\left(rRs\land sRr\rightarrow r=s\right)$.
ii) Theorem. Let $R$ a relation on $A$, $R$ is an equivalence and partial order relation if and only if $R=i_{A}$ and $R\neq\emptyset$.
$\left(\rightarrow\right)$
If $R$ is an equivalence relation, it must necessarily be symmetric, and if $R$ is an partial order relation, it must necessarily be anti-symmetric. So if $R$ is both, necessarily $R$ must be symmetric and anti-symmetric. But we shown that $R$ is symmetric and anti-symmetric if and only if $R\subseteq i_{A}$. Then if $R$ is an equivalence and partial order relation, it must necessarily be true that $R\subseteq i_{A}$. But if $i_{A}\nsubseteq R$ or $R=\emptyset$ are true that means $\exists x\in A\lnot\left(xRx\right)$, that is, $R$ would not be reflexive. Therefore the only solution is that $R=i_{A}$ and $R\neq\emptyset$ are both true .
$\left(\leftarrow\right)$
Our assumption is $R\subseteq i_{A}$, but we know from the theorem i) that $R$ is symmetric and anti-symmetric. We only need to verify that $R$ is transitive and reflexive. Let $x,y,z,a$ be arbitrary elements of A. Suppose $xRy$ and $yRz$. Since $R\subseteq i_{A}$, then $x=y=z$ therefore $xRz$. Since $a=a$ and $i_{A}\subseteq R$, then $aRa$. Therefore since $x,y,z,a$ are arbitrary we can conclude that $\forall x,y,z\in A\left(xRy\land yRz\rightarrow xRz\right)$ and $\forall a\in A\left(aRa\right)$.
Best Answer
The proofs are essentially correct with the sole remarks that:
I for one would go about proving the claims you mention as follows: