How to prove that if R is a partial order then also $R^{-1}$ is also a partial order? My guess was to divide the problem into 3 cases: prove that  $R^{-1}$ is reflexive, antisymmetric and transitive.
To prove that $R^{-1}$ is reflexive, I did:
If  $R$ is reflexive, that follows that if $(a, a) \in R$, then $(a, a) \in R^{-1}$.
For the 2. case, prove that $R^{-1}$ is antisymmetric, I did like this:
Suppose $R$ is antisymmetric, this means that if $(x, y) \in R \land (y, x) \in R$, then $x = y$. But, if $x = y$, then if I replace $y$ by $x$ nothing changes, so $(x, x) \in R$, thus $(x, x) \in R^{-1}$. So, finally, also $(x, x) \in R^{-1}$, and $x = x = y$, hence also $R^{-1}$ is antisymmetric.
For the 3. case:
Suppose $R$ is transitive. This means that if $(x, y) \in R$ and $(y, z) \in R$, then also $(x, z) \in R$.
If $(x, y) \in R$, then $(y, x) \in R^{-1}$
If $(y, z) \in R$, then $(z, y) \in R^{-1}$
If the 2 statements before are true, if $(x, z) \in R$, then $(z, x) \in R^{-1}$
Now we have:
$(y, x) \in R^{-1}$
$(z, y) \in R^{-1}$
$(z, x) \in R^{-1}$
Now, we can observe from the 3 statements just above that also $R^{-1}$ is transitive, in fact:
If $(z, y) \in R^{-1}$ and $(y, x) \in R^{-1}$, we have also $(z, x) \in R^{-1}$. 
Is my proof correct?
 A: Reflexivity: $(a,a)\in R^{-1}$ for all $a$ because $(a,a)\in R$.
Antisymmetry: Let $(a,b)\in R^{-1}$ and $(b,a)\in R^{-1}$. Then $(b,a)\in R$ and $(a,b)\in R.$ Since $R$ is a partial order, $a=b.$
Transitivity: Let $(a,b)\in R^{-1}$ and $(b,c)\in R^{-1}.$ Then $(b,a)\in R$ and $(c,b)\in R.$ Since $R$ is partial order, $(c,a)\in R$. Hence, $(a,c)\in R^{-1}.$
A: The ideas are there, but the wording can be improved. You also are going backwards: you're actually proving that if $R^{-1}$ is a partial order, then $R$ is too.
Try being more systematic. State what you have to prove and apply the hypotheses. Your reasoning about antisymmetry is flawed: you want to deduce that $x=y$; if you start assuming it, then you prove nothing.
I'll denote by $A$ the set on which $R$ is defined.
Reflexive We have to show that $(a,a)\in R^{-1}$, for all $a\in A$.  
Let $a\in A$; then $(a,a)\in R$ by reflexivity, so $(a,a)\in R^{-1}$.
Antisymmetric We have to show that, if $(a,b)\in R$ and $(b,a)\in R$, then $a=b$  
Let $a,b\in A$ with $(a,b)\in R^{-1}$ and $(b,a)\in R^{-1}$. Then $(b,a)\in R$ and $(a,b)\in R$, so $a=b$ because $R$ is antisymmetric.
Transitive We have to show that, if $(a,b)\in R^{-1}$ and $(b,c)\in R^{-1}$, then $(a,c)\in R^{-1}$.
Let $a,b,c\in A$ with $(a,b)\in R^{-1}$ and $(b,c)\in R^{-1}$. Then… [complete]
