Suppose $R$ is partial order, prove that $R^{-1}$ is also a partial order

Question

Suppose $R$ is partial order, prove that $R^{-1}$ is also a partial order.

A partial order is a binary relation that is reflexive, anti-symmetric and transitive.

So if $R$ is a partial order, then it has the following properties:

1. $\forall x \in S$, then $(x, x) \in R$ (reflexivity)

2. $\forall x, y \in S$, then if $((x, y) \in R \land(y, x) \in R) \rightarrow x = y$ (anti-symmetry)

3. $\forall x, y, z \in S$, then if $((x, y) \in R \land (y, z) \in R) \rightarrow (x, z) \in R$ (transitivity)

To prove that $R^{-1}$ is reflexive was not so difficult:

1. $R^{-1}$ is reflexive:

   If $(x, x) \in R$, then $(x, x) \in R^{-1}$ by reflexivity of $R$.

2. $R^{-1}$ is anti-symmetric:

   This is a little bit more complicated, and I am not sure if it's correct.

   The key property I used is: if $(x, y) \in R$, then we know that $(y, x) \in R^{-1}$.

   In this expression: $$\forall x, y \in S, \text{then if} ((x, y) \in R \land (y, x) \in R) \rightarrow x = y$$

we can replace $(x, y) \in R$ with $(y, x) \in R^{-1}$, and $(y, x) \in R$ with $(x, y) \in R^{-1}$, we obtain:

$$\forall x, y \in S, \text{then if} ((y, x) \in R^{-1} \land (x, y) \in R^{-1}) \rightarrow x = y$$

which proves that $R^{-1}$ is anti-symmetric.

3. $R^{-1}$ is transitive:

   I think if I follow the same process and key of the point 2, then I will arrive to say that also $R^{-1}$ is transitive.

Is my proof correct?

Answer 1

Your proof of anti-symmetry of $R^{-1}$ is good.

Concerning transitivity, assume that $(x,y)\in R^{-1}$ and that $(y,z)\in R^{-1}$. Then by definition of $R^{-1}$, you also have $(y,x)\in R$ and $(z,y)\in R$. Since R is transitive, we also have $(z,x)\in R$, and then $(x,z)\in R^{-1}$, proving that $R^{-1}$ is transitive.