If $R$ is a strict partial order prove that is asymmetric Suppose $R$ is a relation on a set $A$, and $R$ is asymmetric if:
$\forall x \in A$ $\forall y \in A$ $((x, y) \in R \rightarrow (y, x)\not \in R)$
The first point of the exercise was to demonstrate that if $R$ is asymmetric then it is also antisymmetric. 
What I am not understanding is a proof solution to the following question:

Show that if $R$ is a strict partial order, then it is also
  asymmetric.

A partial order is a binary relation on a set, say, $A$ that is reflexive, antisymmetric and transitive.
And the proof is the following:

Suppose that R is a strict partial order, and suppose that for some
  $x,y \in A$, $(x,y) \in R$ and $(y,x) \in R$. Then by transitivity of
  $R$, $(x, x) \in R$, which contradicts the fact that $R$ is
  irreflexive. Therefore, $R$ is asymmetric.

What exactly I am not understanding is why if $(x,y) \in R$ and $(y,x) \in R$, since $R$ is antisymmetric. At least, $x = y$. Then I also cannot understand why it follows immediately that it's asymmetric.
 A: I don't know if this will help, but let's try it out.
For every $x\in R$ and $y\in R$, there are only four possibilities:


*

*$(x,y)\in R$ and $(y,x)\in R$ 

*$(x,y)\notin R$ but $(y,x)\in R$

*$(x,y)\in R$ but $(y,x)\notin R$, or

*$(x,y)\notin R$ and $(y,x)\notin R$.


A reflection is asymmetric if only the last three ever occur. The proof shows that it is impossible for the first to occur in $R$, and therefore we conclude that $R$ is asymmetric.
A: A strict partial order is a relation $R$ that is irreflexive (for every $x\in A$, $(x,x)\notin R$) and transitive.
Why is it asymmetric? Suppose $(x,y)\in R$ and $(y,x)\in R$; by transitivity, $(x,x)\in R$, which isn't true because the relation is irreflexive.
Therefore, if $(x,y)\in R$, it holds that $(y,x)\notin R$.

Why did mathematicians invented all that? Simple: strict partial orders and (lax) partial orders are exactly the same thing.
If $A$ is a set, write $\Delta(A)=\{(x,x):x\in A\}$.
If $S$ is a strict partial order on $A$, then $S^*=S\cup\Delta(A)$ is a partial order on $A$ (prove it).
If $R$ is a partial order on $A$, then $R_*=R\setminus\Delta(A)$ is a strict partial order on $A$ (prove it).
Note also that $(S^*)_*=S$ and $(R_*)^*=R$, so there is a very well behaved bijection between the strict and the lax partial order relations on $A$.
Think to the above as the connection between $<$ and $\le$.
