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.
Best Answer
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:
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.