Proof of $\;\text{Asymmetric}(\sqsubset)\rightarrow \text{Antireflexive}(\sqsubset)$ The relation $\;\sqsubset\;\subseteq S\times S$ is asymmetric if 
$$\forall a,b\in S:(a,b)\in\sqsubset\rightarrow (b,a)\notin\sqsubset$$
and it is antireflexive if
$$\forall a\in S:(a,a)\notin\;\sqsubset$$
I want to prove that
$$\text{Asymmetric}(\sqsubset)\rightarrow \text{Antireflexive}(\sqsubset)$$
Now... it seem me obvious that is right, eve using some examples in my mind
but when I try to write the formal proof in this way I'm in confusion at the conclusion.
$(a,b)\in\;\sqsubset\rightarrow (b,a)\notin\;\sqsubset$ 
if $b=a$ I get
$(a,a)\in\;\sqsubset\rightarrow (a,a)\notin\;\sqsubset$
... this is a contraddiction..but I don't understand how it is the proof...
someone can explain me in easy words why this is a proof? I only see that the asymmetry lead to contraddiction when $b=a$ ... there is something I'm missing..
 A: $\text{We take as given that the relation is asymmetric.}  \tag{Premise}$
We are assuming, for the sake of contradiction, that $\sqsubset$ is NOT antireflexive; that is we assume that there is an $a \in S$:
$$(a,a)\in\,\sqsubset\tag{1}.$$ 
Since the relation $\;\sqsubset\;$ is asymmetric, then by definition of asymmetry, $$(a,a)\in\,\sqsubset\rightarrow (a,a)\notin\,\sqsubset\tag{2}.$$
By modus ponens, given $(1),\;\text{and}\; (2)$, we are forced to conclude, therefore $$(a, a) \notin \,\sqsubset\tag{3}$$
$(3)$ contradicts $(1)$, so it cannot be the case that in any asymmetric relation, there exists an $a \in S$ such that $(a, a) \in \sqsubset$. That is, our assumption $(1)$ is false, because it leads to a contradiction. Hence a asymmetric relation must necessarily be antireflexive.

Note, what leads to the contradiction is the assumption that $\sqsubset$ contains at least one element that is related to itself, while also being asymmetric. We've shown an asymmetric relation cannot NOT be antireflexive, i.e., an asymmetric relation is necessarily antireflexive. 
A: Imagine to display the elements in the cartesian plane, then being asymmetric means that if a point is in the relation, then its symmetric with respect to the bisectrix of first quadrant is not. Taking the point on the bisectrix gives antireflexivity
A: Let $\sqsubset$ be an asymmetric relation on $S$.
Let $a\in S$ be arbitrary. Two cases are possible: $a\sqsubset a$ or $a\not\sqsubset a$.


*

*If $a\sqsubset a$, then by asymmetry $a\not\sqsubset a$

*If $a\not \sqsubset a$, then we have mmediately that $a\not\sqsubset a$


In both cases we find $a\not\sqsubset a$. Hence $\sqsubset $ is antireflexive.
