A relation is asymmetric if and only if it's both antisymmetric and irreflexive Prove

A relation is asymmetric if and only if it's both antisymmetric and irreflexive

Given a relation $\mathcal R$ on a set $A$ (a homogeneous binary relation),then:
$\mathcal R$ is antisymmetric if :
$$\forall a,b \in A:(a\mathcal R b \wedge b \mathcal R a)\implies a=b$$
$\mathcal R$ is irreflexive if :
$$\forall a \in A:a\not \mathcal R a $$
$\mathcal R $ is asymmetric if :
$$\forall a,b \in A:a \mathcal R b \implies b\not \mathcal R a$$
$\Longleftarrow$
This direction can be proved using contradiction argument,
Assume $\mathcal R$ is both antisymmetric and irreflexive,but it's not asymmetric ,then:
$$\exists a,b \in \mathcal R :a\mathcal R b \implies b \mathcal R a$$
Therefore:
$$\exists a,b \in \mathcal R  :(a\mathcal R b \wedge b\mathcal R a)$$
On the other hand by the asntisymmetric property of $R$:
$$\forall a,b \in A:(a\mathcal R b \wedge b\mathcal R a)\implies a=b$$
Which means :
$$a\mathcal R b \implies a\mathcal R a$$
Which is not true duo to the irreflexive property of $\mathcal R $,implies if $\mathcal R $ is both antisymmetric and irreflexive,then it's asymmetric.
$\Longrightarrow$
If $\mathcal R$ is asymmetric,then:
$$\forall a,b \in A:a\mathcal R b \implies b\not \mathcal R a$$
One may say take $a=b$ and it follows that $\mathcal R$ is irreflexive,but I don't know how
$$\forall a \in A:a\mathcal R a\implies a\not \mathcal R  a$$
Does make sense.
Also I don't know how to show the asymmetric property of $\mathcal R$ implies the antisymmetric property.
 A: Your proof is correct, but it's a bit hard to read, you should clarify in a sentence that $a, b$ are taken so that they disprove asymmetry, i.e. $aRb\land bRa$.
For the other direction, antisymmetry follows vacuously from asymmetry, as the condition $aRb\land bRa$ is always false.
For irreflexibility, we indeed use asymmetry with $a=b$, yielding
$$\forall a:\ aRa\implies a\not Ra$$
which implies $\forall a:\, a\not Ra$, since if we had $xRx$ for an $x$, it would imply $x\not Rx$, a contradiction.
A: Let's walk through it.
Suppose $R$ is asymmetric.
Suppose we have $aRb$ and $bRa$. Because $R$ is asymmetric and $aRb$, we have $\neg(bRa)$. This is a contradiction; it contradicts the fact that $bRa$. Since everything follows from a contradiction, we have $a = b$. Then $R$ is antisymmetric.
Suppose we have $aRa$. Then we have $\neg (aRa)$. Contradiction. Then $\neg(aRa)$. Then $R$ is irreflexive.
Thus, asymmetric implies antisymmetric and irreflexive.
Now, on the other hand, suppose $R$ is antisymmetric and irreflexive. Suppose that $aRb$. Now suppose that $bRa$. Then $a = b$, by antisymmetry. Then $aRa$. Contradiction, by irreflexivity. Then $\neg(aRb)$. Thus, $R$ is asymmetric.
Thus, antisymmetric and irreflexive implies asymmetric.
I hope this clarifies things.
