How come asymmetry implies antisymmetry It is said that asymmetry implies antisymmetry, but how come so?
If $aRb \Rightarrow \neg(bRa)$, isn't this also the case for $a = b$, which indicates $aRa \Rightarrow \neg(aRa)$, which is a contradiction.
 A: Asymmetry says that given $aRb$, we can't have $bRa$.
Antisymmetry has both $aRb$ and $bRa$ as hypothesis, so in this case it is true by vacuity, since both hypothesis can never be satisfied simultaneously.
A: First some definitions:


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*$R \text{ is antisymmetric in } A \iff (\forall x)(\forall y)((x \in A \land y \in A \land xRy \land yRx) \implies x = y)$

*$R \text{ is asymmetric in } A \iff (\forall x)(\forall y)((x \in A \land y \in A \land xRy) \implies \neg(yRx))$
Second, as you mention if $R$ is asymmetric then $(x \in A \land y \in A \land xRx) \implies \neg(xRx)$ is a contradiction. Therefore if $R$ is asymmetric in $A$, $x \in A$, $y \in A$ and $x = y$ then by reductio ad absurdum$^1$ $\neg(xRy)$


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*In particular if $R$ is asymmetric and $x \in A$ then $\neg(xRx)$
In that sense, if $R$ is asymmetric in $A$, $x \in A$ and $y \in A$ then $x \neq y \iff xRy \lor yRx$ $^2$ 
Third, we can rewrite the definition of antisymmetric as:


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*$R \text{ is antisymmetric in } A \iff (\forall x)(\forall y)(x \neq y \implies (x \notin A \lor y \notin A \lor \neg(xRy) \lor \neg(yRx)))$
Fourth, we have that if $R$ is asymmetric in $A$, $x \in A$, $y \in A$ then $x \neq y \iff xRy \lor xRy$. Let us assume that $xRy$ without loss of generality. Since $R$ is asymmetric $xRy \implies \neg(yRx)$. Therefore we have that


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*$R$ is asymmetric in $A$, $x \in A$, $y \in A$ and $xRy$ then $x \neq y \implies \neg(yRx)$
So $R$ is also antisymmetric $\square$ 
If you want to learn more about relations check out in your local library the book:


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*Mendelson, Elliott. 2008. Number Systems and the Foundations of Analysis. Mineola, N.Y: Dover Publications. ISBN-13: 978-0486457925


In particular check out: Chapter 1 Basic Facts and notions of logic and set theory > 1.16 Reflexivity, symmetry and transitivity > Exercises > 2 > b
$^1$ This means that if $\neg\alpha \implies (\beta \land \neg\beta)$ is true then $\alpha$ is true. In our case assuming $R$ is asymmetric and $xRx$ implies $\neg(xRx)$ meaning that $xRx \land \neg(xRx)$ so $xRx$ can't be true.
$^2$ To be more precise $x \neq y \iff xRy \oplus yRx$ where $\oplus$ refers to the exclusive or.      
