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This question already has an answer here:

Why is the following relation anti-symmetric?

{(1,2), (2,3), (3,4), (1,4), (1,3), (2,4)}

From my understanding, it is anti-symmetric if:

$$ (a, b) \in R, (b, a) \in R, b=a $$

Must hold. Which is it this case not true. I would rather say it is asymmetric since there is no symmetric nor anti symmetric relation at all.

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marked as duplicate by hardmath, Davide Giraudo, M Turgeon, user63181, Najib Idrissi May 19 '14 at 14:39

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The relation is antisymmetric because the implication in the definition (which you have omitted) is vacuously true.

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  • $\begingroup$ Thanks. Haven't heard of this ''vacuously true'' but now it is clear! $\endgroup$ – user149508 May 19 '14 at 12:26
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The relation is anti symmetric. There exist no pairs $(a,b)$ such that $(a,b)\in R$ and $(b,a)\in R$. This means that for every such pair, $a=b$ is true. It is also true that for all such pairs, the pope is an elephant from Mars.

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First of all, I think you're misunderstanding the definition of an antisymmetric relation:

You write $$(a, b) \in R, (b, a) \in R, b=a$$ as though all three conditions separated by commas must hold.

The correct definition is as follows: For all $a, b$ in the given set on which a relation is defined,

$$\text{IF}\;(a, b) \in R \,\text{ AND }\, (b, a) \in R, \;\text{ THEN }\, a= b$$

If there are no cases where $(a, b)\in R$ and $(b, a) \in R$, then the relation is vacuously antisymmetric.


Another way to think about antisymmetry is as follows: A relation fails to be antisymmetric if and only if there exist $a, b$ such that $(a, b) \in R$ and $(b, a) \in R$, AND $a\neq b$

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