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If $R$ is a symmetric relation then:

$$(x,y) \not \in R \rightarrow (y,x) \not \in R$$

Julia
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  • Yes. Perfectly legal. $\neg q\implies\neg p$ is the same statement as $p\implies q$. For $p$ you can take $(y,x)\in R$ and for $q$ you can take $(x,y)\in R$. – drhab Aug 06 '15 at 10:41

1 Answers1

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Yes; the statements are equivalent.

Symmetry is the property that: $\forall x, y: \big((y,x)\in R\to (x,y)\in R\big)$

The contraposition of this is: $\forall x, y:\big( (x,y)\notin R\to (y,x)\notin R\big)$

Graham Kemp
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  • Shouldn't the 'for all' be replaced by 'there exists' – Shailesh Aug 07 '15 at 02:25
  • No, @Shailesh. If it is not "for all" then the relation may not be symmetric. Symmetry is only assured if the implications hold for all pairs from the universe. The existence of one example is not sufficient; there must be no contrary cases. – Graham Kemp Aug 07 '15 at 02:39
  • Additionally, contraposition is an equivalence, so no form of Dual Negation is applicable. If $P(x,y)$ and $Q(x,y)$ are (syntactically) equivalent statements, then $\forall x,y : P(x,y)$ and $\forall x,y: Q(x,y)$ are equivalent statements, but $\exists x,y:Q(x,y)$ is not equivalent (it's weaker). – Graham Kemp Aug 07 '15 at 02:47