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I have the following definition of intransitivity:

R is intransitive iff
for all xyz: xRy & yRz -> ~xRz

Now, given the following: aRb bRc cRd

Can I conclude that ~aRd?

Intuitively, I would say yes*, but I'm having problems with the formal proof.
*For example, if R would be 'is a parent of', and incest is impossible, clearly a is not a parent of d. But how to prove that?

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How about a universe with just two things in it and $R$ the relation "is different from"? –  Arturo Magidin Apr 5 '11 at 21:31
    
@Arturo: That is a nice verbal description of my second example. –  Jonas Meyer Apr 5 '11 at 21:34
    
@Jonas: And in some sense, the same as Ross's (once you go to the quotient set modulo the equivalence relation encoded in the "modulo 2"). –  Arturo Magidin Apr 6 '11 at 0:06

2 Answers 2

up vote 3 down vote accepted

How about a set of naturals, with $aRb \Leftrightarrow a \neq b \pmod 2$?

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So if I understand correctly, that relation would be intransitive, but it does not carry over to the next element. So this is a counterexample to my question. –  JackStoneS Apr 6 '11 at 6:45

The relation $\mathrm{R}$ on $\{1,2,3,4\}$ given by $\mathrm{R}=\{(1,2),(2,3),(3,4),(1,4)\}$ is intransitive.


Added:

$\{(1,2),(2,1)\}$ is also intransitive, and it gives a counterexample with $a=c=1$ and $b=d=2$. This would be the minimal counterexample in a sense, and although I wasn't thinking in those terms, in a way this is a reformulation of Ross's nice example.

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So my conclusion is incorrect.. I think I get it now, thanks! –  JackStoneS Apr 6 '11 at 6:49

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