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I tried to prove that if $R$ is transitive then its inverse is transitive as well. $$\begin{align} & a{{R}^{-1}}b\,\,,\,\,b{{R}^{-1}}c \\ & \Rightarrow bRa\,\,,\,\,cRb \\ & \Rightarrow cRa \\ & \Rightarrow a{{R}^{-1}}c \\ \end{align}$$

is this a correct proof (or am I completely wrong and it's not even true?)

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A note: What you call "complementary" is conventionally called the inverse relation of $R$. (It's interesting that you got the notation right.) Wikipedia says that alternative terms are converse or transpose relation. I am not sure if "complementary" is used by anyone.

True statement and correct proof.

This statement from wikipedia article on inverse relation is relevant:

If a relation is reflexive, irreflexive, symmetric, antisymmetric, asymmetric, transitive, total, trichotomous, a partial order, total order, strict weak order, total preorder (weak order), or an equivalence relation, its inverse is too.

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    Great I confused myself and thought I might be wrong :) Thanks2011-07-30
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    You are welcome, @Jason. When in doubt, check Wikipedia :)2011-07-30
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    The *complement* of a relation $R$ is defined as the relation $S$ (on the same set) which holds iff $R$ does not hold. But I have never heard the term in the wild.2011-07-31
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    @AndreN Oh yes, thanks for pointing it out. But I sure hope OP did not have this complement in mind.2011-07-31
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    @Srivatsan Narayanan: Luckily not, the OP's proof could not be written by someone who was thinking complement.2011-07-31
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    @Andre Yes, thank you.2011-07-31
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    Hehe, such a discussion about my incorrect translation in a hurry from my language :)2011-07-31
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Yes, that proof looks fine to me.

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    And Thanks to you 2 :)2011-07-30