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I'm getting fairly confused with two exercises related to proving a relation's inverse's properties by knowing the original's. I couldn't do either. Any hint is appreciated.

If $R$ is a symmetric relation over $A$ with $A \not = \emptyset$, prove that $R^{-1}$ is also symmetric.

$R$ is symmetric, so we know that $(aRb \implies bRa)$.

We want to prove that $R^{-1}$ is symmetric. If we use the same $a,b$ as before, we could write it like this: $(bRa \implies aRb)$

So we want to prove $(bRa \implies aRb)$.

The rightmost part of that is $aRb$, which, according to our first premise, it implies $bRa$. I think we can replace it:

$$bRa \implies (aRb \implies bRa)$$

Uh, maybe we can simplify this with implication/disjunction:

$$b\not R a \lor (a\not Rb \lor bRa)$$

$$a\not Rb \lor V_0$$

This isn't quite right... But I'm way too lost - what should I have done?

If $R$ is a transitive relation over $A$ with $A \not = \emptyset$, prove that $R \cap R^{-1}$ is also transitive.

Frankly, I'm lacking even the slightest clue here.

The only think I could observe was that since $R$ is transitive, $R^{-1}$ is transitive as well (is this okay to state, or do I have to prove it?). Not sure how would that help me though.

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up vote 2 down vote accepted

I'm not sure what definitions you are using, but the idea should be something like this:

$R^{-1}=\{(b,a):(a,b)\in{R}\}$. Let $(a,b)\in{R^{-1}}$. As a result ${(b,a)\in{R}}$. Since $R$ is symmetric, $(a,b)\in{R}$. So by definition $(b,a)\in{R^{-1}}$. Hence we have that $R^{-1}$ is symmetric.

As for the other question: Suppose $(a,b),(b,c)\in{R\cap{R^{-1}}}$. We would like to show that $(a,c)\in{R\cap{R^{-1}}}$. Now $(a,b),(b,c),(b,a),(c,b)\in{R}$. Since $R$ is transitive we have $(a,c),(c,a)\in{R}$ by using transitivity on $R$. So we have $(a,c)\in{R^{-1}}$ which gives us what we need.

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For the second question, we proved in the first question that $R^{-1}$ is symmetric, so $(a,c) \in R^{-1}$, therefore the intersection $R \cap R^{-1}$ also contains $(a,c)$ since $(a,c) \in R$ as well, right? However, I'm not sure if the previous question should be treated as a separate scenario, or does it not matter? – Zol Tun Kul Nov 5 '13 at 0:24
Sorry. The argument I had in mind, which you wrote down, is not fully correct (since we do not know that $R$ is reflexive.). I can't think of an immediate way to fix this, sorry. It would help to know what definitions you are working with though. – UserB1234 Nov 5 '13 at 1:43
I'm not sure what do you mean by definitions (lol). I just quoted the exercise :(.. But thanks anyway, the insight was very helpful! – Zol Tun Kul Nov 5 '13 at 1:46
Thank you for catching the error in my previous answer. It has been corrected. If the answer is correct now please do me a favor and accept it. – UserB1234 Nov 5 '13 at 1:51
Yeah, it seems your answer is now correct. Thank you. – Zol Tun Kul Nov 5 '13 at 2:29

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