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.
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.