Basically this is what the problem says: R and S are two are anti-symmetric, prove that $R\:∪\:S$ is anti-symmetric as well. I know it isn't because when I try to simplify it, it ends up getting stuck.
My issue is that I cannot find a way to prove that it is not anti-symmetric. Here's where I ended up at:
$aR∪Sb\:$˄$\:bR∪Sa$
$(aRb\:$˅$\:aSb\:$˄$\:(bRa\:$˅$\:bSa)$
$(aRb\:$˅$\:bRa)\:$˄$\:(aSb\:$˅$\:bSa)$
I need a case in which this proves to be false.
On
This is not true in general. Take $$S\stackrel{\rm def}{=}\{(a,b)\in\mathbb{R}\times\mathbb{R}\mid a \leq b\}$$ and $$R\stackrel{\rm def}{=}\{(a,b)\in\mathbb{R}\times\mathbb{R}\mid a \geq b\}$$both antisymmetric binary relations defined on $\mathbb{R}\times\mathbb{R}$. Then $R\cup S = \mathbb{R}\times\mathbb{R}$, which is not antisymmetric.
"My issue is that I cannot find a way to prove that it is not anti-symmetric". This is a huge issue since I will never be able to prove it is not anti-symmetric. For example, if $R \cup S$, with $R$ anti-symmetric, then $R\cup S$ is anti-symmetric.
What you want to do is to find a counter-example to the proposition "If $R$ and $S$ are anti-symmetric, then $R\cup S$ is also anti-symmetric".
A simple counter-example is to take $E=\{1,2\}$ with $R=\{(1,2)\}$ and $S=\{(2,1)\}$. $R$ and $S$ are clearly anti-symmetric, but $R\cup S=\{(1,2),(2,1)\}$ is not anti-symmetric.