$S=\{0,1,2,3\}, R:SxS, (m,n)\in R \text{ if } m+n=4$.
From the condition of $R$, I found that $R=\{(2,2),(1,3),(3,1)\}$.
Now I have to see if $R$ is reflexive, symmetric, antisymmetric, and transitive.
$R$ is not reflexive because $(0,0), (1,1), (3,3) \not\in R$.
$R$ is symmetric because $(1,3), (3,1)\in R$.
$R$ is not antisymmetric because $(1,3), (3,1)\in R$ and $1\not=3$.
$R$ is not transitive because $(1,1)\not\in R$.
I am not really sure if I did this correctly. Specifically the showing that $R$ is transitive. $R$ is transitive whenever $(a,b)\in R \text{ and } (b,c) \in R$, then $(a,c) \in R \text{ for all } a,b,c \in A$. Does it matter that $(2,2) \in R$? Also did I do the others correctly?
You did find $R$ correctly, and each of the four lines beginning with $R$ is correct. However, I'm not exactly sure what the question is that you are trying to answer; the title of your question made me think you just wanted to know whether or not it was an equivalence relation. If that were so, you only needed to show that it isn't transitive.