Show whether the reflexive closure of R is an equivalence relation or not

Question

Let $R$ be a subset of the set of ordered pairs of integers defined recursively by:

$(0,0) \in R$
$(a,b)\in R \rightarrow (a+2,b+3) \in R \wedge (a+3,b+2)\in R$

I think that the reflexive closure of R is an equivalence relation because if (a,b) and (b,c) are true, then (a,c) is true because a=c and (a,a) is true because it's reflexive.

Is this sufficient to show transitivity?

For symmetry I simply stated that it's true due to the recursive definition. I am mainly unsure about the transitivity part.

Thanks.

Answer 1

Clearly $\langle 2,3\rangle\in R$, and two more applications of the generating rule show that $\langle 4,6\rangle\in R$ and $\langle 6,9\rangle\in R$. If $R$ were transitive, $\langle 4,9\rangle$ would have to be in $R$. How would it get there? It clearly doesn’t come from taking the reflexive closure, so it could only come from $\langle 2,6\rangle$ or from $\langle 1,7\rangle$. Show that neither of these pairs can belong to $R$ and hence that $R$ is not transitive; I suggest using structural induction.