Richardson problem being undecidable means something like: given an arbitrary program (Turing machine) P, you can encode the halting problem for P an an expression in $\mathcal R$. That is you can write down a formula that is identically zero if and only if P halts. Since the halting problem is undecidable, there is no decision procedure for telling if such a formula in $\mathcal R$ is identically zero. That's sort of like Hilbert's tenth problem, where you can encode an arbitary program P as a set of diophantine equations, that has a solution iff P halts. Again since the halting problem is undecidable, there is no algorithm to tell whether an arbitrary diophantine system has a solution.
I think the absolute value function being available in $\mathcal R$ may have something to do with the undecidability. In symbolic algebra, the Risch algorithm is a finite procedure for telling whether a given expression made from elementary functions and composition has a closed-form indefinite integral. But I seem to remember that if you add the absolute value function, the problem becomes undecidable.