Thus, we have \(R_1^2=\{(t, t_1), (m, t_1)\}\). Note that \(s_2\) assigns non-zero probability to the pair \((m,t_2)\) which is not in \(R_1^1\), so \(s_2\) does not believe that \(1\) is rational. Thus, we have \(R_2^2=\{(l,s_1), (c,s_1),(l,s_3)\}\). Continuing with this process, we have \(R_1^2=R_1^3\). However, \(s_3\) assigns non-zero probability to \((b,t_2)\) which is not in \(R_1^2\), so \(R_2^3=\{(l, s_1), (c,s_1)\}\). Putting everything together, we have