$; $(b)$ follows since $A_{i}$ is a function of $M_{b}$ and since $X^{n}$ is i.i.d.; $(c)$ follows since $(Y_{i},Z_{i})\textrm{---}(X_{i},A_{i})\textrm{---}$ $(X^{n\backslash i},Y^{i-0},Z^{i-0},M_{b})$ forms a Markov chain