, it is straightforward to use laws of probability to show that the theorem holds. Consider now the case where
. In particular, let
. Using laws of probability and Lemma
$$\begin{aligned} \Pr&(Y^{a_Y=1,a_D=0, \bar{c}=0}_{K+1}=1) \nonumber \\&\quad = \sum _{\bar{l}_K} \Pr (Y^{a_Y=1,a_D=0, \bar{c}=0}_{K+1}=1 \mid \bar{L}^{a_Y=1,a_D=0, \bar{c}=0}_K = \bar{l}_K) \Pr (\bar{L}^{a_Y=1,a_D=0, \bar{c}=0}_K = \bar{l}_K) \nonumber \\&\quad = \sum _{\bar{l}_K} \Big [ \sum _{s=0}^{K} \Pr (Y^{a_Y=1,a_D=0, \bar{c}=0}_{s+1}=1 \mid D^{a_Y=1,a_D=0, \bar{c}=0}_{s+1}\nonumber \\&\quad = Y^{a_Y=1,a_D=0, \bar{c}=0}_{s}=0, \bar{L}^{a_Y=1,a_D=0, \bar{c}=0}_{s} = \bar{l}_{s}) \nonumber \\&\prod _{j=0}^{s} \big \{ \Pr (D^{a_Y=1,a_D=0, \bar{c}=0}_{j+1}=0 \mid D^{a_Y=1,a_D=0, \bar{c}=0}_{j}= Y^{a_Y=1,a_D=0, \bar{c}=0}_{j}=0,\nonumber \\&\quad \bar{L}^{a_Y=1,a_D=0, \bar{c}=0}_{j} = \bar{l}_{j}) \nonumber \\&\qquad \times \Pr (Y^{a_Y=1,a_D=0, \bar{c}=0}_{j}=0 \mid D^{a_Y=1,a_D=0, \bar{c}=0}_{j}= Y^{a_Y=1,a_D=0, \bar{c}=0}_{j-1}=0,\nonumber \\&\quad \bar{L}^{a_Y=1,a_D=0, \bar{c}=0}_{j} = \bar{l}_{j})\nonumber \\&\qquad \times f(L^{a_Y=1,a_D=0,\bar{c}= 0}_{Y,j} = l_{Y,j} \mid Y^{a_Y=1,a_D=0,\bar{c}= 0}_{j} = D^{a_Y=1,a_D=0,\bar{c}= 0}_{j} = 0\nonumber \\&\quad \bar{L}^{a_Y=1,a_D=0,\bar{c}= 0}_{j-1} = \bar{l}_{j-1}, L^{a_Y=1,a_D=0,\bar{c}= 0}_{D,j} = l_{D,j}) \nonumber \\&\qquad \times f(L^{a_Y=1,a_D=0,\bar{c}= 0}_{D, j} = l_{D,j} \mid Y^{a_Y=1,a_D=0,\bar{c}= 0}_{j} = D^{a_Y=1,a_D=0,\bar{c}= 0}_{j} = 0,\nonumber \\&\quad \bar{L}^{a_Y=1,a_D=0,\bar{c}= 0}_{j-1} = \bar{l}_{j-1}) \big \} \Big ] \nonumber \\&\quad = \sum _{\bar{l}_K} \Big [ \sum _{s=0}^{K} \Pr (Y^{a=1, \bar{c}=0}_{s+1}=1 \mid D^{a=1, \bar{c}=0}_{s+1}= Y^{a=1, \bar{c}=0}_{s}=0,\nonumber \\&\quad \bar{L}^{A_Y = A_D = a1, \bar{c}=0}_{s} =\overline{l}_s) \nonumber \\&\prod _{j=0}^{s} \big \{ \Pr (D^{a=0, \bar{c}=0}_{j+1}=0 \mid D^{a=0, \bar{c}=0}_{j}= Y^{a=0, \bar{c}=0}_{j}=0, \bar{L}^{a=0, \bar{c}=0}_{j} = \bar{l}_{j}) \nonumber \\&\qquad \times \Pr (Y^{a=1, \bar{c}=0}_{j}=0 \mid D^{a=1, \bar{c}=0}_{j}= Y^{a=1, \bar{c}=0}_{j-1}=0, \bar{L}^{a=1, \bar{c}=0}_{j}= \bar{l}_{j}) \nonumber \\&\qquad \times f(L^{a=1, \bar{c}=0}_{Y,j} = l_{Y,j} \mid Y^{a=1, \bar{c}=0}_{j} = D^{a=1, \bar{c}=0}_{j} = 0, \bar{L}^{a=1, \bar{c}=0}_{j-1} = \bar{l}_{j-1},\nonumber \\&\quad L^{a=1,\bar{c}= 0}_{D,j} = l_{D,j}) \nonumber \\&\qquad \times f(L^{a=0, \bar{c}=0}_{D,j} = l_{D,j} \mid Y^{a=0, \bar{c}=0}_{j} = D^{a=0, \bar{c}=0}_{j} = 0, \bar{L}^{a=0, \bar{c}=0}_{j-1} = \bar{l}_{j-1}) \big \} \Big ], \end{aligned}$$
(53)
are empty sets. Using Lemma
$$\begin{aligned}&= \sum _{\bar{l}_K} \Big [ \sum _{s=0}^{K} \Pr (Y^{a=1, \bar{c}=0}_{s+1}=1 \mid D^{a=1, \bar{c}=0}_{s+1}= Y^{a=1, \bar{c}=0}_{s}=0, \bar{L}^{A_Y = A_D = a_Y, \bar{c}=0}_{s} = \overline{l}_s ) \\&\quad \prod _{j=0}^{s} \big \{ \Pr (D^{a=0, \bar{c}=0}_{j+1}=0 \mid D^{a=0, \bar{c}=0}_{j}= Y^{a=0, \bar{c}=0}_{j}=0, \bar{L}^{a=0, \bar{c}=0}_{j} = \bar{l}_{j}) \\&\qquad \times \Pr (Y^{a=1, \bar{c}=0}_{j}=0 \mid D^{a=1, \bar{c}=0}_{j}= Y^{a=1, \bar{c}=0}_{j-1}=0, \bar{L}^{a=1, \bar{c}=0}_{j-1}= \bar{l}_{j-1}) \\&\qquad \times f(L^{a=1, \bar{c}=0}_{Y,j} = l_{Y,j} \mid Y^{a=1, \bar{c}=0}_{j} = D^{a=1, \bar{c}=0}_{j} = 0, \bar{L}^{a=1, \bar{c}=0}_{j-1} = \bar{l}_{j-1}, L^{a=1,\bar{c}= 0}_{D,j} = l_{D,j}) \\&\qquad \times f(L^{a=0, \bar{c}=0}_{D,j} = l_{D,j} \mid Y^{a=0, \bar{c}=0}_{j} = D^{a=0, \bar{c}=0}_{j} = 0, \bar{L}^{a=0, \bar{c}=0}_{j-1} = \bar{l}_{j-1}) \big \} \Big ] \\&\quad = \sum _{\bar{l}_K} \Big [ \sum _{s=0}^{K} \Pr (Y_{s+1}=1 \mid C_{s+1}= D_{s+1}= Y_{s}=0, \bar{L}_{s} = \bar{l}_{s}, A = a_Y) \\&\quad \prod _{j=0}^{s} \big \{ \Pr (D_{j+1}=0 \mid C_{j+1}=D_{j}= Y_{j}=0, \bar{L}_{j} = \bar{l}_{j}, A = a_D) \\&\qquad \times \Pr (Y_{j}=0 \mid C_{j}=D_{j}= Y_{j-1}=0, \bar{L}_{j-1} = \bar{l}_{j-1}, A = a_Y) \\&\qquad \times f(L_{Y,j} = l_{Y,j} \mid C_{j}= Y_{j} = D_{j} = 0, \bar{L}_{j-1} = \bar{l}_{j-1}, L_{D,j} = l_{D,j}, A = a_Y) \\&\qquad \times f(L_{D,j} = l_{D,j} \mid C_{j}= Y_{j} = D_{j} = 0, \bar{L}_{j-1} = \bar{l}_{j-1}, A = a_D) \big \} \Big ]. \end{aligned}$$