Clarification - DeGroot Proof on Transitivity Property of Subjective Probability In developing axiomatic foundation for subjective probability DeGroot (Optimal Statistical Decision, 2004, p71) gives two axioms/assumptions:
SP1: For any two events A and B, exactly one of the following three relations must hold:$A \preceq B, A \succeq B, A \sim B $. By definition, $A \preceq B$ is when P(A) $\leq$ P(B)
SP2: If $ A_{1}, A_{2}, B_{1}, B_{2}$ are four events such that $A_{1} \bigcap A_{2}= \emptyset$ and $B_{1} \bigcap B_{2}=\emptyset$ and $A_{i} \preceq B_{i}$ for i=1,2, then $A_{1}\bigcup A_{2} \preceq B_{1} \bigcup B_{2}$. In addition, if either $A_{1}\prec A_{2}$ or $B_{1}\prec B_{2}$ then $A_{1}\bigcup A_{2} \prec B_{1} \bigcup B_{2}$
Based on these two assumptions, he gives a lemma as a direct consequences of the above two assumptions: Lemma 1:Suppose that A, B, D are events such that A $\bigcap$ D = B $\bigcap$ D = $\emptyset$ then A $\preceq$B if and only if $A\bigcup D \preceq B \bigcup D$.
DeGroot continues to give a theorem of transitivity property: Theorem I: If A,B,D are event such that A $\preceq$ B and B$\preceq$D then A$\preceq$D
I need some help understanding the proof of this theorem which he gives as follow:
Consder the seven disjoint events shown in the figure bellow, whose union is $A\bigcup B \bigcup D$. Since A $\preceq$B it follows from Lemma 1 that
(1) $( A\bigcap B ^{c} \bigcap D^{c} ) \bigcup ( A\bigcap B ^{c} \bigcap D ) \preceq ( A^{c}\bigcap B  \bigcap D^{c} ) \bigcup ( A^{c}\bigcap B \bigcap D ) $.
Similarly, since B$\preceq$D it follows from lemma 1 that
(2) $( A\bigcap B  \bigcap D^{c} ) \bigcup ( A^{c}\bigcap B \bigcap D^{c}) \preceq ( A\bigcap B ^{c}\bigcap D) \bigcup ( A^{c}\bigcap B^{c} \bigcap D ) $.
The above lemma doesn't seem to justify this. What i understand from the lemma is that because A $\preceq$B then $( A \bigcup D ) \preceq (( B \bigcup D )  $. How did exactly he go from A $\preceq$B into $( A\bigcap B ^{c} \bigcap D^{c} ) \bigcup ( A\bigcap B ^{c} \bigcap D ) \preceq ( A^{c}\bigcap B  \bigcap D^{c} ) \bigcup ( A^{c}\bigcap B \bigcap D ) $ ?

Next, he argues that since the left sides of the relations (1) and (2) are disjoint and the right sides are also disjoint, it follows from assumption SP2 that
$( A\bigcap B ^{c} \bigcap D^{c} ) \bigcup ( A\bigcap B ^{c} \bigcap D )\bigcup (A\bigcap B  \bigcap D^{c} ) \bigcup ( A^{c}\bigcap B \bigcap D^{c}) \preceq$
$( A^{c}\bigcap B  \bigcap D^{c} ) \bigcup ( A^{c}\bigcap B \bigcap D ) \bigcup ( A\bigcap B ^{c}\bigcap D) \bigcup ( A^{c}\bigcap B^{c} \bigcap D ) $.
If the common event $( A\bigcap B^{c}  \bigcap D ) \bigcup ( A^{c}\bigcap B \bigcap D^{c}$ ) is eliminated from both sides of this relation, it follows from lemma 1 that:
$( A\bigcap B ^{c} \bigcap D^{c} ) \bigcup ( A\bigcap B \bigcap D^{c} ) \preceq ( A^{c}\bigcap B  \bigcap D ) \bigcup ( A^{c}\bigcap B^{c} \bigcap D ) $, he argues that it can now be seen from the above figure and lemma 1 that A $\preceq$D. What I dont understand is how we can justify the elimination of the common event as how he does here.
Any help would be very much appreciated, cheers:)
 A: 
How did exactly he go from $A\preceq B$ into $( A\bigcap B ^{c} \bigcap D^{c} ) \bigcup ( A\bigcap B ^{c} \bigcap D ) \preceq ( A^{c}\bigcap B  \bigcap D^{c} ) \bigcup ( A^{c}\bigcap B \bigcap D ) $ ?

\begin{eqnarray*}
A\preceq B &\iff & AB^c \cup AB \preceq A^cB \cup AB \qquad\text{since $A=AB^c \cup AB$ and $B=A^cB \cup AB$} \\
&\iff& AB^c \preceq A^cB \qquad\qquad\qquad\qquad\text{by Lemma 1} \\
&\iff& AB^cD^c \cup AB^cD \preceq A^cBD^c \cup A^cBD \qquad\text{since $AB^c=AB^cD^c \cup AB^cD$, etc.}
\end{eqnarray*}

What I dont understand is how we can justify the elimination of the common event as how he does here.

Lemma 1 allows for elimination of a common event, $D$. Apply Lemma 1 with:
\begin{eqnarray*}
A &=& AB^cD^c \cup ABD^c \\
B &=& A^cBD \cup A^cB^cD \\
D &=& AB^cD \cup A^cBD^c.
\end{eqnarray*}
Then the lemma, in saying $A\cup D \preceq B\cup D \implies A \preceq B$, eliminates our common event.
The final step to reach $A\preceq D$ is the same as the "expansion" of $A\preceq B$ shown above but in reverse.
