\bigwedge_{c_i \in e} b_i $, as the following derivation shows: $$\begin{aligned}
\bigvee_{e \in E(q) } \, \bigwedge_{c_i \in e} b_i & = & \bigvee_{e \in E(q) } \left( \bigwedge_{c_i \in e} b_i \wedge \bigwedge_{c_i \in L_T \backslash e} (b_i \vee \neg b_i)\right)\\
& = & \bigvee_{e \in E(q) } \bigvee_{L \subseteq L_T\backslash e }  \left( \bigwedge_{c_i \in e} b_i \wedge \left(\bigwedge_{c_i \in L} b_i  \wedge \bigwedge_{c_i \in L_T \backslash (L\union e)} \neg b_i\right)\right)\\
& = & \bigvee_{e \in E(q) , L \subseteq L_T\backslash e }  \left( \bigwedge_{c_i \in  L \union e} b_i  \wedge \bigwedge_{c_i \in L_T \backslash (L\union e)} \neg b_i\right)\\
& = & \bigvee_{ L \subseteq L_T, \exists\theta L\union BK \models q\theta }  \left( \bigwedge_{c_i \in  L } b_i  \wedge \bigwedge_{c_i \in L_T \backslash L} \neg b_i\right)\end{aligned}$$ We first add all possible ways of extending a proof $e$ to a full sampled program by considering each fact not in $e$ in turn. We then note that the disjunction of these fact-wise extensions can be written on the basis of sets. Finally, we rewrite the condition of the disjunction in the terms of Equation (\[eq:p\_suc\]). This is possible as each subprogram that is an extension of an explanation of $q$ 