Consider the following example for two experts on Θ = {**A, B, C**}:

| ∅        | A    | B    | A ∪ B   | A ∪ C   | Θ    |      |
|----------|------|------|---------|---------|------|------|
| Expert 1 | 0    | 0.3  | 0       | 0.4     | 0    | 0.3  |
| Expert 2 | 0    | 0    | 0.2     | 0       | 0.5  | 0.3  |
| mConj    | 0.06 | 0.44 | 0.14    | 0.12    | 0.15 | 0.09 |

$$\begin{array}{|c|c|c|c|}\hline\delta&A&A\cup B&\Theta\\ \hline B&1&0&0\\ \hline A\cup C&0&1/2&0\\ \hline\Theta&0&0&0\\ \hline\end{array}$$

where the columns are the focal elements of the basic belief assignment given by the expert 1 and the lines are the focal elements of the basic belief assignment given by expert 2. The mass 0.06 on ∅ is transfered on A ∪ B and the mass 0.2 on A given by the responses A ∪ B and A ∪ C is transfered on A with a value of 0.1 and on Θ with the same value. For δ1 = 1 − d we have:
1 − d A A ∪ B Θ
B 1 1/2 2/3 A ∪ C 1/2 2/3 1/3 Θ 2/3 1/3 0 Note that δ can be used when the experts are considered reliable. In this case we consider the more precise response. With d, we have the conjunctive rule only when the experts give the same response, else we consider the doubtful responses and we transfer the masses in proportion of the imprecision of the responses (given by the cardinality of the responses) on the part in agreement and on the partial ignorance.

For more than two experts, if the intersection of the responses of the experts is not empty, we can still transfer on the intersection and the union, and the equation (18) becomes:

$$\delta(Y_{1},...,Y_{M})=1-\frac{|Y_{1}\cap...\cap Y_{M}|}{\operatorname*{min}_{1\leq i\leq M}|Y_{i}|}.$$
|Yi|. (20)
From the equation (19), we can define δ by:

$$\delta(Y_{1},...,Y_{M})=1-\frac{|Y_{1}\cap...\cap Y_{M}|}{|Y_{1}\cup...\cup Y_{M}|}.$$
δ(Y1, ..., YM ) = 1 − |Y1 ∪ ... ∪ YM | . (21) as: mDPCR(X) = mConj(X) +X M Finally, the rule is given by: i=1 mi(X) 2 mMix(X) = X Y1∪...∪YM=X δ(Y1, ..., YM)Y M M Y−1 j=1 mσi(j)(Yσi(j)) α   mi(X)+ M X−1 j=1 mσi(j)(Yσi(j))   j=1 mj (Yj ) j=1 mj (Yj ). (22) X +X Y1∩...∩YM=X (1 − δ(Y1, ..., YM )) Y M (24) M−1 ∩ k=1 Yσi (k)∩X=∅ (Yσi (1),...,Yσi (M−1))∈(2Θ)M−1 This formulation can be interesting according to the coherence of the responses of the experts. However, this formulation does not allow the repartition of the partial conflict in an other way than the Dubois and Prade's rule. In the later (31) and (32) equations, we will simply write δ instead of δ(Y1, ..., YM ). Y1∪...∪YM=X Y1∩...∩YM=∅ (1 − α)Y M +X j=1 mj (Yj ),
B. A discounting proportional conflict repartition rule The **PCR6** redistributes the masses of the conflicting focal elements proportionally to the initial masses on these elements. For instance, consider three experts expressing their opinion on Θ = {**A, B, C, D**}:

 $\begin{array}{|c|c|c|c|c|c|}\hline&A&B&A\cup C&\Theta\\ \hline\text{Expert1}&0.7&0&0&0.3\\ \hline\text{Expert2}&0&0.5&0&0.5\\ \hline\text{Expert3}&0&0&0.6&0.4\\ \hline\end{array}$
The global conflict is given here by 0.21+0.14+0.09=0.44, coming from A, B and A ∪ C for 0.21, A, B and Θ for 0.14 and Θ, B and A ∪ C for 0.09. With the generalized **PCR6**
rule (12) we obtain:

$$m_{\mathrm{PCR6}}(A)=0.14+0.21+0.21\,\frac{7}{18}+0.14\,\frac{7}{16}\simeq0.493,$$ $$m_{\mathrm{PCR6}}(B)=0.06+0.21\,\frac{5}{18}+0.14\,\frac{5}{16}+0.09\,\frac{5}{14}\simeq0.194,$$ $$m_{\mathrm{PCR6}}(A\cup C)=0.09+0.21\,\frac{6}{18}+0.09\,\frac{6}{14}\simeq0.199,$$ $$m_{\mathrm{PCR6}}(\Theta)=0.06+0.14\,\frac{4}{16}+0.09\,\frac{3}{14}\simeq0.114.$$
$$\begin{array}{c|c|c}{{\bot\cup B}}&{{\spadesuit\,\,}}\\ \hline1/2}&{{2/3}}\\ \hline2/3}&{{1/3}}\\ \hline1/3}&{{0}}\end{array}$$

First of all, the repartition is only on the elements given the partial conflict. We can apply a discounting procedure in the combination rule in order to transfer a part of the partial conflict on the partial ignorance. This new discounting PCR (noted **DPCR**) can be expressed for two basic belief assignments m1 and m2 and for all X ∈ 2 Θ, X 6= ∅ by:

$$m_{\rm DPCR}(X)=m_{\rm Conj}(X)$$ $$+\sum_{Y\in2^{\Theta}}\alpha\left(\frac{m_{1}(X)^{2}m_{2}(Y)}{m_{1}(X)+m_{2}(Y)}+\frac{m_{2}(X)^{2}m_{1}(Y)}{m_{2}(X)+m_{1}(Y)}\right)$$ $$X_{\cap}Y=\emptyset\tag{23}$$ $$+\sum_{Y_{1}\cup Y_{2}=X}(1-\alpha)m_{1}(Y_{1})m_{2}(Y_{2}),$$ $$Y_{1}\cap Y_{2}=\emptyset$$
$$(20)$$
$$(21)$$

with α ∈ [0, 1], the discounting factor.

In a general case for M experts, we could write this rule