It is easy to check that

$$x_{1,A}+y_{1,B}+x_{2,A}+y_{2,B}=13/36=m_{\land}(A\cap B)$$

Summing, we get:

$m_{\rm PCR5}(A)=(13/36)+(1/72)+(4/21)=285/504\simeq0.57$  $m_{\rm PCR5}(B)=(8/36)+(1/72)+(1/7)=191/504\simeq0.38$  $m_{\rm PCR5}(A\cup B)=2/36\simeq0.05$  $m_{\rm PCR5}(A\cap B=\emptyset)=0$
PCR6 *for combining more than two sources* A generalization of **PCR5** fusion rule for combining altogether more than two experts has been proposed by Smarandache and Dezert in [22]. Recently Martin and Osswald [14], [16] studied and formulated a new version of the **PCR5** rule, denoted PCR6, for combining more than two sources, say M sources with M ≥ 2. Martin and Osswald have shown that **PCR6**
exhibits a better behavior than **PCR5** in specific interesting cases. **PCR6** rule is defined as follows: mPCR6(∅**) = 0** and for all X ∈ GΘ, X 6= ∅,

 mi(X)+ M X−1 j=1 mσi(j)(Yσi (j))   ,(23) M Y−1 j=1 mσi(j)(Yσi (j)) mPCR6(X) = m∧(X) +X M  i=1 mi(X) 2 X M−1 ∩ k=1 Yσi (k)∩X≡∅ (Yσi (1),...,Yσi (M−1))∈(GΘ)M−1
where Yj ∈ GΘ is the response of the expert j, mj(Yj ) the associated belief function and σi counts from 1 to M avoiding i:

$$\left\{\begin{array}{l l}{\sigma_{i}(j)=j}&{\quad\mbox{if$j<i$,}}\\ {\sigma_{i}(j)=j+1}&{\quad\mbox{if$j\geq i$.}}\end{array}\right.\tag{1}$$
$$(24)$$

The idea is here to redistribute the masses of the focal elements giving a partial conflict proportionally to the initial masses on these elements.

In general, for M ≥ 3 sources, one calculates the total conflict, which is a sum of products; if each product is formed by factors of masses of distinct hypothesis, then **PCR6** coincides with **PCR5**; if at least a product is formed by at least two factors of masses of same hypotheses, then **PCR6** is different from **PCR5**:
- for example: a product like m1(A)m2(A)m3(B), herein we have two masses of hypothesis A;
- or m1(A ∪ B)m2(B ∪ C)m3(B ∪ C)m4(B ∪ C), herein we have three masses of hypothesis B ∪ C from four sources.

Example 6: For instance, consider three experts expressing their opinion on Θ = {**A, B, C, D**} in the Shafer's model:
The global conflict is given here by 0.21+0.14+0.09=0.44, coming from:
- A, B and A ∪ C for the partial conflict 0.21,
- A, B and A ∪ B ∪ C ∪ D for 0.14,