Example 4: Taking back example 1 and assuming Shafer's model for Θ, the quantitative Dubois & Prade's rule gives the same result as quantitative Yager's rule since the conflicting mass, m∧(A ∩ B**) = 13**/36, is transferred to A ∪ B, while the other quantitative masses remain unchanged.

## - Proportional Conflict Redistribution Rules (Pcr):

PCR5 *for combining two sources* Smarandache and Dezert proposed five proportional conflict redistribution (PCR) methods [21], [22] to redistribute the partial conflict on the elements implied in the partial conflict. The most efficient for combining two basic belief assignments m1(.)
and m2(.) is the **PCR5** rule given by mPCR5(∅**) = 0** and for all X ∈ GΘ, X 6= ∅ by:

$$m_{\rm PCHS}(X)=m_{\wedge}(X)+\sum_{\begin{subarray}{c}X\in\Theta\\ X\cap Y=\emptyset\end{subarray}}\left(\frac{m_{1}(X)^{2}m_{2}(Y)}{m_{1}(X)+m_{2}(Y)}+\frac{m_{2}(X)^{2}m_{1}(Y)}{m_{2}(X)+m_{1}(Y)}\right)\,,\tag{22}$$  within rule given by the equation (17).  
where m∧(.) is the conjunctive rule given by the equation (17).

Example 5: Let's consider the 2D frame and quantitative masses as given in example 1 and assume Shafer's model (*i.e.*
A ∩ B = ∅), then the conflicting quantitative mass k = m∧(A ∩ B**) = 13**/36 is redistributed only to elements involved in conflict, A and B (not to A ∪ B). We repeat that:
m∧(A ∩ B) = m1(A)m2(B) + m2(B)m1(B) = (1/6) · (1/6) + (4/6) · (3/**6) = 13**/36.

So (1/6) · (1/**6) = 1**/36 is redistributed to A and B proportionally to their quantitative masses assigned by the sources (or experts) m1(A**) = 1**/6 and m2(B**) = 1**/6:

$$\mathrm{hence}$$
$$\operatorname{and}$$
$${\frac{x_{1,A}}{1/6}}={\frac{y_{1,B}}{1/6}}={\frac{1/36}{(1/6)+(1/6)}}=1/12,$$
$$x_{1,A}=(1/6)\cdot(1/12)=1/72,$$

$$y_{1,B}=(1/6)\cdot(1/12)=1/72.$$

Similarly (4/6)·(3/**6) = 12**/36 is redistributed to A and B proportionally to their quantitative masses assigned by the sources
(or experts) m2(A**) = 4**/6 and m1(B**) = 3**/6:

$${\mathrm{hence}}$$
$$\operatorname{and}$$
$${\frac{x_{2,A}}{4/6}}={\frac{y_{2,B}}{3/6}}={\frac{12/36}{(4/6)+(3/6)}}=2/7,$$
$$x_{2,A}=(4/6)\cdot(2/7)=4/21,$$
$$y_{2,B}=(3/6)\cdot(2/7)=1/7.$$
