
$$\left[\begin{array}{l}\frac{x_{12,4}}{(0.1)(0.4)}=\frac{y_{12,B}}{0.1}=\frac{(0.4)(0.4)(0.7)}{0.16+0.7}=\frac{0.112}{0.86}=\frac{11.2}{86}\\ x_{12,4}\equiv0.020837\\ y_{12,B}\equiv0.091163\end{array}\right.$$
$$\begin{array}{c c c}{{}}&{{\mathrm{A}}}&{{\mathrm{B}}}&{{\mathrm{A\cupB}}}\\ {{}}&{{}}&{{}}\\ {{m_{122}^{P C R5}}}&{{0.345262}}&{{0.505522}}&{{0.149216}}\end{array}$$

If we didn't double *m2(.)* in the fusion rule, we'd get a different result. 

Let's suppose we only fusion *m1(.)* with *m2(.):*

|              | A     | B     | A     | B   | A   | B=   |
|--------------|-------|-------|-------|-----|-----|------|
|              | 0.1   | 0.7   | 0.2   |     |     |      |
|              | 0.4   | 0.1   | 0.5   |     |     |      |
|              | 0.17  | 0.44  |       |     |     |      |
| m m m 1 2 12 | 0.322 | 0.668 | 0.100 | 0   |     |      |
| PCR          |       |       |       |     |     |      |
| m            | 5     |       |       |     |     |      |
| 12           |       |       |       |     |     |      |

∪ ∩ Φ 0.10 0.29 
And now we compare the fusion results: 
 A B A B
5 122 5 12 m two sources; importance of
− −
∪
 0.322 0.668 0.100 - 
PCR PCR
m three sources ond source doubled ; importance of sources considered;

 0.506  0.149 - *three*  0.668  0.100 - *two*.  
 0.345 0.506 0.149 - (sec )
 sources not considered.

The more times we repeat *m2(.)* the closer 5 12...2 PCR m (A)Æ *m2(A)=0.4,* 5 12...2 PCR m (B)Æ *m2(B)=0.1,* and 5 12...2 PCR m (A ∪ B)Æ m2(A ∪ B*)=0.5.* Therefore, doubling, tripling, etc. a source, the mass of each element in the frame of discernment tends towards the mass value of that element in the repeated source (since that source is considered to have more importance than the others). 

For the readers who want to do the previous calculation with a computer, here it is the *mPCR*5 Formula for 3 Sources:

$$\begin{array}{l}{{m_{P C R5}(A)=m_{123}+\sum_{\begin{array}{l}{{X,Y\in G^{0}}\\ {A\neq X\times Y\times A}\\ {A[1](X)\cap Y=\emptyset}\end{array}}\left(\frac{m_{1}\left(A\right)^{2}m_{2}\left(X\right)m_{3}\left(Y\right)}{m_{1}\left(A\right)+m_{2}\left(X\right)+m_{3}\left(Y\right)}+\right.}}\\ {{\left.+\frac{m_{1}\left(Y\right)m_{2}\left(A\right)^{2}m_{3}\left(X\right)}{m_{1}\left(Y\right)+m_{2}\left(A\right)+m_{3}\left(X\right)}+\frac{m_{1}\left(X\right)m_{2}\left(Y\right)m_{3}\left(A\right)^{2}}{m_{1}\left(X\right)+m_{2}\left(Y\right)+m_{3}\left(A\right)}\right)+}}\end{array}$$
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