
$$(29)$$

$$(30)$$
$$(31)$$

1−δ2(Y1, Y2) = 1lY1∩Y2=∅ we retrieve Dubois and Prade's rule and the partial conflict can be considered, whereas the rule (26).

The choice of δ1(Y1, Y2) = 1 − δ2(Y1, Y2) can be done by a dissimilarity such as:

$$\delta_{1}(Y_{1},Y_{2})=\delta(Y_{1},Y_{2})\triangleq1-\frac{\mathcal{C}(Y_{1}\cap Y_{2})}{\operatorname*{min}\{\mathcal{C}(Y_{1}),\mathcal{C}(Y_{2})\}},$$
, (29)
or

$$\delta_{1}(Y_{1},Y_{2})=\eta(Y_{1},Y_{2})\triangleq1-\frac{{\mathcal{C}}(Y_{1}\cap Y_{2})}{\operatorname*{max}\{{\mathcal{C}}(Y_{1}),{\mathcal{C}}(Y_{2})\}},$$
, (30)
where C(Y1) is the cardinality of Y1. In the case of the DST framework, C(Y1) is the number of distinct elements of Y1. In the case of the DSmT, C(Y1) is the DSm cardinality given by the number of parts of Y1 in the Venn diagram of the problem [18]. δ(**., .**) in (29) is actually not a proper dissimilarity measure (*e.g.* δ(Y1, Y2**) = 0** does not imply Y1 = Y2), but η(**., .**) defined in (30) is a proper dissimilarity measure. We can also take for δ2(Y1, Y2), the Jaccard's distance, *i.e.* δ2(Y1, Y2) = d(Y1, Y2)
given by:

$$d(Y_{1},Y_{2})={\frac{{\mathcal{C}}(Y_{1}\cap Y_{2})}{{\mathcal{C}}(Y_{1}\cup Y_{2})}},$$

, (31)
used by [10] on the belief functions. Note that d is not a distance in the case of DSmT. Thus, if we have a partial conflict between Y1 and Y2, C(Y1 ∩ Y2**) = 0** and the rule transfers the mass on Y1 ∪ Y2. In the case Y1 ⊂ Y2 (or the contrary),
Y1 ∩ Y2 = Y1 and Y1 ∪ Y2 = Y2, so with δ1(., .) = δ(**., .**) the rule transfers the mass on Y1 and with δ1(., .) = 1 − d(**., .**) it transfers the mass on Y1 and Y2 according to the ratio (C(Y1)/C(Y2)) of the cardinalities. In the case Y1 ∩ Y2 6= Y1, Y2 and ∅,
the rule transfers the mass on Y1 ∩ Y2 and Y1 ∪ Y2 according to δ(**., .**) and d(**., .**).

Example 9: (on the derivation of the weights)
Let's consider a frame of discernment Θ = {**A, B, C**} in Shafer's model (*i.e.* all intersections empty).

a) We compute the first similarity weights δ2(., .) = 1 − δ(**., .**): since

| δ2(., .) = 1 − δ(., .)   | A   | B   | C   | A ∪ B   |
|--------------------------|-----|-----|-----|---------|
| A                        | 1   | 0   | 0   | 1       |
| B                        | 0   | 1   | 0   | 1       |
| C                        | 0   | 0   | 1   | 0       |
| A ∪ B                    | 1   | 1   | 0   | 1       |
| TABLE V                  |     |     |     |         |

VALUES FOR 1 − δ(**., .**)

$$\delta_{2}(A,A)={\frac{{\mathcal{C}}(A\cap A)}{\operatorname*{min}\{{\mathcal{C}}(A),{\mathcal{C}}(A)\}}}={\frac{{\mathcal{C}}(A)}{{\mathcal{C}}(A)}}=1$$ $$\delta_{2}(A,B)={\frac{{\mathcal{C}}(A\cap B)}{\operatorname*{min}\{{\mathcal{C}}(A),{\mathcal{C}}(B)\}}}=0$$
because $A\cap B=\varnothing$ and $\mathcal{C}(\varnothing)=0$. 
$$\delta_{2}(A,A\cup B)={\frac{{\mathcal{C}}(A\cap(A\cup B))}{\operatorname*{min}\{{\mathcal{C}}(A),{\mathcal{C}}(A\cup B)\}}}={\frac{{\mathcal{C}}(A)}{{\mathcal{C}}(A)}}=1$$
