conditional mass, qmQBCR1(.**) is also quasi-normalized since**
L4 + L0 + L1 + L0 + L1 + L0 = L6 = Lmax In summary, one gets the following qualitative conditioned masses with QBCR12:
qmQBCR1(B ∩ D|A ∪ B) = L4 qmQBCR1(A ∪ B|A ∪ B) = L1 qmQBCR1(A|A ∪ B) = L1 Analogously to QBCR1, with QBCR2 the qualitative mass of the element D is transferred to D ∩(A∪B) =
B ∩ D according to the model, since D is in D2 ∪ D3 and the largest element X in D1 **which is contained** by D is B ∩ D. Whence qmQBCR2(B ∩ D|A ∪ B) = L4, while qmQBCR2(D|A ∪ B) = L0**. But, differently** from QBCR1, the qualitative mass of C, which is in D2 ∪ D3, but C has no intersection with A ∪ B **(i.e. the** intersection is empty), is transferred A only since A ∈ A ∪ B and qm1(A**) is different from zero (while other** sets included in A ∪ B have the qualitative mass equal to L0). Whence qmQBCR2(A|A ∪ B) = L1 + L1 = L2, while qmQBCR2(C|A ∪ B) = L0. Similarly, the resulting qualitative conditional mass, qmQBCR2(.**) is also**
quasi-normalized since L4 + L0 + L2 + L0 = L6 = Lmax**. Therefore the result obtained with QBCR2 is:**
qmQBCR2(B ∩ D|A ∪ B) = L4 qmQBCR2(A|A ∪ B) = L2

## 6.2 Example 2

Let's consider a more complex example related with military decision support. We assume that the frame Θ = {A, B, C, D} **corresponds to the set of four regions under surveillance because these regions are known to** potentially protect some dangerous enemies. The linguistic labels used for specifying qualitative masses belong to L = {L0, L1, L2, L3, L4, L5, L6}. Let's consider the following prior qualitative mass qm(.**) defined by:**
qm(A) = L1, qm(C) = L1, qm(D) = L4 All other masses take the value L0. This qualitative mass is quasi-normalized since L1 + L1 + L4 = L**1+1+4** =
L6 = Lmax.

We assume that the military headquarter has decided to bomb in priority region D **because there was a high** qualitative belief on the presence of enemies in zone D according to the prior qbba qm(.**). But let's suppose** that after bombing and verification, it turns out that the enemies were not in D**. The important question the** headquarter is now face to is on how to revise its prior qualitative belief qm(.**) knowing that the absolute truth**
is now not in D, i.e. D¯ (the complement of D**) is absolutely true. The problem is a bit different from the** previous one since the conditioning term D¯ in this example does not belong to the hyper-power set DΘ**. In such** case, one has to work actually directly on the super-power set3 as proposed in [4] (Chap. 8). D¯ **belongs to** DΘ
only if Shafer's model (or for some other specific hybrid models - see **case 2 below) is adopted, i.e. when region** D has no overlap with regions A, B or C. The truth is not in D **is in general (but with Shafer's model or with**
some specific hybrid models) not equivalent to the truth is in A ∪ B ∪ C but with the truth is in D¯ **. That's why**
the following two cases need to be analyzed:

## - **Case 1**: D¯ 6= A ∪ B ∪ C.

If we consider the model represented in Figure 2, then it is clear that D¯ 6= A ∪ B ∪ C.

The Super-Power Set Decomposition (SPSD) is the following:
- if the truth is in A, then D1 **is formed by all non-empty parts of** A;