
## - Qualitative Yager'S Rule:

With Yager's rule, the qualitative conflicting mass L13 6is entirely transferred to the total ignorance A ∪ B, so:

qmY (A ∪ B) = L26
$Y(A\cup B)=L_{\frac{2}{6}}+L_{\frac{13}{6}}=L_{\frac{1}{2}}$
6
= L15
6 and

$$q m_{Y}(A\cap B)=q m_{Y}(\varnothing)=L_{0}$$

while the others remain the same:

$$q m_{Y}(A)=L_{\frac{13}{6}}$$

qmY (B) = L8 6 qmY (.) is quasi-normalized since:

$L_{\frac{13}{6}}+L_{\frac{8}{6}}+L_{\frac{15}{6}}=L_{\frac{36}{6}}=L_6=L_{\max}$. 
If we approximate the linguistic labels L13 6
, L8 6 and L15 6
, still qmY (.) happens to remain quasi-normalized since:
qmY (A) ≈ L[
13 6
] = L2

$\epsilon\approx L_{101}=L_1$ . 

qmY (B) ≈ L[
8 6
] = L1 qmY (A ∪ B) ≈ L[
15 6
] = L3 and L2 + L1 + L3 = L6 = Lmax.

- **Qualitative Dubois & Prade's rule**: In this example the Qualitative Dubois & Prade's rule gives the same result as qualitative Yager's rule since the conflicting mass, qm∧(A∩B) = L13 6
, is transferred to A∪B, while the other qualitative masses remain unchanged.

- **Qualitative Smets' TBM rule**: Smets' TBM approach allows keeping mass on the empty set. One gets:
qmT BM (A) = L13 6 qmT BM (B) = L86 qmT BM (A ∪ B) = L2 6 qmT BM (∅) = L13 6 Of course qm**T BM** (.) is also quasi-normalized.