(B.6)
Applying this result and the additivity of GH, we also have for an arbitrary
basic assignment m the following:
(B.7)
This means that we can replicate all focal elements of m the same number of
times, splitting their weights in a fixed proportion a, b, g,...,and the value of
GH does not change.
(iii) Next, we prove that GH(m
1
) = GH(m
2
), where m
1
= (
a
a,
b+g
b) and m
2
= (
a
a,
b
b,
g
b). Since the actual weights are not essential in this proof, we omit
them for the sake of simplicity; if desirable, they can be easily reconstructed.
The proof is accomplished by showing that GH(m
1
) £ GH(m
2
) and, at the same
time, GH(m
2
) £ GH(m
1
). To demonstrate the first inequality, let us view the
focal elements m
1
and m
2
as collections of intervals of lengths a, b, and a, b,
b, respectively. Furthermore, let us place both intervals of m
1
side by side, the
first and second interval of m
2
side by side, and the third interval of m
2
above
the second interval of m
2
. According to this arrangement, the two projections
of m
2
consist of m
1
and a pair of singletons with appropriate weights, say (
x
1,
y
1). It then follows from the subadditivity that
The last term in this inequality is 0 by Eq. (B.6) and, consequently, GH(m
2
) £
GH(m
1
).
To prove the opposite inequality, let m
3
= (
b
1,
g
1) so that m
1
·m
3
assigns the
same weights to the b’s as does m
2
.We select a sufficiently large integer n, tem-
porarily fixed. Let s and s¢ denote squares n ¥ n and (n + 1) ¥ (n + 1), respec-
tively. We can view m
1
·m
3
·s as a collection of four parallelepipeds, two with
edges a, n, n, and the other two with edges b, n, n. Similarly, m
2
·s¢ can be
viewed as a collection of three parallelepipeds, one with edges a, n + 1, n + 1,
and two with edges b, n + 1, n + 1. We now place two blocks with edges a, n,
n of m
1
·m
3
·s inside one block with edges a, n + 1, n + I of m
2
so that they cover
the main diagonal. Furthermore, we place two blocks with edges b, n, n of
m
1
·m
3
·s inside the separate blocks with edges b, n + 1, n + 2 of m
2
·s¢ so
that they again cover the diagonals. Using additivity and Eq. (B.4), the
construction results in the equation
Using subadditivity and the projections of the construction, we obtain
,.