14.11. References 615
Now to start proving (14.15), letMSbe themembershipfunction for any setS:
MS.x/D
(
1 ifx 2 S;
0 ifx...S:
LetS 1 ;:::;Snbe a sequence of finite sets, and abbreviateMSiasMi. Let the
domain of discourse,D, be the union of theSi’s. That is, we let
DWWD
[n
iD 1
Si;
and take complements with respect toD, that is,
TWWDD T;
forT D.
(b)Verify that forTDandIf1;:::ng,
MTD 1 MT; (14.16)
M.Ti 2 ISi/D
Y
i 2 I
Mi; (14.17)
M.Si 2 ISi/D 1
Y
i 2 I
.1 Mi/: (14.18)
(Note that (14.17) holds whenIis empty because, by convention, an empty product
equals 1, and an empty intersection equals the domain of discourse,D.)
(c)Use (14.15) and (14.18) to prove
MDD
X
;¤If1;:::;ng
. 1/jIjC^1
Y
j 2 I
Mj: (14.19)
(d)Prove that
jTjD
X
u 2 D
MT.u/: (14.20)
(e)Now use the previous parts to prove
jDjD
X
;¤If1;:::;ng
. 1/jIjC^1
ˇˇ
ˇˇ
ˇ
\
i 2 I
Si