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
[niD 1Si;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/DY
i 2 IMi; (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 proveMDDX
;¤If1;:::;ng. 1/jIjC^1Y
j 2 IMj: (14.19)(d)Prove that
jTjDX
u 2 DMT.u/: (14.20)(e)Now use the previous parts to provejDjDX
;¤If1;:::;ng. 1/jIjC^1ˇˇ
ˇˇ
ˇ
\
i 2 ISi