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On generalized Boolean functions II
Discrete Mathematics 40 (1982) 277-284 North-Holland Publishing Company
ON GENERALIZED
277
BOOLE
FUNCTIONS
Nicolae T~D~REANU Faculty of Muthematics, Uniuersiry of Craiova. Craioua, Romania Received 29 June 1981 This cycle of papers is b%ed on the concept of generalized Boolean functions introduced by the author in the fhst article of the series. Every generalized Boolean function f : B” + B can be written in a manner similar to the canonical disjunctive form using some function defined on A x B, where A is a finite subset of B containing 0 and 1. The set of those functions f is denoted by GBF,,[A]. In this paper the following questions ar. presented: (1) What is the relationship between GBF,[A,J and GBF,[A,] when A, c A,. (2) What can be said about GBF,.,[A, n AA and GBF,[A, U A,] in comparison with GBF,,[A,]nGBF,[A,j and GBF,,[A,] U GBF,[A,], respectively.
1.. Inlrodw!tion Let B#{O, 1) be a Boolean algebra and A a finite subset with the property that (0,1) c A c B (by = we denote proper inclusion). Two sets of functions, denoted by %(A, B) and GBFJA], are defined in [3]: %(A, B) is the set of those functions g : A x J3 + B which satisfy the conditions: g(O,O) = g(L 1) = 1, I
g(W)=
n E@, x), beA\W
while GBFJA] is the set of those functions f : B” + B, called generalized Bookan functions, for which there are ge %(A, B) such that for any (x,, . . . , G)E I?“,
f(x l--r~)=a
1.
.~~EAf(al,...,4)g(a,,x,)...g(4,~). .
.
The concept of generalized Boolean functions is not trivial since there are functions f : B” + B which are not in GBFJA] and the set of Boolean functions is included properly in GBFJA] (see [3, Theorem 41). In this paper we clarify the following questions: (1) What is the relationship between GBFJA,] and GBFJA,] when A, c AZ. (2) What can be said about GBF,[A, f7 A,] and GBF,[A, U A,] in comparison with GBFJAJ nGBF,,[A,] and GBFJA,] UGBF,[Az], respectively. For prerequisites on Boolean algebra as well as for the notation which is used in this paper we refer to [1,2]. 0012-365X/82/0000-0000/$02.75
@j 1982 North-Holland
N. TJTindiireanu
278
The results Rl, R2 and R3 below, obtained in [3], will be used in this paper: I. Let g E%(A, B). For any n 3 1 and (x,, . . . , x,) E B” we have: a,*..lcA g& Particularly,
x1) ’
l
l
Sk,
4
lJoeA g(ar, n) = 1 therefore
R2. For a tixed element gk
y) =
= 1..
JJjaEAg(a, X) = 01.
a0 E A, the function
g : A x B 3 B, defined as follows:
1
if [x = y E A] or [ix = a,) and (y E B\A)],
0
otherwise,
is in $(A, B). .
If f~ GBF,,[A],
then f~ GBFJA].
More precisely,
if
then
2. BdcIemmm The results in this section are used in the theorems
of the next section.
follows from the fact that f(x 1, * * *, &I) = al ..UAfh
- - - ,%I
gh,
%> ’ * * hn,
4
for some g E WA, B) and from g(a, x) d I for any (u, x) E A x B. Now inequality II
n,.....a,,eA
f(% . ..dL)~fh....a~
(2)
from R3 and (1) by the fact that B s b implies li 2 6.
is obtained .
Let a
EB\A
be Q fixcal element and A a finite subset such that
On generalized Boolean f~mctims ZZ
279
(0, 1)~ A c
B. For any finite subset C with property A U{a)c C, the function g :C x B ---*B, defined QSfollows: gtx, Y) =
1
if [x=yeA]
0
otherwise,
or [x=a
and ~EB\A],
is in %(C, B)\S?(A, B). Proof. We have g(0, 0) = g(1, 1) = 1 since (0, 1) E A. The function following relations:
g satisfies the
,IJcgthY) = 19 d%, Y)
gtu,, Y) =
(3)
0, Ul #
4,
for any y E B. We denote by rl, . . . , t,, the elements of A. If y E A, then y = ti for some iE{l,..., n}, hence g(& y) = 1 and g(u,: y) g(up, y) = 0 since at least one of ~1 or 2.42is not 4. If y E B\A, then g(a, y)= 1 and g(ui, y) g(u2, y) = 0 since at least one of u1 or up is not a. Therefore, g E %(C. B). The function g f&fi;lns the relation Urcl g(h, a) = 0, therefore g$ %(A, B). In the next lemma it is shown that by introducing generalized Boolean function, we obtain a generalized Lermma 3. If f E GBF,[A],
fictitious arguments Boolean function.
in a
then the following function h : Bn+p --PB, defined as
hh, . -.,~,Y1,...,Yp)=f(XT,...,X) for any x1,. . . , q,, yl, . . . , yp E B, is in GBF,+,[A]. Proof.
We have
for some g E %(A, B). Taking into account Rl, we obtain: LJ,
.
h(al, ....12n.h,....b,)gh, xJ. m- gh,,d
gb, y,). - -db,,Y,)
b,.....&,,eA u fb,,. -.,a,hh,x~)-al.....%
=
-gb,+,Mb,,y,)--
g(b,,y,)
b,,....b,eA
=
U al....,a,,eA
therefore
f(al,.-.,~)r!(al,x*)...g(~,X)=h(xl,...,
h E GBF,,+,[A].
G,y ,,..., y,),
IV.Wd&eunu
280
3. The redb Everywhere in this section we denote by A1 and A,. two finite sets such that 1, ;!. If Ai c A*, then we will USCthe notation A1 -(al,. . . , ak} {0,1}6A,cB,i= }. If A, c A, and g, E %(A,. B), then we will say that and &\A1 ={&+lr.. . , ak+n the function g2 : A2 x B -s, B is an extension of gl to A2 if gz E %(A2, B) and g2 is an extension of g, ‘to AZ x B. If g E ‘3(A2, B), then we denote by g,\, the restriction of g to A,. We remark now that pi, is not forthwith in $?(A,, B). This problem is treated in the following theorem. then there is Q function g E (e(A2, B) such that rem 1. If AlcAl, gA, &‘%(A,, 8). Every function g, E %(A,, B) has an extemlion to A2 and only one: g2(a,
xl
=( &(a, x), (a, X)F.Al IQ,
x B,
(a, x) E (Az\A,)
x B.
Prsaf. In Lemma 2 we take a := ak+l, C = A2 and A = Al. It follows that g is in VU,, BYMA,. B), i.e. go, #%B(‘A,,B). To prove that g2E %(A2, B) we first remark that
lJ,
g2k
x) = ,g, &(G x) = 1.
Further, if UP b, then either one of them, say a, is in B\A in which case g2ra, x) = 0, or a, 6 E A which implies g2(a, x) g2(b, x) = gl(a, x) g,(b, x) = 0. 'NOW, let h E %(A2, B) be another extension of g, to AZ. i;or k + 1 c i s k + n we have Mai, x1 = bSAn\{&1 @b, x1.s ,II, 5
@4 x) = Fa &Cl+ x) = 0 beA,
1
by Rl. It follows that .
h(u, x) =
(a, xl E Al x B, (a, x) E (&\AJ
therefore
x B,
h = g2.
Because of the fact that a function g E %(Al, B) has only one extension tc AZ;, we can say without any confusion: ‘the extension of g to A2’. 2. If A, c AZ, then GBF,,[A,]c
GBF,[A2].
f. Let f~ GBF,[A1 J-There is g E %((.A,, B) such that for any (x1, . . . , x,,) E B”,
On generalizedBoolean functionsII
If gl is the extension
281
of g to AZ, then
gen; fh
- *- 4l) d% Xl) - * - d4v 4 =fh
=(a ,,.,.
7
* . . , x,),
therefore f E GBFJAJ. Let us prove that the inclusion is proper. We define the function follows:
jr : n + jj as
x E B\Al, XEAl.
We denote
by q the function
defined by:
k+n dx)=
ak+l
u r=k+l
g(%
d9
where g is the function deked in Lemma 2 for a = ak+i, C = AZ and A = A,. If x E Al, then g(u,., x) = 0 for r~(k + 1, . . . , k + n}, therefore q(x) = 0. If x E B\A1, then g(a k+l,x)‘I and g(a,,x)=O fur rE{k+? ‘, . . . , k + n), therefore It follows that for any x E B we have q(x) = Qk+l. k+n h(x)
=
q(J) =
u r=k+l
k+n uk+l
f&h
x)
= rv,
h(&)
d%
x),
therefore h E GBF1[A2]. But h$ GBF,[AJ, else we should have h = 0. By introducing the fictitious arguments x2, . . . , x,., in h we obtain a function which is in GBF,,[A2]\GBF,[A1] (see Lemma 3). C~r~lIary.
For any X and Y such that (0, 1) E Xc B and (0,l) c Y c B, we have: GBF,,[Xn
Y]EGBF,,[X]~GBF,[Y],
GBF,,[Xj U GBFJY]
E GBFJX
(4) u Y].
(5)
The proof is immediate by Theorem 2 since X17 Y c_X E X U Y anld X 17 Y c YEXUY. In the following two theorems we show that (4) and (5) always hold with proper inclusion except in the case when X E Y or YE X. remi GBFJAJ,
then GBF,,[A1 nA,]=G’BF,[A,]!n 3. If A,c_A, 01 A,G Al, eke we have GBF,[AInAJcGBF,,[A1]nGBF,[A,].
N. ~iindtireanu
282
Proof. If AicAj therefore
(i, jE{?,2),
GBF,[Ai]
if j), then GBF,[A~]E
n GBF,[Aj]
by Theorem
2,
GBFm[Ai I3 Aj].
= GBF,[Ai]=
Suppose that A1\A2 # fl and A2\A1 # 8. As G13F,[A1 tl A,]E GBF,,[AJ (i = 1,2), it remains to prove that the inclusion is proper. Take y1 E AI\AZ and y2 E Az\A,. We define the func+,ions gi : Ai X B + i3 (i:=1,2) as follows: g&4 .w)=
1
if [U = x E Ai] or [a = yi and x E B\Ai],
0
otherwise.
By R2 it follows that gi E %(Ai, B). We now define the function relation:
For i = 1,2 we denote
f : B” -+ B by the
by b the foiiowirpg function:
Let us prove that f = h, = hz. For (x1, . . . , x,,) E B” we have the following cases: (a) If x1,. . . . xn E A1 Cl AZ, then .,&)=h2(%,. ..J?t,). Ml, . . .A)=f(%.. (b) If there is some 4 E A1\A2 (1 S i G n), then we denote element:
zj =
xi,
1
by Zj the following
i=L
y2, i=2.
We have
X tdl,...,
n
i-l.i+l,...,
n)
gj(%
We remark now tha; Zj$ At n AZ, therefore &L
* - - 7 &J =
u al,...&-I.
Q,+x~..-r(In+ = I = Ax,, (cJ
if
there is some q E A2\Al,
the prec%oles relation
tE(l,___, iQi+l,....
. . .
&)a
n) “”
“)
, 4,).
then interchange
the roles of 1 and 2 in (a).
(d) If x1,. . . , x,,E B\(A, U A,) then bzii(x1,. - * 7 hi = f(yi, - * * 9 Vi)
yields:
= I=
f(Xl,
* * * 9 X, ).
-_,
,,,I ,..,,.
II
On generalized Boolean functions I1
283
It follows that f~ GBF,[A*] n GBF,,[A2]. Obviously f$ GBF,,[A1 n AJ, else f should be identically zero.
Theorem 4. If Al c A2 or A, E Al, then GBF,,[A,]IJ
GBF,[A2] = GBF,[Ar
GBF,[A,]U
GBF,,[A,]c
U AZ];
else GBF,,[Ar 1; A*].
E%of. We have GBF,CAr]U GBF,,[Az]~ GBF,[A, U AS] and it remains to prove that if A1\A2 # $3P A2\A1, then the inclusion is proper. We consider a eAl\A2 and b EA~\A~. Let g :(A, UA,)xB + B be the function defined by: !dYi x) =
1
if [y=:rEAIUAz]
0
otherwise.
or [y=O and XEB\(A~UA,)],
By R2 we have g E %(A, U A*, B). Let us consider the function f : B” + B defined as follows: x1 = a, x1= b, ~1E B\ju, b); x2,
- * * , x,,
being arbitrary elements
in B. We have:
r
U =
U
~~,....cz,,EA,UA~
a 0, ~1)de.
a gb, x1)u ii 0, x1)= fh,
~2)
- - - gbn,
G,)
1
. . . , al,
from the definition of g and by the fact that from OE Al nA2 it follows that u # 0 # b and therefore g(u, x) = 0 = g(b, x) for any x# {a, b}. It follows that f~ GBF,[Ar U A,]. If f~ GBF,[A,], then by Lemma 1 we must have ii =
f(h x2,
. . . ,d
scl
,... lJA,
fh,
. . . , c,,)
=
a
It follows that 1 = a U ii = a which is not true since cl$ ,A1 tl A;. If f E GBF,[A,] follows that a = 0 which is again false. Therefore f$GBF,[A,]UGBF,[A,].
. _. .._ I ..j..,...,,._.^,..,..,,.,
”
..I
.,,,.
it
284
N. ‘J%nd&anu
Tk author wishes to thank Professor S. Rudeanu for his valuable remarks which improved the manuscript.
/l] C.P. Popovici. L. Livovschi and N. Tbdgreanu, Switching circuit:;, autom?ia and formal languages, Univ. of Bucharest (1980) (in Roumanian). /2] S. Rudeanu, Boolean ‘Functions and Equations (North-Holland, Anlaterdam, 1974). [3] N. TsndHreanu, On generalized Boolean functions I, Discrete Math. 34 (l!XO). 293-299.
ON GENERALIZED
277
BOOLE
FUNCTIONS
Nicolae T~D~REANU Faculty of Muthematics, Uniuersiry of Craiova. Craioua, Romania Received 29 June 1981 This cycle of papers is b%ed on the concept of generalized Boolean functions introduced by the author in the fhst article of the series. Every generalized Boolean function f : B” + B can be written in a manner similar to the canonical disjunctive form using some function defined on A x B, where A is a finite subset of B containing 0 and 1. The set of those functions f is denoted by GBF,,[A]. In this paper the following questions ar. presented: (1) What is the relationship between GBF,[A,J and GBF,[A,] when A, c A,. (2) What can be said about GBF,.,[A, n AA and GBF,[A, U A,] in comparison with GBF,,[A,]nGBF,[A,j and GBF,,[A,] U GBF,[A,], respectively.
1.. Inlrodw!tion Let B#{O, 1) be a Boolean algebra and A a finite subset with the property that (0,1) c A c B (by = we denote proper inclusion). Two sets of functions, denoted by %(A, B) and GBFJA], are defined in [3]: %(A, B) is the set of those functions g : A x J3 + B which satisfy the conditions: g(O,O) = g(L 1) = 1, I
g(W)=
n E@, x), beA\W
while GBFJA] is the set of those functions f : B” + B, called generalized Bookan functions, for which there are ge %(A, B) such that for any (x,, . . . , G)E I?“,
f(x l--r~)=a
1.
.~~EAf(al,...,4)g(a,,x,)...g(4,~). .
.
The concept of generalized Boolean functions is not trivial since there are functions f : B” + B which are not in GBFJA] and the set of Boolean functions is included properly in GBFJA] (see [3, Theorem 41). In this paper we clarify the following questions: (1) What is the relationship between GBFJA,] and GBFJA,] when A, c AZ. (2) What can be said about GBF,[A, f7 A,] and GBF,[A, U A,] in comparison with GBFJAJ nGBF,,[A,] and GBFJA,] UGBF,[Az], respectively. For prerequisites on Boolean algebra as well as for the notation which is used in this paper we refer to [1,2]. 0012-365X/82/0000-0000/$02.75
@j 1982 North-Holland
N. TJTindiireanu
278
The results Rl, R2 and R3 below, obtained in [3], will be used in this paper: I. Let g E%(A, B). For any n 3 1 and (x,, . . . , x,) E B” we have: a,*..lcA g& Particularly,
x1) ’
l
l
Sk,
4
lJoeA g(ar, n) = 1 therefore
R2. For a tixed element gk
y) =
= 1..
JJjaEAg(a, X) = 01.
a0 E A, the function
g : A x B 3 B, defined as follows:
1
if [x = y E A] or [ix = a,) and (y E B\A)],
0
otherwise,
is in $(A, B). .
If f~ GBF,,[A],
then f~ GBFJA].
More precisely,
if
then
2. BdcIemmm The results in this section are used in the theorems
of the next section.
follows from the fact that f(x 1, * * *, &I) = al ..UAfh
- - - ,%I
gh,
%> ’ * * hn,
4
for some g E WA, B) and from g(a, x) d I for any (u, x) E A x B. Now inequality II
n,.....a,,eA
f(% . ..dL)~fh....a~
(2)
from R3 and (1) by the fact that B s b implies li 2 6.
is obtained .
Let a
EB\A
be Q fixcal element and A a finite subset such that
On generalized Boolean f~mctims ZZ
279
(0, 1)~ A c
B. For any finite subset C with property A U{a)c C, the function g :C x B ---*B, defined QSfollows: gtx, Y) =
1
if [x=yeA]
0
otherwise,
or [x=a
and ~EB\A],
is in %(C, B)\S?(A, B). Proof. We have g(0, 0) = g(1, 1) = 1 since (0, 1) E A. The function following relations:
g satisfies the
,IJcgthY) = 19 d%, Y)
gtu,, Y) =
(3)
0, Ul #
4,
for any y E B. We denote by rl, . . . , t,, the elements of A. If y E A, then y = ti for some iE{l,..., n}, hence g(& y) = 1 and g(u,: y) g(up, y) = 0 since at least one of ~1 or 2.42is not 4. If y E B\A, then g(a, y)= 1 and g(ui, y) g(u2, y) = 0 since at least one of u1 or up is not a. Therefore, g E %(C. B). The function g f&fi;lns the relation Urcl g(h, a) = 0, therefore g$ %(A, B). In the next lemma it is shown that by introducing generalized Boolean function, we obtain a generalized Lermma 3. If f E GBF,[A],
fictitious arguments Boolean function.
in a
then the following function h : Bn+p --PB, defined as
hh, . -.,~,Y1,...,Yp)=f(XT,...,X) for any x1,. . . , q,, yl, . . . , yp E B, is in GBF,+,[A]. Proof.
We have
for some g E %(A, B). Taking into account Rl, we obtain: LJ,
.
h(al, ....12n.h,....b,)gh, xJ. m- gh,,d
gb, y,). - -db,,Y,)
b,.....&,,eA u fb,,. -.,a,hh,x~)-al.....%
=
-gb,+,Mb,,y,)--
g(b,,y,)
b,,....b,eA
=
U al....,a,,eA
therefore
f(al,.-.,~)r!(al,x*)...g(~,X)=h(xl,...,
h E GBF,,+,[A].
G,y ,,..., y,),
IV.Wd&eunu
280
3. The redb Everywhere in this section we denote by A1 and A,. two finite sets such that 1, ;!. If Ai c A*, then we will USCthe notation A1 -(al,. . . , ak} {0,1}6A,cB,i= }. If A, c A, and g, E %(A,. B), then we will say that and &\A1 ={&+lr.. . , ak+n the function g2 : A2 x B -s, B is an extension of gl to A2 if gz E %(A2, B) and g2 is an extension of g, ‘to AZ x B. If g E ‘3(A2, B), then we denote by g,\, the restriction of g to A,. We remark now that pi, is not forthwith in $?(A,, B). This problem is treated in the following theorem. then there is Q function g E (e(A2, B) such that rem 1. If AlcAl, gA, &‘%(A,, 8). Every function g, E %(A,, B) has an extemlion to A2 and only one: g2(a,
xl
=( &(a, x), (a, X)F.Al IQ,
x B,
(a, x) E (Az\A,)
x B.
Prsaf. In Lemma 2 we take a := ak+l, C = A2 and A = Al. It follows that g is in VU,, BYMA,. B), i.e. go, #%B(‘A,,B). To prove that g2E %(A2, B) we first remark that
lJ,
g2k
x) = ,g, &(G x) = 1.
Further, if UP b, then either one of them, say a, is in B\A in which case g2ra, x) = 0, or a, 6 E A which implies g2(a, x) g2(b, x) = gl(a, x) g,(b, x) = 0. 'NOW, let h E %(A2, B) be another extension of g, to AZ. i;or k + 1 c i s k + n we have Mai, x1 = bSAn\{&1 @b, x1.s ,II, 5
@4 x) = Fa &Cl+ x) = 0 beA,
1
by Rl. It follows that .
h(u, x) =
(a, xl E Al x B, (a, x) E (&\AJ
therefore
x B,
h = g2.
Because of the fact that a function g E %(Al, B) has only one extension tc AZ;, we can say without any confusion: ‘the extension of g to A2’. 2. If A, c AZ, then GBF,,[A,]c
GBF,[A2].
f. Let f~ GBF,[A1 J-There is g E %((.A,, B) such that for any (x1, . . . , x,,) E B”,
On generalizedBoolean functionsII
If gl is the extension
281
of g to AZ, then
gen; fh
- *- 4l) d% Xl) - * - d4v 4 =fh
=(a ,,.,.
7
* . . , x,),
therefore f E GBFJAJ. Let us prove that the inclusion is proper. We define the function follows:
jr : n + jj as
x E B\Al, XEAl.
We denote
by q the function
defined by:
k+n dx)=
ak+l
u r=k+l
g(%
d9
where g is the function deked in Lemma 2 for a = ak+i, C = AZ and A = A,. If x E Al, then g(u,., x) = 0 for r~(k + 1, . . . , k + n}, therefore q(x) = 0. If x E B\A1, then g(a k+l,x)‘I and g(a,,x)=O fur rE{k+? ‘, . . . , k + n), therefore It follows that for any x E B we have q(x) = Qk+l. k+n h(x)
=
q(J) =
u r=k+l
k+n uk+l
f&h
x)
= rv,
h(&)
d%
x),
therefore h E GBF1[A2]. But h$ GBF,[AJ, else we should have h = 0. By introducing the fictitious arguments x2, . . . , x,., in h we obtain a function which is in GBF,,[A2]\GBF,[A1] (see Lemma 3). C~r~lIary.
For any X and Y such that (0, 1) E Xc B and (0,l) c Y c B, we have: GBF,,[Xn
Y]EGBF,,[X]~GBF,[Y],
GBF,,[Xj U GBFJY]
E GBFJX
(4) u Y].
(5)
The proof is immediate by Theorem 2 since X17 Y c_X E X U Y anld X 17 Y c YEXUY. In the following two theorems we show that (4) and (5) always hold with proper inclusion except in the case when X E Y or YE X. remi GBFJAJ,
then GBF,,[A1 nA,]=G’BF,[A,]!n 3. If A,c_A, 01 A,G Al, eke we have GBF,[AInAJcGBF,,[A1]nGBF,[A,].
N. ~iindtireanu
282
Proof. If AicAj therefore
(i, jE{?,2),
GBF,[Ai]
if j), then GBF,[A~]E
n GBF,[Aj]
by Theorem
2,
GBFm[Ai I3 Aj].
= GBF,[Ai]=
Suppose that A1\A2 # fl and A2\A1 # 8. As G13F,[A1 tl A,]E GBF,,[AJ (i = 1,2), it remains to prove that the inclusion is proper. Take y1 E AI\AZ and y2 E Az\A,. We define the func+,ions gi : Ai X B + i3 (i:=1,2) as follows: g&4 .w)=
1
if [U = x E Ai] or [a = yi and x E B\Ai],
0
otherwise.
By R2 it follows that gi E %(Ai, B). We now define the function relation:
For i = 1,2 we denote
f : B” -+ B by the
by b the foiiowirpg function:
Let us prove that f = h, = hz. For (x1, . . . , x,,) E B” we have the following cases: (a) If x1,. . . . xn E A1 Cl AZ, then .,&)=h2(%,. ..J?t,). Ml, . . .A)=f(%.. (b) If there is some 4 E A1\A2 (1 S i G n), then we denote element:
zj =
xi,
1
by Zj the following
i=L
y2, i=2.
We have
X tdl,...,
n
i-l.i+l,...,
n)
gj(%
We remark now tha; Zj$ At n AZ, therefore &L
* - - 7 &J =
u al,...&-I.
Q,+x~..-r(In+ = I = Ax,, (cJ
if
there is some q E A2\Al,
the prec%oles relation
tE(l,___, iQi+l,....
. . .
&)a
n) “”
“)
, 4,).
then interchange
the roles of 1 and 2 in (a).
(d) If x1,. . . , x,,E B\(A, U A,) then bzii(x1,. - * 7 hi = f(yi, - * * 9 Vi)
yields:
= I=
f(Xl,
* * * 9 X, ).
-_,
,,,I ,..,,.
II
On generalized Boolean functions I1
283
It follows that f~ GBF,[A*] n GBF,,[A2]. Obviously f$ GBF,,[A1 n AJ, else f should be identically zero.
Theorem 4. If Al c A2 or A, E Al, then GBF,,[A,]IJ
GBF,[A2] = GBF,[Ar
GBF,[A,]U
GBF,,[A,]c
U AZ];
else GBF,,[Ar 1; A*].
E%of. We have GBF,CAr]U GBF,,[Az]~ GBF,[A, U AS] and it remains to prove that if A1\A2 # $3P A2\A1, then the inclusion is proper. We consider a eAl\A2 and b EA~\A~. Let g :(A, UA,)xB + B be the function defined by: !dYi x) =
1
if [y=:rEAIUAz]
0
otherwise.
or [y=O and XEB\(A~UA,)],
By R2 we have g E %(A, U A*, B). Let us consider the function f : B” + B defined as follows: x1 = a, x1= b, ~1E B\ju, b); x2,
- * * , x,,
being arbitrary elements
in B. We have:
r
U =
U
~~,....cz,,EA,UA~
a 0, ~1)de.
a gb, x1)u ii 0, x1)= fh,
~2)
- - - gbn,
G,)
1
. . . , al,
from the definition of g and by the fact that from OE Al nA2 it follows that u # 0 # b and therefore g(u, x) = 0 = g(b, x) for any x# {a, b}. It follows that f~ GBF,[Ar U A,]. If f~ GBF,[A,], then by Lemma 1 we must have ii =
f(h x2,
. . . ,d
scl
,... lJA,
fh,
. . . , c,,)
=
a
It follows that 1 = a U ii = a which is not true since cl$ ,A1 tl A;. If f E GBF,[A,] follows that a = 0 which is again false. Therefore f$GBF,[A,]UGBF,[A,].
. _. .._ I ..j..,...,,._.^,..,..,,.,
”
..I
.,,,.
it
284
N. ‘J%nd&anu
Tk author wishes to thank Professor S. Rudeanu for his valuable remarks which improved the manuscript.
/l] C.P. Popovici. L. Livovschi and N. Tbdgreanu, Switching circuit:;, autom?ia and formal languages, Univ. of Bucharest (1980) (in Roumanian). /2] S. Rudeanu, Boolean ‘Functions and Equations (North-Holland, Anlaterdam, 1974). [3] N. TsndHreanu, On generalized Boolean functions I, Discrete Math. 34 (l!XO). 293-299.