boolean sets and most general solutions of boolean equations
TRANSCRIPT
Information Sciences 180 (2010) 2440–2447
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Information Sciences
journal homepage: www.elsevier .com/locate / ins
Boolean sets and most general solutions of Boolean equations
Sergiu RudeanuUniversity of Bucharest, Mathematics and Computer Science, Str. Academiei 14, 010014 Bucharest, Romania
a r t i c l e i n f o
Article history:Received 12 August 2009Received in revised form 27 January 2010Accepted 29 January 2010
Keywords:Boolean equationSubsumptive general solutionParametric general solutionBoolean setBoolean transformation
0020-0255/$ - see front matter � 2010 Elsevier Incdoi:10.1016/j.ins.2010.01.029
E-mail address: [email protected]
a b s t r a c t
The aim of this paper is twofold. First we determine the most general form of the subsump-tive general solution of a Boolean equation (Theorems 1 and 2). Then we discuss severalcharacterizations of Boolean sets, meaning sets of zeros of Boolean functions, and provethat every Boolean transformation X ¼ UðTÞ is the parametric general solution of a certainBoolean equation.
� 2010 Elsevier Inc. All rights reserved.
1. Introduction
The study of Boolean equations in arbitrary Boolean algebras began with Boole, Schröder, Whitehead and Poretski. A Bool-ean function f : Bn ! B over a Boolean algebra ðB;_; �; 0;0;1Þ is obtained from variables and constants by superpositions of thebasic operations _; �; 0. We say that f is evanescible if the Boolean equation f ðXÞ ¼ 0 is consistent, i.e., it has solutions X 2 Bn; thishappens if and only if
QA2f0;1gn f ðAÞ ¼ 0. There are two kinds of compact representations of the set of solutions to a consistent
Boolean equation f ðXÞ ¼ 0.
1) Let u1; . . . ;un : Bn ! B be Boolean functions and set U ¼ ðu1; . . . ;unÞ. We say that formulas
xi ¼ uiðt1; . . . ; tnÞ ði ¼ 1; . . . ;nÞ; ð1Þ
or in vector form X ¼ UðTÞ, determine a parametric general solution of equation f ðXÞ ¼ 0 if
f ðXÞ ¼ 0() 9T 2 Bn X ¼ UðTÞ: ð2Þ
2) Let
uj; v j : Bn�j ! B ðj ¼ 1; . . . ;nÞ ð3Þ
be Boolean functions. We say that the system of recurrent inequalities
ujðxjþ1; . . . ; xnÞ 6 xj 6 v jðxjþ1; . . . ; xnÞ ðj ¼ 1; . . . ;nÞ; ð4Þ
. All rights reserved.
S. Rudeanu / Information Sciences 180 (2010) 2440–2447 2441
where un;vn 2 B, determines a subsumptive general solution of equation f ðXÞ ¼ 0 provided the recurrent inequalities (4) be-have in the same way as in the conventional procedure of successive elimination of variables; the exact definition will begiven in the next section.
While parametric general solutions began with Boole himself, McColl [16] and Peirce [17] showed how to construct sub-sumptive general solutions.
Many parametric general solutions have been constructed in the literature; see the monographs [19,23]. Moreover, Ru-deanu [19, Chapter 3, Section 2], Deschamps [11] (see also [19, Theorem 3.4]) and Bankovic [2,3] obtained general forms ofthe parametric general solutions, meaning necessary and sufficient conditions upon the vector function U in order thatX ¼ UðTÞ be a parametric general solution of the equation f ðXÞ ¼ 0. Analogous results for Post equations were obtainedby Bankovic [5,6] (see also [23, Chapter 5, Section 3]) and Rudeanu [23, Theorem 5.3.5]. A similar theorem within a moregeneral framework is due to Bankovic [7]. See also [4,8,9,14,15,1].
The paper [10] by Brown and Rudeanu characterizes subsumptive general solutions in terms of another concept calledrecurrent cover. However it seems natural to look for a direct characterization of subsumptive general solutions as well,meaning necessary and sufficient conditions upon the functions uj;v j in order that inequalities (4) determine a subsumptivegeneral solution of a Boolean equation. This will be done in Theorems 1 and 2 of the present paper.
A related problem is suggested in a paper by Reischer, Simovici, Stojmenovic and Tosic [18]. Motivated by connections tothe study of bio-circuits and of circuits based on frequency multiplexing, these authors introduce the set Cf of zeros of aBoolean function f and determine the internal structure of Cf in terms of ring sum, complementation and the differencex� y ¼ xy0. From this point of view Theorem 2 is a new construction of Cf , while Theorem 1 describes the union of the setsCf when f runs over the set of evanescible Boolean functions of n variables. In the present paper we refer to the sets Cf asBoolean sets (instead of the term Boolean collection used in [18]). In the last section of this paper we discuss several charac-terizations of Boolean sets. In particular we prove that every Boolean transformation X ¼ UðTÞ is the parametric general solu-tion of a certain consistent Boolean equation (Theorem 3).
2. Most general subsumptive general solutions
Recall first the method of successive elimination of variables, based on the fact that a Boolean equation in one unknownax _ bx0 ¼ 0 is consistent if and only if ab ¼ 0, in which case the solutions are given by b 6 x 6 a0. The eliminants of a Booleanfunction f : Bn ! B are defined recursively by
f1ðx1; . . . ; xnÞ ¼ f ðx1; . . . ; xnÞ; ð5:1Þ
fkðxk; . . . ; xnÞ ¼ fk�1ð1; xk; . . . ; xnÞfk�1ð0; xk; . . . ; xnÞ ðk ¼ 2; . . . ;nÞ; ð5:kÞ
fnþ1 ¼ fnð1Þfnð0Þ: ð5:nþ 1Þ
After the eliminants have been determined in this order, if fnþ1 ¼ 0 then one solves in reversed order the equationsfn ¼ 0; fn�1 ¼ 0; . . . ; f1 ¼ 0; if fnþ1–0 then none of these equations has solutions. In the former case, if each equation is solvedin the above form b 6 x 6 a0, then one obtains the set of solutions in the form of recurrent inequalities
fnð0Þ 6 xn 6 f 0nð1Þ; ð6:nÞ
fjð0; xjþ1; . . . ; xnÞ 6 xj 6 f 0j ð1; xjþ1; . . . ; xnÞ ðj ¼ n� 1; . . . ;1Þ: ð6:jÞ
The properties of the representation (6) of the solutions of a consistent Boolean equation f ðXÞ ¼ 0 have been taken in [10]as the definition of the concept of subsumptive general solution. First, an auxiliary definition: if k 2 f1; . . . ;ng andðbk; . . . ; bnÞ 2 Bn�kþ1, then f is said to be evanescible with respect to ðbk; . . . ; bnÞ if the equation f ðx1; . . . ; xk�1; bk; . . . ; bnÞ ¼ 0has solutions ðx1; . . . ; xk�1Þ 2 Bk�1. For k ¼ 1 this reduces to f ðb1; . . . ; bnÞ ¼ 0. Now let (3) be Boolean functions. The systemof recurrent inequalities (4) is said to determine a subsumptive general solution to the equation f ðXÞ ¼ 0 if the following con-ditions are equivalent for every k 2 f1; . . . ;ng and every ðxk; . . . ; xnÞ 2 Bn�kþ1:
(i) f is evanescible with respect to ðxk; . . . ; xnÞ;(ii) relations (6.j) hold for j ¼ k; . . . ;n.
The conventional subsumptive general solution is given by ukðxkþ1; . . . ; xnÞ ¼ fkð0; xkþ1; . . . ; xnÞ andvkðxkþ1; . . . ; xnÞ ¼ f 0kð1; xkþ1; . . . ; xnÞ.
A family of Boolean functions gk : Bn�kþ1 ! B ðk ¼ 1; . . . ;nþ 1Þ (hence gnþ1 2 B) is called a recurrent cover of a Booleanfunction f : Bn ! B provided that
_nþ1j¼k
gj ¼ fk ðk ¼ 1; . . . ;nþ 1Þ: ð7:kÞ
We are going to use the following result ([10,23, Proposition 6.4.4]). Let f : Bn ! B and (3) be Boolean functions. Then f isevanescible and the system (4) determines a subsumptive general solution of the equation f ðXÞ ¼ 0 if and only if the se-quence of functions defined by
2442 S. Rudeanu / Information Sciences 180 (2010) 2440–2447
gkðxk; . . . ; xnÞ ¼ ukðxkþ1; . . . ; xnÞx0k _ v 0kðxkþ1; . . . ; xnÞxk ðk ¼ 1; . . . ;nÞ; ð8:kÞ
gnþ1 ¼ 0; ð8:nþ 1Þ
is a recurrent cover of f.Now we prove the following
Theorem 1. There is an evanescible Boolean function f such that the sequence (4) of Boolean inequalities is a subsumptive generalsolution of the equation f ðXÞ ¼ 0 if and only if the following identities hold:
ukv 0k 6_n
j¼kþ1
ðujx0j _ v 0jxjÞ ðk ¼ 1; . . . ;n� 1Þ; ð9:kÞ
unv 0n ¼ 0: ð9:nÞ
Proof. In view of the Proposition mentioned above, we will prove that there is a function f such that the sequence (8) is arecurrent cover of f if and only if relations (9) hold. In this proof the functions gk are defined by (8) and we take f ¼
Wnj¼1gj;
note that when we assume that (8) is a recurrent cover of a function f, the above choice is obligatory by (7.1).The sequence (8) is a recurrent cover of f if and only if
_nj¼k
gj ¼ fk ðk ¼ 1; . . . ;nÞ ð10:kÞ
and 0 ¼ fnþ1. The latter equality says that fn is evanescible, while gn ¼ fn by (10.n). Therefore the recurrent cover conditionbecomes: (10) holds and gn is evanescible, that is, relations (10) and (9.n) hold.
So it remains to prove that
_nj¼k
gj ¼ fk ðk ¼ 1; . . . ;nÞ () uhv 0h 6_n
j¼hþ1
gj ðh ¼ 1; . . . ;n� 1Þ:
In order to establish this equivalence we prove the following auxiliary result.Take h 2 f1; . . . ;ng and suppose
Wnj¼hgj ¼ fh. Then
_nj¼hþ1
gj ¼ fhþ1 () uhv 0h 6_n
j¼hþ1
gj:
Indeed,
fhþ1ðxhþ1; . . . ; xnÞ ¼ fhð0; xhþ1; . . . ; xnÞfhð1; xhþ1; . . . ; xnÞ ¼ ghð0; xhþ1; . . . ; xnÞ __n
j¼hþ1
gj
!ghð1; xhþ1; . . . ; xnÞ _
_nj¼hþ1
gj
!
¼ uh __n
j¼hþ1
gj
!v 0h _
_nj¼hþ1
gj
!¼ uhv 0h _
_nj¼hþ1
gj;
therefore
_nj¼hþ1
gj ¼ fhþ1 ()_n
j¼hþ1
gj ¼ uhv 0h __n
j¼hþ1
gj () uhv 0h 6_n
j¼hþ1
gj:
SinceWn
j¼1gj ¼ f , a repeated application of this lemma yields the desired euivalence (10) () (9). h
Corollary 1. If the conditions of the above theorem are fulfilled, then the function f is uniquely determined by
f ¼ f1 ¼_nj¼1
gj ¼_nj¼1
ðujx0j _ v 0jxjÞ:
Proof. By the definition of a recurrent cover, taking into account that gnþ1 ¼ 0. h
Theorem 2. The functions (3) determine a subsumptive general solution of a consistent Boolean equation f ðXÞ ¼ 0 if and only ifthey satisfy conditions (9) and
_nj¼1
ðujx0j _ v 0jxjÞ ¼ f : ð11Þ
S. Rudeanu / Information Sciences 180 (2010) 2440–2447 2443
Proof. By Theorem 1 and Corollary 1. h
Example 1. The eliminants f1; . . . ; fnþ1 form the recurrent cover corresponding to the conventional solution by recurrentinequalities. The corresponding functions uk;vk are given by
ukðxkþ1; . . . ; xnÞ ¼ fkð0; xkþ1; . . . ; xnÞ; vkðxkþ1; . . . ; xnÞ ¼ f 0kð1; xkþ1; . . . ; xnÞ:
In this case relations (9) are satisfied as equalities:
ukðxkþ1; . . . ; xnÞv 0kðxkþ1; . . . ; xnÞ ¼ fkð0; xkþ1; . . . ; xnÞfkð1; xkþ1; . . . ; xnÞ¼ fkþ1ðxkþ1; . . . ; xnÞ;_n
j¼kþ1
ðujðxjþ1; . . . ; xnÞx0j _ v 0jðxjþ1; . . . ; xnÞxjÞ
¼_n
j¼kþ1
ðfjð0; xjþ1; . . . ; xnÞx0j _ fjð1; xjþ1; . . . ; xnÞxjÞ
¼_n
j¼kþ1
fjðxj; . . . ; xnÞ ¼ fkþ1ðxkþ1; . . . ; xnÞ:
The evanescible Boolean functions f are characterized by fnþ1 ¼ 0.
3. More on parametric general solutions
In this section we relate parametric general solutions to Boolean sets and subsumptive general solutions. The main resultsolves the inverse problem of interpreting any Boolean transformation X ¼ UðTÞ as the parametric general solution of a suit-ably chosen Boolean equation (Theorem 3).
We begin with the trivial remark that giving a parametric general solution X ¼ UðTÞ of a Boolean equation f ðXÞ ¼ 0 is thesame as providing the set Cf ¼ fUðTÞjT 2 Bng.
Further we recall Löwenheim’s parametric general solution of a Boolean equation f ðXÞ ¼ 0. Löwenheim proved [12,13] (seealso [19], Theorem 2.11) that if N 2 Bn is a particular solution, i.e., f ðNÞ ¼ 0, then a parametric general solution U is providedby
uiðTÞ ¼ nif ðTÞ _ tif 0ðTÞ ðT 2 BnÞ: ð12Þ
A particular solution N can be obtained by a variant of successive elimination of variables, which we illustrate below inthe case n ¼ 2.
Example 2. Set ðx1; x2Þ ¼ ðx; yÞ. Suppose the Boolean equation
a3xy _ a2xy0 _ a1x0y _ a0x0y0 ¼ 0 ð13Þ
is consistent, that is, a3a2a1a0 ¼ 0. Write (13) in the form
ða3y _ a2y0Þx _ ða1y _ a0y0Þx0 ¼ 0 ð13:1Þ
and regard it as an equation in x. The consistency condition is ða3y _ a2y0Þða1y _ a0y0Þ ¼ 0, that is,
a3a1y _ a2a0y0 ¼ 0: ð13:2Þ
The common consistency condition of Eqs. (13.2), (13.1) and (13) is
a3a2a1a0 ¼ 0: ð13:3Þ
Choose e.g. the following particular solution of (13.2):
n2 ¼ a2a0: ð14:2Þ
Taking n2 for y in (13.1) we obtain in turn
ða3a2a0 _ a2a00Þx _ ða1a2a0 _ a0a02Þx0 ¼ 0;a2ða3 _ a00Þx _ a0ða1 _ a02Þx0 ¼ 0:
Choose e.g. the following particular solution of the latter equation:
n1 ¼ a0ða1 _ a02Þ: ð14:1Þ
We have thus obtained the particular solution (14) to Eq. (13).Setting ðt1; t2Þ ¼ ðs; tÞ, the parametric solution (12) of the Boolean Eq. (13) with the particular solution (14) is given by
2444 S. Rudeanu / Information Sciences 180 (2010) 2440–2447
u1 ¼ a0ða1 _ a02Þða3st _ a2st0 _ a1s0t _ s0t0Þ _ sða03st _ a02st0Þ¼ a0ða1 _ a02Þða3st _ a2t0 _ a1s0 _ s0t0Þ _ sða03t _ a02t0Þ; ð15:1Þ
u2 ¼ a2a0ða3st _ st0 _ a1s0t _ s0t0Þ _ tða03st _ a01s0tÞ¼ a2a0ða3s _ a1s0 _ t0Þ _ tða03s _ a01s0Þ: ð15:2Þ
If we wish to determine all Boolean sets Cf , we could find all evanescible Boolean functions f by solving equationQ2n
j¼1aj ¼ 0 with respect to the unknowns a1; . . . ; a2n . Although the latter equation is easy to solve, this procedure does notseem practical; applying Theorem 1 appears to be a better solution.
On the other hand, we may wish to transform a parametric general solution into a subsumptive general solution.
Example 3. Consider again a consistent Boolean equation in two unknowns
a3x1x2 _ a2x1x02 _ a1x01x2 _ a0x01x02 ¼ 0: ð13Þ
Löwenheim’s general parametric solution (12) becomes
u1ðTÞ ¼ n1f ðTÞ _ t1f 0ðTÞ ¼ n1f ðTÞ _ t1ða03t2 _ a02t02Þ; ð16:1Þu2ðTÞ ¼ n2f ðTÞ _ t2f 0ðTÞ ¼ n2f ðTÞ _ t2ða03t1 _ a01t01Þ: ð16:2Þ
Let us transform it into a subsumptive general solution.We are going to determine functions u1;v1 : B! B and u2;v2 2 B such that
u1ðu2Þ 6 u1 6 v1ðu2Þ; ð17:1Þu2 6 u2 6 v2; ð17:2Þ
be a subsumptive general solution (4), for which we will check directly that conditions (9) are satisfied.Using Schröder’s theorem on the range of a Boolean function (see e.g. [19, Theorem 2.4]), it is clear that
u2 ¼min u2 ¼Y
a;b2f0;1guða;bÞ ¼ n2a0ðn2a1 _ a01Þn2a2ðn2a3 _ a03Þ ¼ a0a2n2;
but since ða2n1 _ a0n01Þn
02 ¼ 0, it follows that
a2a0 ¼minða2x _ a0x0Þ 6 a2n1 _ a0n01 6 n2;
therefore u2 and similarly v2 are uniquely determined by
u2 ¼ a2a0; v2 ¼ a03 _ a01: ð18:1Þ
This provides a direct checking of (9.2), which is (9.n): u2v 02 ¼ a2a0a3a1 ¼ 0.Now we will obtain u1 and v1 by solving the functional Eq. (17.1), which can be done by reducing them to ordinary
Boolean equations.Setting u1ðxÞ ¼ Ax _ Cx0, the first inequality (17.1) becomes
Aðn2f ðTÞ _ t2ða03t1 _ a01t01ÞÞ _ Cðn02f ðTÞ _ t02ða02t1 _ a00t01ÞÞ 6 n1f ðTÞ _ t1ða03t2 _ a02t02Þ:
According to the well-known Verification Theorem [13] (see also [19, Theorem 2.13]), the above identity holds if and onlyif it is satisfied for the 0–1 values of the variables:
An2a0 _ Cðn02a0 _ a00Þ 6 n1a0;
Aðn2a1 _ a01Þ _ Cn02a1 6 n1a1;
An2a2 _ Cðn02a2 _ a02Þ 6 n1a2 _ a02 ¼ n1 _ a02;
Aðn2a3 _ a03Þ _ Cn02a3 6 n1a3 _ a03 ¼ n1 _ a03;
which is equivalent to
½An2a0 _ Cðn02 _ a00Þ�ðn01 _ a00Þ ¼ 0;
½Aðn2 _ a01Þ _ Cn02a1�ðn01 _ a01Þ ¼ 0;½An2a2 _ Cðn02 _ a02Þ�n
01a2 ¼ 0;
½Aðn2 _ a03Þ _ Cn02a3�n01a3 ¼ 0;
then to
An2n01a0 _ Cðn01n
02 _ a00Þ ¼ 0;
Aðn2n01 _ a01Þ _ Cn02n
01a1 ¼ 0;
An2n01a2 _ Cn02n
01a2 ¼ 0;
An2n01a3 _ Cn02n
01a3 ¼ 0;
S. Rudeanu / Information Sciences 180 (2010) 2440–2447 2445
which reduces to the single equation
Aðn2n01 _ a01Þ _ Cðn01n
02 _ a00Þ ¼ 0;
whose solutions are A 6 a1ðn1 _ n02Þ and C 6 a0ðn1 _ n2Þ, so that the solutions of the functional equation in u1 are
u1ðx2Þ 6 a1ðn1 _ n02Þx2 _ a0ðn1 _ n2Þx02: ð18:2:1Þ
Setting v1ðxÞ ¼ Dx _ Ex0, the second inequality (17.1) becomes
n1f ðTÞ _ t1ða03t2 _ a02t02Þ 6 Dðn2f ðTÞ _ t2ða03t1 _ a01t01ÞÞ _ Eðn02f ðTÞ _ t02ða02t1 _ a00t01ÞÞ:
This identical inequality is equivalent to the system
n1a0 6 Dn2a0 _ Eðn02 _ a00Þ;n1a1 6 Dðn2 _ a01Þ _ En02a1;
n1 _ a02 6 Dn2a2 _ Eðn02 _ a02Þ;n1 _ a03 6 Dðn2 _ a03Þ _ En02a3;
which is equivalent to
n1a0½D0n2a0 _ E0ðn02 _ a00Þ� ¼ 0;n1a1½D0ðn2 _ a01Þ _ E0n02a1� ¼ 0;ðn1 _ a02Þ½D
0n2a2 _ E0ðn02 _ a02Þ� ¼ 0;ðn1 _ a03Þ½D
0ðn2 _ a03Þ _ E0n02a3� ¼ 0;
then to
n1a0ðD0n2 _ E0n02Þ ¼ 0;n1a1ðD0n2 _ E0n02Þ ¼ 0;D0n2a2n1 _ E0ðn1n
02 _ a02Þ ¼ 0;
D0ðn1n2 _ a03Þ _ E0n02a3n1 ¼ 0;
which reduces to the single equation
D0ðn1n2 _ a03Þ _ E0ðn1n02 _ a02Þ ¼ 0:
The latter equation characterizes the solutions of the functional equation in v1 by their complements:
v 01ðx2Þ 6 a3ðn01 _ n02Þx2 _ a2ðn01 _ n2Þx02: ð18:2:2Þ
Relations (18.2) enable us to check condition (9.1):
u1v 01 6 a1a3n02x2 _ a2a0n2x02 6 a1a3x2 _ a2a0x02 ¼ v 02x2 _ u2x02:
We have thus proved that if the functions uk;vk ðk ¼ 1;2Þ satisfy conditions (18), then the inequalities (4.1), (4.2) deter-mine a subsumptive general solution of (6).
Another aspect is pointed out if we state the problem of finding a parametric general solution to a Boolean equation in thefollowing sketchy way: given a Boolean function f, find (1) such that (2) holds. This leads us to the following inverse problem:given (1), find a Boolean function f such that (2) holds. Here is a precise statement of the latter problem.
Define a Boolean transformation as a function U ¼ ðu1; . . . ; fnÞ : Bn ! B in the following way:
UðTÞ ¼ ðu1ðTÞ; . . . ;unðTÞÞ 8T 2 Bn: ð19Þ
The inverse problem is whether for any Boolean transformation U there is a consistent Boolean equation f ðXÞ ¼ 0 suchthat formulas (1) determine the general reproductive solution of that equation. The answer is affirmative:
Theorem 3. For any Boolean transformation U, formula X ¼ UðTÞ determines the parametric general solution of equation
YA2f0;1gn
_ni¼1
ðxi þuiðAÞÞ ¼ 0; ð20Þ
where þ denotes the ring sum xþ y ¼ xy0 _ x0y.
Proof. Set gðX; TÞ ¼Wn
i¼1ðxi þuiðTÞÞ. Then
X ¼ UðTÞ () xi þuiðTÞ ¼ 0 ði ¼ 1; . . . ; nÞ () gðX; TÞ ¼ 0;
therefore 9TX ¼ UðTÞ ()Q
AgðX;AÞ ¼ 0 and the latter equality is (20). The consistency of Eq. (20) follows from
2446 S. Rudeanu / Information Sciences 180 (2010) 2440–2447
9XY
gðX;AÞ ¼ 0() 9X 9T X ¼ UðTÞ;
where the last equality is true. h
Example 4. For n ¼ 2, the transformation (1) can be written in the form
x1 ¼ at1t2 _ bt1t02 _ ct01t2 _ dt01t02;
x2 ¼ mt1t2 _ nt1t02 _ pt01t2 _ qt01t02;
while Eq. (20) becomes
½ðx1 þ aÞ _ ðx2 þmÞ�½ðx1 þ bÞ _ ðx2 þ nÞ�½ðx1 þ cÞ _ ðx2 þ pÞ�½ðx1 þ dÞ _ ðx2 þ qÞ� ¼ 0:
We refer the reader to the papers [20–22,24], devoted to the study of Boolean transformations. Among other results, sev-eral sets associated with a Boolean transformation were characterized by Boolean equations; so in the present terminologythey are Boolean sets. In particular Corollary 2 of [20] characterizes the range of a Boolean transformation by the equation
_C2f0;1gn
_A2f0;1gn
ðUðAÞÞC0@
1AXC ¼ 1; ð21Þ
where the following notation has been used: x1 ¼ x; x0 ¼ x0;X ¼ ðx1; . . . ; xnÞ 2 Bn;C ¼ ðc1; . . . ; cnÞ 2 f0;1gn;XC ¼ xc1
1 ; . . . ; xcnn
and similarly for ð/ðAÞÞC .Eq. (21) written in dual form is
_C
_A
ðUðAÞÞC" #0
XC ¼ 0; ð210Þ
so that the affirmative answer to the above ‘‘inverse problem” had already been given in Corollary 2 from [20]. However wehave included Theorem 3 for the sake of self-containedness and because the results seem different at first glance. To see theexact relationship between Theorem 3 and its predecessor Corollary 2, we recall one of the theorems known under the com-mon name of Löwenheim’s Verification Theorem (see e.g. [19, Corollary of Theorem 2.14]): two consistent Boolean equationshðXÞ ¼ 0 and kðXÞ ¼ 0 have the same set of solutions if and only if hðCÞ ¼ kðCÞ 8C 2 f0;1gn. By applying this theorem it fol-lows that Eqs. (20) and (21’)) are identical:
YA
_ni¼1
ðci þuiðAÞÞ ¼_
A
ðUðAÞÞC" #0
8C 2 f0;1gn: ð22Þ
Property (22) can also be checked directly:
_A
ðUðAÞÞC" #0
¼Y
A
½ðu1ðAÞÞc1 . . . ðunðAÞÞ
cn �0 ¼Y
A
_ni¼1
uiðAÞÞci
� �0
and ½ðuðAÞc�0 ¼ ðuðAÞ þ c0Þ0 ¼ uðAÞ þ c.4. Conclusions
In this paper we have dealt with four related concepts in an arbitrary Boolean algebra B: parametric general solution of aBoolean equation, subsumptive general solution of a Boolean equation, Boolean transformation and Boolean set. The para-metric general solution goes back to Boole. A subsumptive general solution, which is also an old concept, means the set ofsolutions expressed by recurrent inequalities. Boolean transformations, studied in previous papers of the author, are func-tions from Bn to Bm expressed by m Boolean functions from Bn to B. The concept of a Boolean set, introduced in [18], meansa subset of Bn characterized by a Boolean equation.
Whereas there are several papers dealing with various possibilities of constructing a parametric general solution, the sim-ilar idea of most general subsumptive general solution was introduced much later and characterized in terms of another con-cept, called recurrent cover [10]. The contribution of the present paper in this respect is Theorem 2, which provides anintrinsic characterization of the most general subsumptive general solution.
In connection with the introdution of the set Cf of zeros of a given Boolean function f, the idea in this paper has been toinvestigate the union
SCf where f runs over the set of all evanescible Boolean functions f. More exactly, we have solved the
following problems: (A) determine the systems of recurrent inequalities that are subsumptive general solutions of certainBoolean equations, and (B) determine the Boolean transformations that are parametric general solutions of certain Booleanequations (so the point is that the Boolean equation is not given a priori). The answer to problem A is Theorem 1. The easyanswer to problem B is Theorem 3: every Boolean transformation U : Bn ! Bn is the parametric general solution of a certain
S. Rudeanu / Information Sciences 180 (2010) 2440–2447 2447
Boolean equation. As a matter of fact, it was proved in [20] that the range of every Boolean transfomation U : Bn ! Bm is whatwe call now a Boolean set.
4.1. Open problem
Extend this research to Post algebras.
References
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