equational bases of boolean algebras

11
Equational Bases of Boolean Algebras Author(s): F. M. Sioson Source: The Journal of Symbolic Logic, Vol. 29, No. 3 (Sep., 1964), pp. 115-124 Published by: Association for Symbolic Logic Stable URL: http://www.jstor.org/stable/2271618 . Accessed: 14/06/2014 09:33 Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at . http://www.jstor.org/page/info/about/policies/terms.jsp . JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR, please contact [email protected]. . Association for Symbolic Logic is collaborating with JSTOR to digitize, preserve and extend access to The Journal of Symbolic Logic. http://www.jstor.org This content downloaded from 62.122.73.86 on Sat, 14 Jun 2014 09:33:31 AM All use subject to JSTOR Terms and Conditions

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Page 1: Equational Bases of Boolean Algebras

Equational Bases of Boolean AlgebrasAuthor(s): F. M. SiosonSource: The Journal of Symbolic Logic, Vol. 29, No. 3 (Sep., 1964), pp. 115-124Published by: Association for Symbolic LogicStable URL: http://www.jstor.org/stable/2271618 .

Accessed: 14/06/2014 09:33

Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp

.JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact [email protected].

.

Association for Symbolic Logic is collaborating with JSTOR to digitize, preserve and extend access to TheJournal of Symbolic Logic.

http://www.jstor.org

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Page 2: Equational Bases of Boolean Algebras

THE JOURNAL OF SYMBOLIC LOGIC

Volume 29, Number 3, Sept. 1964

EQUATIONAL BASES OF BOOLEAN ALGEBRAS

F. M. SIOSON

1. Introduction. It is well-known that a Boolean algebra (B, +,., -) may be defined as an algebraic system with at least two elements such that (for all x, y, z e B):

B1. x+y-y+x 31. xy - yx

B2. x+yz (x+y)(x+z) B2. x(y+z) xy+xz

B3. x+y = x B3. X(Y+f) = X

B4. x+(y+9) = B4. x(3i34 ) Y= Y9

These axioms or equations are not independent, in the sense that some of them are logical consequences of the others. B. A. Bernstein [1] thought that the first three and their duals form an independent dual-symmetric definition of a Boolean algebra, but R. Montague and J. Tarski [3] proved later that B1 (or B1) follows from B2, B3, B1, B2, B3 (from B1, B2, B3, B2, B33). Thus it was shown that the system of equations

B1

B2 B2 I I

B3 B3 I

or its dual I forms a complete set of generators for all equations in a Boolean algebra. In fact, we shall show that it is an independent system of generating equations, i.e., an equational basis. We shall also prove that these two together with

31 B1 B1 B1

B2 B2 B2 B2 B2 B2

B3 B3 B3 B3 B3

B4 B4 B4 B4 B4 II III IV V

and their duals II, III, IV, V are the only possible equational bases out of the eight postulates given in the beginning.

Received June 22, 1964.

115

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Page 3: Equational Bases of Boolean Algebras

116 F. M. SIOSON

2. Independence of I.

N1. Independence-Model of B1 in I.

+ 0 1 a b c d .0 1 a b c d Y 0 0 1 a b c d 0 0 0 00 0 0 0 1

liii 11 l l l l O l a b c d 1 0

a a 1 a a a 0 a a 0 0 a a b bb 11 b b 1 b O b 0 b b O b a Cc 11 - c I c O c O c c 0 c d dd 1 d l I d d 0 d d 0 0 d d c

Note, a+d # d+a. B3 and f3 are both easily verified. We shall verify in detail only B2 and omit the analogous verification of 32. First, observe that (l)xx = x, x+x = x, x+O x = O+x, xO Ox = 0, xl = lx x, x+l l+x I; (2) xy O ill yx 0; (3) when x,y # 0, l,xy Oif/ x+y = 1. Inasmuch as O+yz = yz = (O +y)(O+z), l+yz - 1 = 11 I (1 +y)(l +z),andwhenxO, l,x+Oz x+O x =x('7)= (x+O) (x+z),x+lz - x+z 1 (x+z) = (x + l)(x+z), then it suffices to consider only the cases when x, y 0 0, 1. We next treat the cases when two of the elements x, y, z are equal. Thus, x+xz = x+(O) = x = x(') = (x+x)(x+z) and x+yy x+y = (x+y)(x+y). Moreover, if x+y = x, then x+yx = x+(Y) = x - xx (x+y)(x+x); and if x+y = 1 (so that yx = 0), then x+yx x+O x = lx = (x+y) (x+x). Hence, it remains to consider the cases when x, y, and z are distinct from one another. Now, observe that for any four distinct elements x, y, z, u : 0, 1, two and exactly two elements in each of the two sets xu, yu, zu and xy, xz, xu are zero. By virtue of this and earlier observations, then xy = xz 0 (so that xu : 0) implies yz : 0. For, suppose xy = xz = 0, xu : 0, and yz = 0. Then zx = zy 0 and hence zu : 0 contrary to the fact that xu = 0. Whence, if xy = xz = 0, yz 0 (so that x +y =x+z 1, yz y), then x+yz x+y 1I I I (x+y)(x+z). If xy # 0, xz =yz 0 (so that x+y x xy, x+z =y+z l),thenx+yz =x+O =x =x+y = (x+y)I = (x+y)(x+z).

12. Independence-Model of B2 in I.

-F 0 1 . 01 y 0 01 0 00 0 1

1 10 1 0 1 1 0

Note, 1 + 10 # (1 + 1)(I +0). Here and in what follows, verifications of the remaining pertinent equations will always be omitted unless they are nontrivial.

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Page 4: Equational Bases of Boolean Algebras

EQUATIONAL BASES OF BOOLEAN ALGEBRAS 117

13. Independence-Model of B2 in I.

+ 0 1 . 01 y I5 0 01 0 10 0 1

Observe, 0(1+0) =A 0 1+00.

14. Independence-Model of B3 in I.

+ 0 1 . 01 0 01 0 00

1 1 1 1~~~~ 0 1

Note, 0+11 #0.

I5. Independence-Model of B3 in I.

+ 0 1 .01 y5 0 01 0 00 0 0

1 1 1 ~~~1 0 1 1 0

Note, 1(0+0) = 1.

3. Completeness and Independence of II. To prove that II (and

hence II) forms an adequate formulation of a Boolean algebra, it is sufficient to derive B2 (that is to say, I) from it. This derivation is effected as follows:

THEOREM 3.1. x+x = y+y.

Proof. x+x (x+x(y+y)- (y+y(x+x) - y+y (B3, B1, B3). THEOREM 3.2. xx = x and x+x x. Proof. xx - xx+xx = x(x+x) x (B3, B2, B3). Since by B4, (y+y)+(y+y) (y+y), then x+x = x(y+y)+x(y+y)

x((y+y)+(y+y)) = x(y+y) = x (B3, B2, Remark, B3). THEOREM 3.3. (y+y)x 9 x (B1, B3). THEOREM 3.4. (x+y)z xz+yz (B1, B2). THEOREM 3.5. (y+y)+x y+y. Proof. (y+y)+x ((y+y)+x)(x+?) =((y+y)+x)x+((y+y)+x9

((y+y)x+xx)+((y+y)xi+xx)= (x+x)+x x+x- y+y (F33, B2, T3.4, T3.3-T3.2-B3, T3.2, T3.1).

THEOREM 3.6. yx+x = x and y+xy = y. Proof. yx+x = yx+(z+z)x = (y+(z+z))x = (z+z)x = x (T3.3, T3.4,

B4, T3.3). y+xy = (z+z)y+xy = ((z+z)+x)y = (z+z)y = y (T3.3, T3.4, T3.5,

T3.3).

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Page 5: Equational Bases of Boolean Algebras

118 F. M. SIOSON

THEOREM 3.7. x+y (y+x)(x+y). Proof. (y+x)(x+y)- (y+x)x+(y+x)y = (yx+xx)+(yy+xy)

(yx+x)+(y+xy) = x+y (B2, T3.4, T3.2, T3.6). THEOREM 3.8. x+y = y+x. Proof. x+y = (y+x)(x+y) = (x+y)(y+x) = y+x (T3.7, B1, T3.7). THEOREM 3.9. x= yy. Proof. xx xx+yy yy+xx = yy (B3, T3.8, B3). THEOREM 3.10. x+iy = x+y. Proof. X+?y = (x+yx)+(xy+xx) = (xx+yx)+(xx+xy) - (xx+yx)+

(xx+yx) = (x+y)x+(x+y)x = (x+y)(x+x) = x+y (T3.6-B3, T3.2-

T3.8) B1,,T3.4, B2, IT3)-

THEOREM 3.11. x(xy) zz. Proof. x(xy) = (xy)= (xy)x+xx = (xy+x)x = (yx+x)x = =z

(B3, B3, T3.4, B1, T3.6, T3.9). THEOREM 3.12. x+yz = (x+y)(x+z).

Proof. (x+y)(x+z) = (x+y)x+(x+y)z = (xx+yx)+(xz+yz) = (x+yx) +(xz+yz) x+(xz+yz) x+x(xz+yz) = x+(x(xz)+R(yz)) = x+ (zz+x(yz)) = x+(x(yz)+zz) x+x(yz) = x+yz (B2, T3.4, T3.2, T3.6, T3.10, B2, T3.11, T3.8, B3, T3.10).

II1. Independence-Model of B1 in II.

+ 0 1 a b . 1 a b y Y|

0 0 1 a b 0 0 0 0 0 0 1

1 1 1 1 1 1~~~~ 0 1 0 1 a a 1 a 1 a 0 a a 0 a b

b b Ib b 0 b Ob b a

Observe, 1 a =# al. The verification of B2 may be accomplished as follows: x(O+z) xz = O+xz = xO+xz and hence by commutativity of addition x(y+O) xy+xO. Similarly, x(1+z) = xl = x = x+( ) = xl+xz and hence by commutativity x(y+ 1) = xy+xl. Thus, it remains to consider the cases when y, z #& 0, 1. If, in addition, y = z, then x(y+z) = x(y+y)=

xy = xy+xy = xy+xz. Since O(y+z) 0 = 0+0 = Oy+Oz, the only remaining cases are those when x # 0, y # z, y, z # 0, 1. In this instance, y+z = 1, xy 7& xz and xy+xz = x. Hence x(y+z) = xl = x = xy+xz.

II2. Independence-Model of B2 in II.

? 0 1 .01 y ~ 0 01 0 10 0 1 1 1 1 1 0 1 0

Clearly, 0(0+ 1) #00+01.

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Page 6: Equational Bases of Boolean Algebras

EQUATIONAL BASES OF BOOLEAN ALGEBRAS 119

II3. Independence-Model of B3 in II.

+ 0 1 .01 y9 0 01 0 00 0 1 1 0 1 1 0 1 1 1

Note, 0+11 #0.

II4. Independence-Model of B3 in II.

+ 0 1 .01 y9 0 01 0 00 0 1

1 1 1 1~~~~ 0 0 1 0

Observe, I(y+9) = 1.

II5. Independence-Model of B4 in II.

+ 0 1 .01 y 0 01 0 00 0 1 1 1 0 1 0 1 1 0

In this case, l+(y+y) #4y+9.

4. Completeness and Independence of III. To prove the complete- ness of III, it is sufficient to derive B3 (and hence I) from it

THEOREM 4. 1. xX = y9 (B3, B1, B3). THEOREM 4.2. x(yq) =y. Proof. x(yq) = x(yq)+xx = x(y9+9) = x(x+yq) x= y9 (B3,

B2, B1, B3, T4.1). THEOREM 4.3. xx = x. Proof. x = x+y9 = x+(y9)(y9) = (x+y9)(x+y9) xx (B3, T4.2,

B2, B3). THEOREM 4.4. x+X = y+9 (B4, B1, B4). THEOREM 4.5. x(y+9) = x. Proof. x(y+9) = x(x+x) = xx+xx = xx = x (T4.4, B2, B3, T4.3).

III1. Independence-Model of B1 in III.

The independence-model of B1 in I can be used as an independence-model of B1 in III. Note that B2, B2, and B3 are also equations in III. The re- maining axiom B4 in III obviously holds in the model.

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Page 7: Equational Bases of Boolean Algebras

120 F. M. SIOSON

III2. Independence-Model of B2 in III.

+ 0 1 .0 1 0 01 0 0 0 1 1 1 10

Note, 1+11 #& (1+1)(1+1).

III3. Independence-Model of B2 in III.

+ 0 1 . 01 0 00 0 0 1 1 01 1 1 I0I

In this model notice that 1(1+0) # 11+10.

I1i4. Independence-Model of B3 in III.

+ 01 . 0 Y 000 0 00 101 101

In this case note that 1 +yy $ 1.

1115. Independence-Model of B4 in III.

? 0 1 .01 y5 0 00 0 01 0 1 10 1 1 1 1 1 1

Observe that 0+(1+ l) # 1 + 1.

5. Completeness and Independence of IV. To prove the complete- ness of IV, it is sufficient to derive B1 (and therefore III) from IV.

THEOREM 5.1. (x+y)z = xz+yz (B1, B2, B1). THEOREM 5.2. xx = x and x+x = x. Proof. First note that (a) x(x+x) - xx+x+X= xx (B2, B3), (b) (y+ )+

(y+f= y+? (B4). Thus, (x+x)(x+x) = x(x+x) +x(x+x) =x((x+)+ (x+X)) X (x+X) = xx (T5. 1, B32, (b), (a)). On the other hand, x = x+xx= (x+x)(x+x9) = xx (B3, B2, previous result). Similarly, x+x xx+xx x(x+x)+x(x+x) x((x+x)+(x+x)) = x(x+x) = xx = x (Previous result,

(a), B2, (b), (a), previous result). THEOREM 5.3. y(Y+Y) = y (Proof of T5.2.). THEOREM 5.4. xy+y = y = yx+y.

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Page 8: Equational Bases of Boolean Algebras

EQUATIONAL BASES OF BOOLEAN ALGEBRAS 121

Proof. y = yy+?) y(x+(y+y)) yx+yy+h) = yx+y = xy+y (T5.3, B4, B2, T5.3, B1).

THEOREM 5.5. xy = x(xy).

Proof. First observe, (x+z)(y+z) x(y+z)+z(y+z) = (xy+xz)+

(zy+zz) = (xy+xz) +(zy+z) = (xy+xz)+z = (xy+x)(xy+z)+z = x(xy+z)+z (T5.1, B2, T5.2, T5.4, B2, T5.4). Setting z = z2, one obtains xy - x(xy+z2)+zz x(xy) (B3, B3).

THEOREM 5.6. y+xy = y = y+yx. Proof. y = = y+x xx) = (y+x)(y+xx) = (y+x)y = yy+xy

y+xy = y+yx (B3, T5.5, B2, B3, T5.1, T5.2, B). THEOREM 5.7. (x+y)(y+x) = x+y. Proof. (x+y)(y+x) = x(y+x)+y y+x) (xy+xx)+(yy+yx)

(xy+x)+(y+yx) = x+y (T5.1, B2, T5.2, T5.4-T5.6). THEOREM 5.8. x+y = y+x (T5.7, B1, T5.7). Since ad # da in the independence-model of B1 in III and the independ-

ence-models of B2, B2, B3, B4 are all commutative with respect to the

.-operation, it is clear that the independence-models of III will prove the independence of IV.

6. Completeness and Independence of V. We shall derive the dual

II of II from V by proving B3. THEOREM 6.1. x+x = y+y (B4, B1, B4).

xx = yy (B3, B1, B3). THEOREM 6.2. (y+9)x = x. Proof. (y+g)x = (x+x)(x+yq) = x+x(yy) = x+y = x (T6.1-B3,

B2, B4, B3).

THEOREM 6.3. x(y+y) = x(yy).

Proof. x(y+9) = (x+y9)(y9+y9) = (yy+x) (y9+y) yy+x(yy) x (y9) + yy= x(yq) (B3-T6.1, B1, B2, 1, B3).

THEOREM 6.4. x(y+y) = x. Proof. Let x = yy in Theorem 6.3. so that (yq)(y+9) (yy)(yy) = zz.

Then x = x+zz = x+(y9)(y+9) = (x+y9)(x+(y+9)) x(y+y) (B3, Remark, B2, B3-B4).

The same independence-models of B1, B2, B3, and B4 in III may be used

for proving the independence of these equations in V. This is evident inasmuch as these four models all satisfy the equation B4 which is the only one in V not present in III. Thus what remains to be shown is the inde-

pendence of B4. If, however, B4 were dependent on the rest of V, then

B1, B2, B3,B4 would be a system of generating equations, contrary to the

fact that III is a basis.

7. Non-Existence of Other Equational Bases. By a systematic process of elimination, we shall now show that I-V and their duals l-V

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Page 9: Equational Bases of Boolean Algebras

122 F. M. SIOSON

are the only possible equational bases one can obtain from equations B1-B4, IB1-14

First, observe that each of the (8) 8 possible subsystems of B1-B4, ]1-B4 with seven equations contains properly at least one of the already known bases I-V, I-V. None of them could therefore be a basis.

Next, of the (8) 28 subsystems of the original eight that consist of six equations, four are self-dual:

B1 -1 1 1 B1 1 B2 B2

B2 B2 B2 B2 B3 B3 B3 B3

B3 B3 B4 B4 B4 B4 B4 B4 (a) (b) (c) (d)

The remaining systems are those given by the following enumeration and their duals.

B1 B1 B1 B1

B2 B2 B2 B2 B2 B2

B3 B3 B3 B3 B3 B3 I I I I/ \/ \

B4 - 4 B4 B4 B4 B4 B4 B4 (e) M) (g) (h)

B1 B1 B1 B1 B1 B

B2 B2 B2 B2 B2 B2

B3 B3 B3 B3 B3 B3

B4 B4 B4 B4 B4 B4 Wi (i) (k)(I

B1 31 B1 31 B1 31 B1 -31 \ / \ / I ~ ~ ~ ~ ~~~~~~~~~~~~~~~~~~~~~~I II

B2 B2 B2 B2 B2 B2

B3 B3 B3 B3 B3 B3

B4 B4 B4 B4 (m) (n) (o) (P)

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EQUATIONAL BASES OF BOOLEAN ALGEBRAS 123

Now, observe that the equational systems (a), (e), (g), (h), (i), (j), (k), (n), and (o) each contain one of the already known bases I-V, I-V. They are therefore non-independent.

The rest of the six-element systems, (b), (c), (d), (1), (1), (in), and (P), are incomplete in the sense that at least one known property of Boolean algebras is independent of each of them. The models necessary to prove these are given below.

1. Independence-Model of B3 from (b).

? 01- .01 yy 001 001 01 1 1 1 1 1 1 1 1

2. Independence-Model of 32 from (c).

+ 0 1 a .0 1 a Y y 0 0 1 a 0 0 0 0 0 1

1 1 1 1 1~~I0 1 a 1 0

a a 1 1 a 0 a 0 a a

3. Independence-Model of B1 from (d). The independence-model of B1 in I satisfies all the axioms of the system

of equations (d), without satisfying B1.

4. Independence-Model of Th1 from (I). It is also easy to see that the independence-model of I31 in II satisfies

all the equations of (/), but not Bl1.

5. Independence-Model of B3 from (1).

? 0 1 .01 y 01 0 00 0 1

6. Independence-Model of B4 from (m).

+ 0 1 01 Y 0 00 0 01 0 1

101 1I 0

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Page 11: Equational Bases of Boolean Algebras

124 F. M. SIOSON

7. Independence-Model of 133 from (p).

Consider the (pseudo-complemented) lattice T of all open subsets of the real line (under its natural topology) with x defined as the complement of the closure of x. Relative to the operations of intersection and union denoted respectively by * and +, both the distributive and commutative laws hold in T. Moreover, y9 = 0 (the empty set) and hence x+yy = x and x(yy)

yq. On the other hand, if x _ (1,3) and y (2,4), then (1 ,3) * ((2,4) + (2,4)) # (1,3) since 2? (2,4) + (2,4) = (2,4) + (- oo,2) + (4, oo).

By dint of the above proofs of independence, it follows that none of the systems of equations (b), (c), (d), (1), (1), (in), and (p) includes an equational basis of a Boolean algebra. A complete enumeration of all five-element and duals of five-element subsystems of these seven systems of equations will reveal that it contains every subset of B1, B2, B3, B4, B1, B2, B3, B4 with five equations, except for the already known equational bases I, II, III, IV, V and their duals. This completes the demonstration that there are no other equationally basic subsystems of our original eight than the ten we already know. Particularly, this also shows that no dual-symmetric basis exists out of the eight. It remains an open question whether there exists a dual-symmetric equational basis of a Boolean algebra based on the three operations, join, meet, and complementation.

The author wishes to thank the referee for detecting a number of errors in the original manuscript.

REFERENCES

[1] B. A. BERNSTEIN, A simplification of the Whitehead- Huntington set of postulates for Boolean algebras, Bulletin of the American Mathematical Society, vol. 22

(1916), pp. 458-459. [2] B. A. BERNSTEIN, A dual-symmetric definition of Boolean algebra free from

postulated special elements, Scripta mathematica, vol 16 (1950), pp. 157-160. [3] R. MONTAGUE and J. TARSKI, On Bernstein's self-dual set of postulates for Boolean

algebras, Proceedings of the American Mathematical Society, vol. 5 (1954), pp. 310-311.

UNIVERSITY OF FLORIDA

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