A SET OF INDEPENDENT AXIOMS FOR BOOLEAN ALGEBRA

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1 A SET OF INDEPENDENT AXIOMS FOR BOOLEAN ALGEBRA By R. M. DICKER [Received 17 November 1961] 1. Introduction IT is well known that there are a number of sets of axioms for Boolean algebra. In this paper, I give a set of five independent axioms using conditioned disjunction as the ternary operation which is postulated. The ternary operation is most naturally defined on the set 2 n this is done in 2. The theorem is stated in 3 together with a slight generalization of this result; the existence of a unit element is omitted and the result is a set of axioms for a 'locally' Boolean algebra. A generalization of Stone's representation theorem for Boolean algebras is given in 10. This deals with the representation of semi-lattices as subsets of a set. Finally, this result is used to give an alternative proof of my first theorem. 2. The ternary operation Let 2 be a set containing just two elements, 0 and 1 say; then 2 n is the set of n-tuples in 0 and 1. The ternary operation a is defined component-wise on 2"; thus it is only necessary to define it on 2, as follows:... / i \ 77 n <x(0,y,z) = y, <x(l,y,z) = z for all y,z in 2. Now, 2 n may be interpreted as the Boolean algebra which is the set of all subsets of a set of n elements, and then a(x, y, z) is a Boolean function of x, y, z. We find that a{x,y,z) = {x'ny)u{xnz) in the usual notation for Boolean algebra. This operation is called conditioned disjunction by Church (2). The element (0,0,..., 0) is the zero, 0, of the Boolean algebra and (1,1,..., 1) is the unit, 1. It is obvious that a satisfies the following identities: (i) oc{a{x, y, z), u, v) = oc(x, oc(y, u, v), a(z, u, v)); (ii) oc(0,x,y) = x\ (iii) <x{x,0,y) = a(y,0,x); (iv) <x(x,x,0) = 0; (v) a(l,x,y) = y. These identities are in fact the axioms that we assume later. Proc. London Math. Soc. (3) 13 (1963) 20-30

2 AXIOMS FOR BOOLEAN ALGEBRA Statement of the theorems In order to unify the numerous axiom systems for Boolean algebra, it is simplest to regard a Boolean algebra as a set B together with a binary relation, is contained in, and various operations, complement, union, intersection, plus any others which may be useful. Then an axiom system will postulate the existence of certain of these operations and some identical relations, with the property that all the other operations may be defined in terms of the given operations and all the identical relations which hold in a Boolean algebra can be deduced therefrom. To the above, we now add a ternary operation, which we shall write x(y,z) instead of <x(x,y,z). This notation is economical and makes the application of the law (i) easier; it is analogous to writing xy for the product of two elements in a group. THEOREM 1. The set S together with a ternary operation, written x{y,z), is a Boolean algebra if (i) x(y,z){u,v) = x(y(u,v),z(u,v)), (ii) there exists 0 in S such that 0(x, y) = x, (iii) there exists 8 in S such that x(6,y) = y(6,x), (iv) there exists (p in S such that x(x, <p) = <p> (v) there exists 1 in S such that l{x,y) = y, for all x, y, z, u, v in S. 0 = 6 = <pis the zero element, 1 is the unit, and the other Boolean operations can be deduced from the definition: x^y if and only if x(0,y) = x. Each of the axioms (i)-(v) is independent of the others and x(y, z) = (x' n y) U (x n z) is identically true, where n, u, ', are intersection, union, and complement, respectively, relative to the partial ordering ^. Theorem 1 may be generalized by omitting axiom (v). The result is THEOREM 2. The set S together with a ternary operation which satisfies axioms (i), (ii), (iii), and (iv) of Theorem I is a relatively complemented distributive lattice with a zero. The partial ordering is defined as in Theorem 1, 0 = 6 = <p is the zero element, and x(y, z) = (a; n z) U ^(y, x) identically, where ^(y, x) is the element such that ^{y, x) n x = 0, and ^(y, x) U x = x u y. A relatively complemented distributive lattice with a zero is better described as a locally Boolean algebra. For, if a is any element of the lattice S, the set {x x ^ a} is a sub-lattice which is a Boolean algebra. Note that, if S is finite, the union of all elements of S is a unit element. Thus, as a corollary of Theorem 2 we have THEOREM 3. A finite set which satisfies the conditions of Theorem 2 is a Boolean algebra.

3 22 R. M. DICKER We give a simple, independent proof of this in 9. The method of proof may be generalized, to give a second proof of Theorem 1; this is done in The independence of the axioms Let F be a field, and define a ternary operation on F, written x(y,z), in each of the following four ways: (a) 0(y,z) = y, l(y,z) = z for all y,z in F, and x(y,z) = 0 otherwise; (b) x(y, z) = xz for all x, y, z in F; (c) 0(y,z) = y, l(y,z) = z for all y,z in F, and x(y,z) = z otherwise; (d) x(y, z) y xy + xz for all x, y, z in F. The examples (a), (b), (c), and (d) show the independence of the axioms (i), (ii), (iii), and (iv) respectively, provided that F is not the integers mod 2. By Theorem 3, we cannot find a finite example which shows the independence of axiom (v); for this we must use the set F(J) of all finite subsets of the positive integers (including the empty set) together with the operation defined by: for each x, y, z in F(J), x{y, z) = (y\x) U (z n x). 5. Deductions from the axioms (i)-(iv) Throughout this section we assume that the set S satisfies the conditions of Theorem 2. Also, in the statements and proofs of the lemmas we omit the quantifier 'for all a,b,c,... in S\ LEMMA 1. The three elements 0, 6, and <p, whose existence is postulated by (ii), (iii), and (iv), are all equal. Proof. 0(0,9?) = cp by (iv), and 0(0,95) = 0 by (ii); hence 0 = <p. Thus 6(6,0) = 0 by (iv), but 0(0,6) = 6{6,0) and 0(6,6) = 6; hence 0 = 6. 0 is used for the element 0 = 6 = <p in all later deductions from the axioms. LEMMA 2. The following identities hold in S: (vi) a(0,a) = a; (vii) a(b,b) = b; (viii) a(b, 0) = 0 if and only if a(0, b) = b; (ix) a(b, 0) = b if and only if a(0,b) = 0. Proof. For (vii): a(0,0) = 0; hence a(0,0)(6,0) = 0(6,0) and so a(0(6,0), 0(6,0)) = 6, that is a(b, b) = b. For (vi): a(a, 0) (a, 0) = 0(a, 0) by (iv); hence a(0, a) = a.

4 AXIOMS FOR BOOLEAN ALGEBRA 23 For (viii) and (ix): multiply on the right by (6,0). For example, if a{b, 0) = 0, then a(6, 0) (6,0) = 0(6,0) and so a(0,6) = 6. COROLLARY. The statements a^b, a(0,6) = a, 6(0, a) = a, and b(a, 0) = 0 are all equivalent. At this point, some lattice-theoretic ideas are needed; we refer to Birkhoff (1) for the definitions and notation. LEMMA 3. The relation ^ is a partial ordering on S. Proof. By (vi) a^a. If a^6 and b^c, that is, if a(0,6) = a and 6(0, c) = 6, then a(0, c) = a(0,6) (0, c) = a(0,6(0, c)) = a. If a < 6 and 6 ^ a, then obviously a = b. LEMMA 4. Relative to the partial ordering ^ cm $, the greatest lower bound of a and b, af)b, exists and adb = a(0,6); also the least upper bound au 6 exists and aub = a(b,a). Proof. Obviously a(0,6)^6 and a(0,6) ^ a. Now if c < a and c ^ 6 then a(0,6) (0, c) = a(0,6(0, c)) = a(0, c) = c, that is, c ^ a(0,6). For the upper bound, 6^a(6,a) is true if a(6,a)(6,o) = 0, this is true if a(0, a(6,0)) = 0 and so if a(b, 0) (0, a) = 0 if a(6(0, a), 0) = 0 if 6(0, a) ^ a, which is true. Next, a^a(b,a) if a(b,a)(a,0) = 0 if a(b(a,0),0) = 0 if a(0,6(a, 0)) = 6(a, 0) if 6(a, 0)(0,a) = 6(a, 0), which is true. Suppose that a^cand6 ^c; thena(b,a) (0,c) = a(6(0, c),a(0,c)) = a(6,a). This completes the proof. 6. Deductions from the axioms (i)-(v) Throughout this section we assume that S is a set which satisfies the conditions of Theorem 1. LEMMA 5. The following identities hold in S: (x) a(0,1) = a; (xi) b(a, 0) = 0 if and only if a(l,b) = 1; (iii)' a(6,l) = 6(a,l); (iv)' o(l,o) = l; (vi)' a{a, 1) = a; (viii)' a(l,6) = 1 if and only if a(b, 1) = 6. Proof. For (x): a(0,1) = 1(0, a) = a. The primed statements are the duals of the unprimed statements already given; the dual of the statement F(a, 6> = G(a, 6> is

5 24 R. M. DICKER For example, from a(0,b) = 6(0, a) it follows that which reduces to a(b, 1) = b(a, 1). For (xi), if 6(a,0) = 0 then 6(1, 0) (0,a) = 0; hence a(0,6(l,0)) = 0. Thus a(0,6(1,0)) (1,0) = 0(1,0), which reduces to a(l,6) = 1. Similarly a(l,6) = 1 implies that 6(a, 0) = 0. COROLLARY. The statements a^b, a(l,b) = 1, a(b, 1) = 6, and b(a, 1) = 6 are all equivalent. LEMMA 6. The following results hold: (xii) all6 = a(b, 1); (xiii) (a U c) n (6 U c) = (a n 6) u c; (xiv) c(0,a)(a(6,0), 1) = a(6,c). Proof. The dual of the proof of a n 6 = a(0,6) proves that a U 6 = a(b, 1). Then o(c, 1) (0, b(c, 1)) = o(c(0,6(c 5 1)), 6(c, 1)) = o(c, 6(c, 1)) = o(0,6) (c, 1). This proves (xiii). Finally, to prove (xiv), Thus Also, Hence a(b, c) (0, c(0, a)) = a(6(0, c(0, a)), c(0, a)) = a(6,1) (0, c(0, a)) = 6(a,l)(0,c(0,a)) = c(0,a). c(0,a)^a(6,c). a(6, c) (a(6,0), 0) = a(6(a(6 5 0), 0), c(a(6,0), 0)) = o(0, c(a(6,0), 0)) = a(l,c)(a(6,0),0) = c(a(l,0), 1) (a(6,0),0) = 0. a(6,o)^a(6,c) and c(0,a) (a(6,0), 1) ^a(6,c). Now, if x^y then 2/(1,0) ^ic(l,0), since from x(0,y) = x it follows that &(l,o)(y(l,o),l) = a;(l,o). From c(0,a)(a(6,0), 1) = c(a(6,0),a(6,1)), it follows that c(a(6,0),a(6, l))^a(6,c). If we replace b,c by 6(l,0),c(l,0) in this expression and multiply by (1,0), is obtained, and this reduces to a(b,c) ^c(a(6,0),a(6,1)). This completes the proof. 7. Proof of Theorem 1 We have obtained almost all of the results which are needed, and it is only necessary to string them together.

6 AXIOMS FOR BOOLEAN ALGEBRA 25 First, it is necessary to show that, given a Boolean algebra (in the usual sense), the operation given by x(y,z) = (x'n y) V (x n z) satisfies the laws (i)-(v). This is obvious in the light of 2; equally, using a direct calculation, (ii)-(v) are obtained immediately and (i) follows easily from the fact that iff(x) is a Boolean polynomial in x, then f(x) = (/(0) n x') u (/(I) n x). Also, the partial ordering, x^y if and only if x(0,y) = x, which follows from this ternary operation, is identical with the usual partial ordering of the Boolean algebra; hence there is no circularity in the definition. Next, it is necessary to show that a set S which satisfies the conditions of Theorem 1 is a Boolean algebra (in the usual sense) relative to the partial ordering ^. By Lemma 4 and (xiii), it is a distributive lattice; obviously O^a^l for all a; and a(l,0) is the complement of a, since a("ltt(l,0) = 0 and aua(l,0) = 1. These results are sufficient to show that S is a Boolean algebra. Finally, there is no circularity in these definitions if the new ternary operation, say a[b,c], which is defined by a[b, c] = (a' n 6) U (a n c), is identical with the old. This follows from (xiv), and the proof is completed. 8. Proof of Theorem 2 We assume now that S is a set which satisfies the conditions of Theorem 2. We have to show that S is a locally Boolean algebra; in terms of the ternary operation, the equivalent result is LEMMA 7. For any element s of S, the set sr\s = {xes\x(0,s) = x) is a subalgebra of S which satisfies also axiom (v), with 1 = s. Proof. If x,y,zesc\s, then x(y,z) (0,s) = x(y,z) and so x(y,z)esr\s. Hence, s n S is a subalgebra of S. If x,yesn8, then s{0,x{y,0)) = x{y,o). Hence, s{x{y,0),0) = 0, s(x, y) (y, 0) = 0, y ^ s{x, y). Similarly, from s(0, y(x, 0)) = y{x, 0), it follows that s(y(x, 0), 0) = 0, s(y(s, 0) (0, a;), y(s, 0) (0, y)) = 0, s(x{0, y(s, 0)), 2/(0, y(s, 0))) = 0, s(x, y) (0, y(s, 0)) = 0, y(s, 0) (0, s(x, y)) = 0, y{s{x, y), 0) = 0, s(x, y) ^ y. Thus s(x,y) = y. The proof of Theorem 2 is deduced from the proof of Theorem 1 by using 5 and Lemma 7. By Lemmas 3 and 4, S is certainly a lattice, and it is distributive since, for any a, b,c in S, the distributive law for these elements is obtained by putting s = a{b(c,b),a) in Lemma 7 and using Theorem 1. Similarly, by putting s = a(b,a), it can be shown that ( {a,b) = b(a,0). As before, 0 is the zero element and

7 26 R. M. DICKER a(b, c) = (a n c) U #(6, a) is true for the elements a, 6, and c. This completes the proof of Theorem 2 as stated; it is clear from Lemma 7 that a slightly stronger theorem could have been stated: axioms (i)-(iv) are a set of independent axioms for a locally Boolean algebra. 9. Proof of Theorem 3 We need the results of 5 up to Lemma 3; this proof is independent of the later deductions. Since the set S is now finite, minimum non-zero elements exist relative to the partial ordering on S. Let M be the set of all such minimum elements. For each a in S, put /x(a) = {m e M m ^ a}. Then /x is a map of S into 2 M, the set of all subsets of M. The following argument proves that /i(o(6, c)) = (/x(a) n fi(c)) u ((M V(a)) n /.(&)) for each a, b, c in S: if and only if a(b, c) (0, m) = m if and only if a(6(0, m), c(0, m)) = m if and only if either 6(0, m) = m and c(0, m) = m, or 6(0, m) = 0, c(0,m) = m and a(0,m) = m, or 6(0, ra) = m, c(0,ra) = 0 and a(0,ra) = 0 if and only if me (/u,(a) n/x(c)) U ((ikf\/u,(a)) n /x(6)). From this it follows that /x must map # onto 2 M, since /x(a(6,a)) = /u(a)u/i(6). Also, ju. is a one-one map, for if /x(a) = /A(6), then {x(a(b, 0)) is the empty set; hence a(6,0) = 0 and so 6 ^a. Similarly, a^b, so that a = b. Thus /x maps $ isomorphically onto the set of all subsets of M. The isomorphism can be regarded either as taking ^ into set inclusion, or as taking the ternary operation into the operation a defined on 2 M in 2. This completes the proof of Theorem 3. This proof can be generalized by using a method due to Stone. The result is an alternative proof of the above theorems. We turn to the generalization now, and this is used in 13 to give the alternative proof. 10. Semi-lattices and their representations By a semi-lattice we shall mean a commutative semi-group with a zero, in which every element is idempotent. It must be pointed out that the

AXIOMS FOR BOOLEAN ALGEBRA 27 standard terminology is semi-lattice with zero. The set 2 A ' of all subsets of the set N, with intersection as the binary operation, is a semi-lattice.

8 AXIOMS FOR BOOLEAN ALGEBRA 27 standard terminology is semi-lattice with zero. The set 2 A ' of all subsets of the set N, with intersection as the binary operation, is a semi-lattice. Suppose that A is a homomorphism of a semi-lattice S into the semi-lattice 2-^; then we say that X(S) is a homomorphic representation of S as subsets of the set N. It is well known that we can find an isomorphic representation of S by taking N = S and putting A (a) = as. Stone's representation theorem (3) for a Boolean algebra B is an isomorphic representation of B as subsets of (a subset of) 2^. In the more general situation above, we have a simpler representation than this but the next theorem shows that there is an equivalent representation as subsets of 2 s. THEOREM 4. Given any representation X(S) of a semi-lattice S as subsets of the set N, there is a representation /x($) of S as subsets of 2 s with /x(#) isomorphic to X(S). Proof. The map A: S->2 N is given. For each n in N, let S n be the set of all a in S such that nex(a). (In fact, S n is a dual-ideal of S. In the above theorem we could replace 2 s by the set of all dual-ideals.) Put T = {S n n e N}, making T a subset of 2 s, and define /x: S-+2 T by ix(a) = {S n \nex(a)}. Thus S n en(a) if and only if nex{a) and from this it is easily deduced that /x(a6) = /x(a)dju,(6), and that ju(a) = /A(6) if and only if X{a) = A(6). Our generalization of Stone's theorem is a representation of a semi-lattice as subsets of the set of maximal dual-ideals. This representation has the advantage that, if the semi-lattice is in fact a lattice, unions are properly represented (Theorem 6), and the disadvantage that the representation is not an isomorphism unless an extra condition is satisfied. 11. Maximal dual-ideals In order to state and prove Lemma 8, we make the following definitions. A domain D of a semi-lattice S is a subset of the set S\{0} such that if a,bed then abed. A dual-ideal of a semi-lattice S is a domain J) such that, if aed, bes, and ab = a, then 6eD. The domains of S are partially ordered by inclusion, and the word maximal refers to this ordering. By using Zorn's lemma, we may prove that, if D is any domain of a semi-lattice S, there exists a maximal domain of S containing D. Further, a domain D is maximal if and only if, for all aes, either aed or there exists CGD such that ac = 0. This follows from the fact that if aes\d and ad^q for all d e D, then D u ad U {a} is a domain. LEMMA 8. Let S be a semi-lattice. If aes and a#(), then there is a maximal dual-ideal of S containing a. Also, a domain D of S is a maximal

9 28 R. M. DICKER dual-ideal if and only if, for all bes, either bed or there exists c in D such that be = 0. Proof. The first part is obvious since {a} is a domain. The second part follows from the results given above by showing that a maximal domain is a dual-ideal. Suppose that D is a maximal domain of S, a ed, ab = a and also, if possible, that 6 is not an element of D. Then there exists c in D such that be = 0; hence ac = abc = 0, which is impossible. COROLLARY. If m is a maximal dual-ideal of a semi-lattice S, the elements a i (1 ^i ^2 } ) are i n m ana% th e dements bj (l^j ^q) are not in m, then there is an element cinm such that fyc = c (1 ^i^p) and bjc = 0 (1 ^j^q). Proof. For each j there exists Cy in m such that c^bj = 0. Put c = a 1 a 2...a p c 1 c 2...c q. It is this result which enables us to work with maximal dual-ideals in the same way as with minimum elements of the semi-lattice. 12. The representation theorems THEOREM 5. Let M be the set of all maximal dual-ideals of a semi-lattice 8. Define /x: 8^2 M by ix(a) - {mem\aem} for each a in S. Then n(s) is a representation of S as subsets of the set M and is an isomorphism if and only if, for any a,b in S such that ab = a and a^b, there exists c in S such that c i 0, ac = 0 and be = c. Proof. From m e /u.(a6) if and only if ab m if and only if a em and bem if and only if m e /x(a) n /x(6) it follows that fx is a homomorphism and (x(s) is a representation of S. Suppose that /x is an isomorphism and that a, b are elements of 8 such that ab = a and a^b. Then n(ab) = ft(6) implies that ab = b which is impossible; hence /x(ab) is strictly contained in ^(b). Let m be an element of /x(6)\/x(a6); then there is an element cinm such that ac = 0 and be = c. Conversely, suppose that, for any a,b in S such that ab = a and a^b, there exists c in S such that c#0, ac = 0, and be = c; also suppose, if possible, that there exist x and y in 8 such that /x(x) = fx(y) and x^y. If xy^y, then, since xyy = xy, there exists c such that c#0, xyc = 0, and yc = c. Let m be a maximal dual-ideal containing c. Then me[x(y), but m is not an element of fx(xy) = /x(x) = fi(y) which is impossible. This proves the theorem. The advantage of this representation lies in the fact that, whenever what we should naturally call the union of a and 6 exists in S, then it is mapped onto /x(a) U fx(b). This is stated precisely in

10 AXIOMS FOR BOOLEAN ALGEBRA 29 THEOREM 6. For any a, b, c, d in the semi-lattice S, if ac = a, be = 6, and ad = 0, bd = 0, cd = d imply that d = 0, then fx(c) = /x(a)u/tx(6), where /x is the map defined in Theorem 5. Proof. From ac = a and 6c = 6 it follows that /x(a) U /x(6) <= JU.(C). Assume, if possible, that fx(c) strictly contains /x(a) u fx(b). Let m be an element of /x(c)\(/x(a)u/x(6)); then cera, a$m, and 6^m. Hence, there exists dem such that ad = 0, bd = 0, and cd = d; but this contradicts the assumption of the theorem. Theorems 5 and 6 are a generalization of the method that Stone used to prove his theorem. The essential difference is that we confine our attention to one binary operation only. We can prove Stone's theorem easily by using the above theorems as follows. Let B be a Boolean algebra. Then B is a semi-lattice under the operation a D 6 of B. Let M be the set of maximal dual-ideals of B, and construct /x as in Theorem 5. Then fx is an isomorphism, since if an 6 = a and a#6, then a' n 6 is a suitable element to use as c in Theorem 5. The hypothesis of Theorem 6 is true if c = a U 6; hence fx(a U 6) = fx{a) U fx(b). By definition, JU(O) is the empty set and /x(l) = M. The proof is completed by showing that ix(a') = ifcf\/x(a). If meyi{a') then a'em, and since d'no = 0, a$m; hence mem\^(a). Conversely, if mem\fx(a), then a$m and there exists 6 in m such that a n b = 0. In this case, a' n 6 = b and so a' em and me/x(a'). 13. Second proof of Theorems 1 and 2 This is a straightforward generalization of the proof of Theorem 3, using maximal dual-ideals instead of minimum elements. The results of 5 are needed in order to show that S is a semi-lattice under the binary operation aft = a(0,6). Theorem 5 is used to obtain an isomorphic representation of S from which all the necessary results can be deduced to provide an alternative proof for the most tedious part of 6-8. For a complete proof we need the independence of the axioms and the elementary calculation from a Boolean algebra (in the usual sense), which have already been given. For convenience, we use ab and a(0,6) interchangeably in this section. Let S be a set which satisfies the conditions of Theorem 2; then, by Lemmas 1-4, 8 is a semi-lattice under the operation ab = a(0,6). Construct M and fx as in Theorem 5, then /x is one-one. For, if a(0,6) = a and b^a, then a(6,0)#0 since a(6,0) = 0 implies that a(0,b) = b, and also a(b, 0) (0,6) = a{b, 0) and a{b, 0) (0, a) = a(a(0,6), 0) = a{a, 0) (0, b) = 0; the result follows by Theorem 5.

11 30 AXIOMS FOR BOOLEAN ALGEBRA If me fm(a(b, c)) then a(b, c) em, and from this there arise four possibilities for 6 and c: (a) m^ju(6) and m^ju-(c). Then there exists x in m such that xb = 0, xc = 0 and, x{0,a(b,c)) = x; hence a(b,c) (0,x) = a(0,0) = 0, which is impossible. (b) mefx{b) and ra /x(c). Then there exists x in m such that #6 = x, xc = 0, and a:(0,a(6,c)) = a:; hence a; = a(6,c)(0,x) = a(z,0); hence a{q,x) = 0 and a^m. (c) m^/lt(6) and me/x(c). Then there exists a; in m such that a:c = x, xb = 0, and x(0,a{b,c)) = x; hence x = a(b,c)(0,x) = a(0,x) and so a em. (d) me/x(6) and me/it(c). Then there exists a; in m such that xb = x and xc = x; hence a(6, c) (0, a;) = a(x, x) = x and there is no restriction on a. This proves that fx(a{b, c)) (^(a) n /x(c)) u ((Jf\/i(a)) n /x(6)). Conversely, if m e/x(a) n /u,(c), there arise two possibilities for 6: either 6 em; hence be em and a(b,c)(0,bc) = 6c and so a(b,c)em, or b$m; and there exists a; in m such that xa = x, xc = x, and xb = 0; hence a(6, c) (0,a;) = a; and a(6, c)em. If m e (M\fx(a)) n p(b), there arise two possibilities for c: either cem; and there exists x in m such that a;a = 0, a;6 = a;, and xc = x; hence a(6, c) (0, a;) = x and so a(6, c) em, or c$m; and there exists x in m such that xa = 0, a;6 = a;, and ax = 0, hence a(b, c) (0, a;) = a(x, 0) = x and so a(6, c)em. This proves that (/u(a) n /n(c)) U ((if\/x(a)) n fi{b)) g ia(a(6, c)). Thus, there exists a one-one mapping ^ of $ into subsets of a set if such that /x(a(6, c)) = (/x(a) n ^(c)) U ((Jf\/Lt(o)) n /*(&)). From this mapping, it is obvious that Lemmas 4-7 hold. Thus this section can be used to replace Lemmas 4-7. I wish to thank Dr K. A. H. Gravett who, as my supervisor, has given me considerable help. Sections arose directly out of a problem which he suggested to me. REFERENCES 1. GARRETT BIRKHOFF, Lattice theory, revised edition, American Math. Soc. Colloquium Publications XXV (New York, 1948). 2. A. CHURCH, Introduction to mathematical logic (Princeton, 1956). 3. M. H. STONE, 'The theory of representations for Boolean algebras', Trans. American Math. Soc. 40 (1936) Department of Mathematics The University of Glasgow

REGULAR IDENTITIES IN LATTICES

TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 158, Number 1, July 1971 REGULAR IDENTITIES IN LATTICES BY R. PADMANABHANC) Abstract. An algebraic system 3i = is called a quasilattice