• TABLE OF CONTENTS
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 Title Page
 Acknowledgement
 Table of Contents
 List of Figures
 Index of symbols
 Index of terminology
 Definitions and elementary...
 The paste job
 The representation of orthocomplemented...
 Bibliography
 Biographical sketch














Title: Orthomodular lattices
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Title: Orthomodular lattices
Physical Description: viii, 94 leaves : illus. ; 28 cm.
Language: English
Creator: Greechie, Richard Joseph, 1941-
Publication Date: 1966
Copyright Date: 1966
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Subject: Lattice theory   ( lcsh )
Mathematics thesis Ph. D
Dissertations, Academic -- Mathematics -- UF
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non-fiction   ( marcgt )
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Thesis: Thesis - University of Florida.
Bibliography: Bibliography: leaves 93.
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Table of Contents
    Title Page
        Page i
    Acknowledgement
        Page ii
    Table of Contents
        Page iii
    List of Figures
        Page iv
    Index of symbols
        Page v
    Index of terminology
        Page vi
        Page vii
        Page viii
    Definitions and elementary results
        Page 1
        Page 2
        Page 3
        Page 4
        Page 5
        Page 6
        Page 7
        Page 8
        Page 9
        Page 10
        Page 11
        Page 12
        Page 13
        Page 14
        Page 15
        Page 16
        Page 17
        Page 18
        Page 19
        Page 20
    The paste job
        Page 21
        Page 22
        Page 23
        Page 24
        Page 25
        Page 26
        Page 27
        Page 28
        Page 29
        Page 30
        Page 31
        Page 32
        Page 33
        Page 34
        Page 35
        Page 36
        Page 37
        Page 38
        Page 39
        Page 40
        Page 41
        Page 42
        Page 43
        Page 44
        Page 45
        Page 46
        Page 47
        Page 48
        Page 49
        Page 50
        Page 51
        Page 52
        Page 53
        Page 54
        Page 55
        Page 56
        Page 57
        Page 58
        Page 59
        Page 60
        Page 61
        Page 62
        Page 63
        Page 64
        Page 65
        Page 66
        Page 67
        Page 68
        Page 69
    The representation of orthocomplemented posets by sets
        Page 70
        Page 71
        Page 72
        Page 73
        Page 74
        Page 75
        Page 76
        Page 77
        Page 78
        Page 79
        Page 80
        Page 81
        Page 82
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        Page 84
        Page 85
        Page 86
        Page 87
        Page 88
        Page 89
        Page 90
        Page 91
        Page 92
    Bibliography
        Page 93
    Biographical sketch
        Page 94
        Page 95
        Page 96
Full Text










ORTHOMODULAR LATTICES
























By
RICHARD JOSEPH GREECHIE










A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF
THE UNIVERSITY OF FLORIDA
IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE
DEGREE OF DOCTOR OF PHILOSOPHY












UNIVERSITY OF FLORIDA

APRIL, 1966












ACKNOWLEDGEMENT


The author wishes to acknowledge the helpful

suggestions, encouragement, and superior example of his

advisor, Professor David Foulis.












TABLE OF CONTENTS


page

ACKNOWLEDGEMENT . . . . .. . . . ii

LIST OF FIGURES . . . .. . . . . . iv

INDEX OF SYMBOLS . . . .. . . . . v

INDEX OF TERMINOLOGY . . . . . . ... . vi

CHAPTER

1. DEFINITIONS AND ELEMENTARY RESULTS . . .. 1

1. From Relations to Orthomodular Lattices 1

2. Standard Results in the Theory
of Orthomodular Lattices . . . . 9

3. Morphisms, Ideals, Filters, and Sections 16

2. THE PASTE JOB . . . . . . . .. 21

1. New Lattices from Old . . . . .. 21

2. A Partial Converse . . . . .. 47

3. Variation on a Theme . . . . .. 54

3. THE REPRESENTATION OF ORTHOCOMPLEMENTED
POSETS BY SETS . . . . . . ... 70

1. F-sequences . . . . . . ... 70

2. Order Ortho-homomorphisms . . . .. 79

3. Travis Matrices . . . . . .. 83

4. Examples . . . . . . . .. 87

BIBLIOGRAPHY . . . . . . . ... . 93

BIOGRAPHICAL SKETCH . . . . . . . .. 94












LIST OF FIGURES


page

Figure 1 . . . . . . . . 10

Figure 2 . . . . . . . ... 40

Figure 3 . . . . . . . ... 42

Figure 4 . . . . . . ... 42

Figure 5 . . . . . . . ... 43

Figure 6 . . . . . . . ... 45

Figure 7 . . . . . . . ... 46

Figure 8 . . . . . . . ... 65

Figure 9 . . . . . . . ... 65

Figure 10 . . . . . . . ... 66

Figure 11 . . . . . . . ... 66

Figure 12 . . . . . . ... 87

Figure 13 . . . . . . . ... 88

Figure 14 . . . . . . . ... 88

Figure 15 . . . . . . . ... 89

Figure 16 . . . . . . . ... 90

Figure 17 . . . . . . . ... 91

Figure 18 . . . . . . ... 92













INDEX OF SYMBOLS


page

M . . 1

2, 2 . 1

E . . . 1

C, 3 . .. 1

M-N . . 1

T.A.E. . 1

XXX . . 1

R-1 . . .




Aly . . 2

2 3
R2, R . 2

X/R . . 3

<, < . . 3

(X,R) . . 3

U(M), T(M) 4

sup M, inf M 4

, A . . 5

0, 1 . . 6


P(0,x), P(x,1

':P -> P

(P,<,')

J_ . . .

OMI . .

DOMI . .

Cp . . ..

C . . .

C(M) . .

Pol(M) . .

D(a,b,c) .

M(b,c) . .



S . . .
x



[x] . .

[z]:[x]R[y]

[z]:[x] r [y]


page

). 6

7

S 7

S 9

. 9

S. 10

S. 11

S. 11

S. 11

S. 13

S. 15

S. 15

S. 17

S. 19

S. 22

S. 22

S 24


[0,1)i .

HS . .

D 16 .

N, R

[0,1]


1 2

fM . .
x

B . .
B
Y

Y
bi, B



[y] . .
. .



H, E, F

(2*,p*, ')

5* . .


page

30

* 54

66

71

S71

71

. 71

S71

. 72

. 72

S72

. 74

S74

. 74

80

S83

S83












INDEX OF TERMINOLOGY


age


adjunction of a crown .

anti-symmetric . . .

atom . . . . .

atomic . . . . .

Boolean . . . . .

center . . . . .

chain . . . . .

co-atom . . . . .

commutes . . . .

comparable . . . .

complement . . . .

complemented poset . .

complete join
homomorphism . . .

complete join-meet
homomorphism . . .

complete lattice . .

complete meet
homomorphism . . .

composition . . . .

Condition I . . . .


. . . 75


pa


p;


ge

1

36


Condition II


67 converse . . . .

2 corresponding sections

6 deterministic
Travis matrix . .
6
diagonal . . . .
12
diagonal ...
12 restricted to .

4 disjoint sum ... .54,

6 distributive lattice

11 distributive triple

4 domain . . . .

7 dominate . . . .

7 dual . . . . .

epimorphism . . .
16
equivalence class

17 equivalence relation

5 Foulis-Holland Theorem

greatest (element)
16
r-sequence . . .
2
F-state . . .
75
r-state space . .








page

horizontal sum .. .54,55

implements . . .. .24

infimum . . . . 4

irreducible . . .. .12

isomorphism . . .. 17

join . . . . 4

join dense . . . 7

join homomorphism .16

join-meet homomorphism 16

largest (element) .. 4

lattice . . . . 5

least (element) . . 4

linearly ordered subset 4

lower bound . . . 4

meet . . . . 4

meet homomorphism . 16

modular lattice . 15

modular pair ... .15

monomorphism ... .17

one . . . . . 6

order filter ... .18

order homomorphism 16

order ideal . . .. .18

orthocomplementation 7


orthocomplemented lattice

orthocomplemented poset

orthogonal . . .

orthogonal family .

ortho-homomorphism . .

orthomodular identity

orthomodular lattice

orthomodular poset .

partially ordered set .

partially orders . .

partial ordering . .

polynomial in M . . .

poset . . . . .

poset with zero and one

principal order filter

principal order ideal

principal section . .

proper order filter .

proper order ideal .

range . . . . .

reducible . . . .

reflexive . . . .

relation . . . .

representative . . .


page

9

7

9

9

16

9

9

9

3

3

3

13

3

6

18

18

19

18

18

1

12

2

1

3









section . . . .

simple (section) . .

smallest (element) .

subcomplete . ...

sublattice . . .

sub-orthomodular lattice

supremum . . . .

symmetric . . . .


page

19

56

4

5

5

9

4

2


transitive . . .

Travis matrix . .

trivial order filter

trivial order ideal

unit . . . .

upper bound .....

"weak" partial order
induced on P* by Z*

zero . . .


viii


page

2

83

18

18

6

4


83

6












CHAPTER 1


DEFINITIONS AND ELEMENTARY RESULTS


1. From Relations to Orthomodular Lattices


1.1.1 Notation. We assume that the reader has a

knowledge of the concepts of set theory (including the

various forms of the axiom of choice) and cardinal number

theory. The cardinal number of a set M is denoted by the
# M
symbol #M. The power set of a set M is denoted by 2 If

M = m, then 2 is sometimes written 2m.

The symbols E, c, and n are used in their usual

set-theoretic sense. The complement of N in M is denoted

by M-N. Finally, the symbol T.A.E. stands for "the

following statements are equivalent."



1.1.2 Definition. Let X be any set. We define XXX to

be the set [(x,y):x,yEX). By a relation on X we mean a

subset of XXX. If R is a relation then we call the

relation [(x,y):(y,x)ER) the converse of R and write it as
-1
R-1. {x:(x,y)ER) is called the domain of R and fy:(x,y)ER}

is called the range of R. Note that the domain of R equals
the range of R- and the range of R equals the domain of
the range of R and the range of R equals the domain of








-l
R-1. Let M c X and xEX, then MR denotes

(y:(z,y)ER for some zEM) and xR denotes {x)R. (Here, as in

the sequel, relations operate on the right. In keeping

with this convention, most functions operate on the right;

however, in analytically flavored contexts functions may

operate on the left.) If R and S are two relations on X,

then we define the composition of R and S, denoted by RS,

to be the relation [(x,y):there exists zEX such that

(x,z)ER and (z,y)ES). By the diagonal of XXX, symbolized

by AX or simply A if there can be no confusion' as to what

set is under consideration, we mean ((x,x):xEX). If Y c X,

then the diagonal of XXX restricted to Y, written AIy, is

defined to be [(x,x):xEYI.

Let R be a relation on X. Then R is said to be
-l
reflexive if A c R; R is said to be symmetric if R c R; R

is said to be anti-symmetric if (x,y)ER and (y,x)ER imply

x = y; R is said to be transitive if RR c R. We sometimes
2 3
write RR as R (RR)R as R etc.



1.1.3 Remark. (1) Any relation of the form RUR- is a

symmetric relation.

(2) A is a symmetric relation.

(3) The union of two symmetric relations is again a

symmetric relation.

(4) Let R a in some index set I, and T be relations








on X. Then ( U R )T = U (R T) and
aEI a aE a

T(U R ) = U (TR).
a6I aEI a

The proof follows immediately from the

definitions.



1.1.4 Definition. If R is a reflexive, symmetric, and

transitive relation on X, then R is called an equivalence

relation on X. If R is an equivalence relation, then a set

of the form xR is called an equivalence class, the set of

all such equivalence classes is denoted by X/R, and an

element of such an equivalence class is called a

representative of the equivalence class. The following

statement is an immediate consequence of the above

definition: If R is an equivalence relation, then x is a

representative of the equivalence class yR if and only if

xR = yR.

If R is reflexive, anti-symmetric, and transitive

on X, then R is called a partial ordering on X; in this

case (X,R) is called a partially ordered set or simply a

poset.

If s is a relation on the set P such that (P,s)

is a poset, then we say that S partially orders P. If

x,yEP and x S y, then y is said to dominate x, x is said to

be dominated by y. If at least one of x 9 y, y s x holds,









then x and y are said to be comparable. If x and y are

comparable for all x,y in a subset M of P, then M is called

a linearly ordered subset of P or a chain in P. If x is

dominated by y and x X y, then we sometimes write x < y.

Let M c P; if uEP is such that xEM implies that x & u, then

u is called an upper bound for M; if VEP is such that xEM

implies that v & x, then v is called a lower bound for M.

The set of all upper bounds for M is denoted by

U(M), i.e.,

U(M) = (zEP:x s z for all xEMJ.

The set of all lower bounds for M is denoted by

T(M), i.e.,

T(M) = fzEP:z S x for all xEM].

An element z6P is called the greatest (largest)

element of a subset M of P if zEM and zEU(M). An element

zEP is called the least (smallest) element of a subset M of

P if zEM and zET(M). An immediate consequence of the fact

that s is anti-symmetric is the fact that the greatest and

least elements of a subset M of P are unique (if they

exist). An element z of P is called the supremum (join) of

the subset M of P, written sup M, if z is an upper bound

for M, and, whenever u is an upper bound for M, z E u. An

element z of P is called the infimum (meet) of the subset M

of P, written inf M, if z is a lower bound for M, and,

whenever v is a lower bound for M, v S z.








A poset (P,t) is called a lattice in case, for

each non-empty finite subset M of P, there exist z,wEP such

that z = sup M and w = inf M. In case M = (x,y), then

sup M is sometimes written xvy and inf M is sometimes

written xAy. Also if M = IxI, ..., xn], then sup M and

inf M are sometimes written xl1...vxn and x A...AXno

respectively. It may be shown that sup and inf satisfy the

usual (generalized) associative and commutative laws.

The statement that the supremum (resp., infimum)

of a subset M of P exists is clearly equivalent to the

statement that U(M) has a least element (resp., T(M) has a

greatest element). To say that z = sup M (resp.,

w = inf M) is to say that U(M) = U((z}) (resp.,

T(M) = T([w))).

A poset (P,g) is called a complete lattice if,

for every M c P, there exist z,wEP such that z = sup M and

w = inf M; if such is the case we say simply that sup M and

inf M exist.

A subset M of a lattice (L,A) is called a

sublattice of L in case (M,|IM) is a lattice. A sublattice

M of the lattice L is called subcomplete in case the

following obtains: if N c M and sup N exists as computed

in L, then sup N exists as computed in M, the two are

equal, and the common value is in M. A sublattice M of a

lattice L may be (1) complete but not subcomplete, (2)

subcomplete but not complete, (3) neither, or (4) both.









Let L be the lattice of all subspaces of an infinite dimen-

sional Hilbert space H (with set-theoretic inclusion as the

partial order), and let M be the sublattice of L consisting

of all closed subspaces of H. Then M is an example of (1).

Any lattice which is not complete, considered as a

sublattice of itself, is an example of (2). The lattice of

all finite subsets of an infinite set X, considered as a

sublattice of the power set of X, is an example of (3).

Any complete lattice, considered as a sublattice of itself,

is an example of (4).

An element 0 (resp., 1) of a poset P with the

property that U(0) = P (resp., T(1) = P) is called the zero

(resp., unit) of P. The uniqueness of the zero (resp.,

unit), if it exists, is an immediate consequence of the

fact that the partial ordering is anti-symmetric. The unit

is sometimes called the one of the poset. If there exists

a zero and a one in P, then P is called a post with zero

and one. In a poset P with zero and one, P(0,x) denotes

(zEP:0 S z s xJ and P(x,l) denotes IzEP:x & z s 1).

If P contains a zero element O, then an element x

in P is said to be an atom of P in case 0 < x and if

0 < y < x for some yEP, then y = 0 or y = x. If P contains

a one element 1, then an element x in P is said to be a

co-atom of P in case x < 1, and if x : y S 1 for some yEP,

then y = x or y = 1. P is said to be atomic in case every









non-zero element of P dominates an atom of P. If every

non-zero element of P is the join of the atoms it

dominates, then the atoms of P are said to be loin dense.

If x is an element of a poset P with zero and

one, then an element y of P is called a complement of x in

P in case xvy exists, xAy exists, xvy = 1, and xAy = 0. If

every element of a poset P has a complement in P, then P is

called a complemented poset. If P is a complemented poset

and if, moreover, there is a mapping ':P -> P such that

(1) x 9 y implies y' s x',

(2) x'' = x (where, by definition, x'' = (x')'), and

(3) x' is a complement for x,

then P is called an orthocomplemented poset. ':P -> P is

said to be an orthocomplementation on P.



1.1.5 Notation. We reserve the right to use any of the

symbols P, (P,t), or (P,Z,') to represent the (orthocom-

plemented) poset P. By judicious utilization of this

standard abuse of terminology, we will single out the

salient feature of P under discussion, distinguish between

two such structures if necessary, and maintain a minimum of

notational overweight.



1.1.6 Definition. Because of the symmetry in the

definitions of supremum and infimum, zero and one, we have









the following principle of duality for posets with zero and

one:

If, in any valid statement which holds for all

posets with zero and one, we interchange the symbols for

supremum and infimum, interchange zero and one, and reverse

all inequalities, then we obtain another valid statement

(called the dual of the original statement) which holds for

all posets with zero and one.

Similar principles of duality hold for general

posets and for orthocomplemented posets.



1.1.7 Lemma. Let P be an orthocomplemented poset, then

P satisfies the Generalized DeMorgan Laws, i.e., for

x EP (aEI), (1) if sup (xa:a6I) or inf [x':alEI exists in

P, then they both exist and (sup {x :aEI])'= inf (x':aEI);

and (2) if inf (x :aEIj or sup (x:alEI} exists in P, then

they both exist and (inf (x :aEI})' = sup ({x:aEI].

Proof. If sup (x :aEI) exists and equals x, then

x x for all aEI, so that x' x' for all aEI. If y is

any other lower bound for x', then y' x for all aEI,

y' ; x, and finally y & x'. Hence x' = inf {x':aEI).

Similarly, if inf (x':aEI exists and equals x, then

x' = sup (x :aEI). Hence (1) is valid. (2) follows from

(1) by duality.








1.1.8 Definition. Two elements x,y of an ortho-

complemented poset are said to be orthogonal, written

x _L y, in case x t y'. Note that if x __ y if and only if

y x. A family (x :aEA} of elements of P is said to be

an orthogonal family in case a 7 B implies that x xg.

An orthocomplemented poset P is called an orthomodular

post in case P satisfies the following two properties:

(1) if x,yEP are such that x y, then xvy exists

in P, and

(2) for x,yEP, x s y implies y = xv(y'vx)'.

The latter condition is called the orthomodular identity,

abbreviated OMI.

An orthomodular (resp., orthocomplemented) poset

which is a lattice is called an orthomodular (resp., ortho-

complemented) lattice. A sublattice L1 of an orthomodular

lattice L is said to be a sub-orthomodular lattice of L,

written L1 < L, in case the restriction of the orthocom-

plementation on L to L makes L1 an orthomodular lattice.

In case L1 < L and L1 L, we sometimes write L1 S L.



2. Standard Results in the Theory
of Orthomodular Lattices

1.2.1 Theorem. Let L be an orthocomplemented lattice.

Then T.A.E.

(1) L is an orthomodular lattice, i.e., L satisfies

the OMI,








e a f implies f = ev(fAe'),

for all e,fEL,

(2) L satisfies the dual orthomodular identity,

abbreviated DOMI,

e f implies e = fA(evf'),

for all e,fEL,

(3) *eCf implies fCe, for all e,frL,

(4) e < f and fAe' = 0 implies e = f,for all e,fEL,

(5) L does not contain a sublattice of the form


1


Figure 1


Proof. The equivalence of (1), (2), and (3) is

proved in [4, Theorem 1, p. 68]. The equivalence of (4)

and (5) is obvious. Moreover, it is clear that (1) implies

(4). Hence it suffices to prove that (5) implies (1).

Suppose that (5) is valid but that (1) is

invalid. Then there exist a,fEL such that a f and

f > av(fAa'). Let e = av(faa'). We claim that

{0,l,e,e',f,f'} is a sublattice of L of the type given in








Figure 1. For, f a e implies fAa' a eAa'. Moreover,

e = av(fAa') implies e 2 fAa'; hence eAa' a (fAa')Aa' = fAa'

and consequently fAa' = eAa'. Since

f've e ; eAa' = fAa' f fAe' = (f've)', it follows that

f've = 1, and hence fAe' = 0. Therefore fve' = 1 and

eAf' = 0. But fAf' = eAe' = 0 and fvf' = eve' = 1. Hence

(0,l,e,e',f,f'] is a sublattice of L of the type given in

Figure 1 which contradicts the assumption that (5) is

valid, i.e., L contains no such sublattice. Consequently

(5) implies (1).



1.2.2 Corollary. The atoms of an atomic orthomodular

lattice are join dense.

Proof. The result follows from an application of

part (4) of Theorem 1.2.1.



1.2.3 Definition. Let L be an orthomodular lattice.

Corresponding to every element fEL, we define a mapping

pf:L -> L as follows:

For eEL, ecpf = (eVf')Af.

For e,fEL, we say that e commutes with f if and only if

ecf = eAf in which case we write eCf. Also, for any subset

M of L we define

C(M) = {eEL:eCf for all fEM).

By CC(M) we mean C(C(M)). The set C(L) is called the








center of L. Note that (0,1) c C(L) c L always holds. If

(0,1) = C(L), then L is said to be irreducible; if
c
(0,1) f C(L), then L is said to be reducible. If

C(L) = L, then L is said to be Boolean.



1.2.4 Theorem. Let L be an orthomodular lattice, and

let M,N c L.

(1) C(M) is a subcomplete sub-orthomodular lattice

of L.

(2) M c N implies C(N) c C(M).

(3) M c C(C(M)).

(4) C(M) = C(C(C(M))).

(5) M = C(C(M)) if and only if there exists N c L

such that M = C(N).

(6) Let M be a sub-orthomodular lattice of L.

Then T.A.E.

(a) M is a Boolean lattice,

(b) M c C(M),

(c) C(C(M)) C C(M).

(7) If M c C(M), then C(C(M)) is a subcomplete

Boolean sub-orthomodular lattice of L.

Proof. (1) is essentially proved in

[2, Lemma 3, p. 67). (2) and (3) follow immediately from

Definition 1.2.3; (4), (5), and (6) are immediate conse-

quences of (2) and (3). (7) follows from (1), (4), and (6).








1.2.5 Definition. Let L be an orthomodular lattice,

and let M c L. By a polynomial in M, written Pol(M), is

meant any finite combination of supreme and infima of

elements of M or primes of such elements, or any finite

combination of supreme and infima of such combinations of

elements of M or primes of such combinations of elements of

M, etc.



It is to be emphasized that, although a poly-

nomial in M may be represented in different ways as

combinations of elements of M or the primes of such

elements, in any given representation only a finite number

of elements of M appear and that the symbols v, A, and

appear only a finite number of times.



The following proposition contains the elementary

properties of polynomials which will be utilized in the

sequel.

1.2.6 Proposition. Let L be an orthomodular lattice,

and let M,N c L.

(1) If M c N, then Pol(M) c Pol(N).

(2) Pol(Pol(M)) = Pol(M).

(3) If M c Pol(N), then Pol(M) c Pol(N).

Proof. These results follow immediately from

Definition 1.2.5.








1.2.7 Proposition. Let L be an orthomodular lattice,

and let e,f,gEL.

(1) T.A.E.

(a) eCf,

(b) fCe,

(c) e'Cf.

(2) If e Cf for all aEI, then (sup (e :aEI})Cf.

(3) If e f, then eCf.

(4) eEC(L) if and only if e has exactly one

complement in L.

Proof. (1), (2), and (3) are proved in

[2, Lemmata 1, 2, and 3, p. 67]. (4) is proved in

[4, Corollary to Theorem 4, p. 71].



1.2.8 Theorem (Foulis-Holland Theorem). Let L be an

orthomodular lattice, and let e,f,gEL. If any two of the

three relations eCf, eCg, fCg hold, then

(eVf)Ag = (eAg)v(fAg) and (eAf)vg = (evg)A(fvg).

Proof. The proof is found in [2, Theorem 5,

p. 68] and in [4, Theorem 3, p. 69].



An immediate consequence of Proposition 1.2.7 and

the Foulis-Holland Theorem is the following

1.2.9 Corollary. Let L be an orthomodular lattice, let

M c L be any family of mutually commuting elements of L,








let eEL be such that eCf for all fEM, and let pEPol(M).

Then eCp.

Proof. poe = (pve')Ae = (pAe)v(e'Ae) = pAe.


1.2.10 Definition. Let L be a lattice, and let a,b,cEL.

Then we say that (a,b,c) is a distributive triple and we

write D(a,b,c) in case (avb)Ac = (aAc)V(bAc). We say that

(b,c) is a modular pair and we write M(b,c) in case

D(c,a,b) holds whenever c 9 b. L is said to be a

distributive (resp., modular) lattice in case D(a,b,c)

holds for all a,b,cEL (resp., M(b,c) holds for all b,cEL).



1.2.11 Remark. An orthomodular lattice is a Boolean

lattice if and only if it is a distributive lattice.

Proof. This is an immediate consequence of the

Foulis-Holland Theorem.

1.2.12 Lemma. Let L be an orthomodular lattice, and let

M c L be any family of mutually commuting elements of L.

Then Pol(M) is a Boolean sub-orthomodular lattice of L.

Proof. Let p,q,rEPol(M). Apply Proposition 1.2.7

to the components of p, q, and r to obtain that p, q, and r

commute. Then apply the Foulis-Holland Theorem to obtain

D(p,q,r).



1.2.13 Lemma. Let B be a Boolean lattice, let xEB, and

let a be an atom of B. Then either a s x or x s a'.









Proof. a = aAl = aA(xvx') = (aAx)v(aAx'). Since

a is an atom of B, 0 < aAx < a and 0 s aAx' s a. If both

aAx = 0 and aAx' = 0, then a = 0 which is a contradiction.

Hence aAx = a or aAx' = a. If both obtain then a S x and

a x', hence a = 0 which is a contradiction. Hence

exactly one of a x, x s a' obtains.



3. Morphisms, Ideals, Filters,and Sections


1.3.1 Definition. Let P and Q be posets. Let M c P

and N c Q. Then a mapping f from M into N is said to be

an order homomorphism in case x,yEM and x s y imply that

xf t yf. If f has the property that, for all x,yEM, xvy

exists in M implies xfvyf exists in N and (xvy)f = xfvyf,

then f is called a join homomorphism. A meet homomorphism

is defined dually. If f is both a join homomorphism and a

meet homomorphism, then f is said to be a join-meet

homomorphism.

If M and N are sub-orthocomplemented posets of P

and Q, respectively, then a function f:M -> N is said to

be an ortho-homomorphism in case x'f = (xf)'. If f has the

property that, for all K c M such that sup K exists and is

in M, sup (Kf) exists and (sup K)f = sup (Kf), then f is

said to be a complete join homomorphism. A complete meet

homomorphism is defined dually. A homomorphism which is

both a complete join homomorphism and a complete meet








homomorphism is said to be a complete join-meet

homomorphism.

If f is a homomorphism of any type, then f is
-1
said to be a monomorphism from M to N in case f :Mf -> M

is a homomorphism of the same type as f, f is said to be an

epimorphism from M to N in case Mf = N, and f is said to be

an isomorphism from M to N in case f is both a monomorphism

and an epimorphism.



1.3.2 Lemma. Let (LLl1) and (L2, 2) be orthomodular

lattices, let S, and S2 be sub-orthomodular lattices of L

and L2, respectively, and let 9:S1 S2 be an order ortho-

isomorphism. Then 9 is a join-meet ortho-isomorphism. If,

in addition to the above hypotheses, L1 and L2 are complete

lattices and S1 and S2 are subcomplete sublattices, then 9

is a complete join-meet ortho-isomorphism.

Proof. We assume that Sl and S2 are subcomplete

and prove that 8 is a complete join-meet ortho-isomorphism.

The proof of the other statement is merely a simplification

of the proof given.

Let M c Sl, m = sup1 M, and n = sup2 (MG) (where

supi indicates the supremum in Li). Since S1 and S2 are

subcomplete, mESL and nES2; hence there exists wES1 such

that w8 = n. We must show that w = m. From x 5l m for all

xfM, it follows that x8 <2 m6 for all x8EM9; hence

suP2 (MS) = n = we <2 me, and therefore w 9l m. But from








x8 S n = we for all xARM9, it follows that x sl w for all

xEM; hence sup1 M = m i1 w. Therefore w = m and conse-

quently 8 is a complete join homomorphism. By a dual

argument, 9 is a complete meet homomorphism. A similar

argument proves that -1 has these properties; the result

follows.



1.3.3 Definition. A subset I of an orthocomplemented

poset P is said to be an order ideal of P if and only if

mEI and n < m imply nEI. A subset F of P is said to be an

order filter of P if and only if mEF and m S n imply nEF.

If I (resp., F) is an order ideal (resp., order

filter) on P with I # P (resp., F X L) then I (resp., F) is

said to be a proper order ideal (resp., proper order

filter) on P. If I = [0,1) or I = P (resp., F = {0,1) or

F = P), then I (resp., F) is said to be a trivial order

ideal (resp., trivial order filter). An order ideal I

(resp., order filter F) is said to be a principal order

ideal (resp., principal order filter) in case there exists

xEP such.that I = P(0,x) (resp., F = P(x,l)).



Note. Let I & L. I is an order ideal of P if

and only if the set F = (xEL:x'EI) is an order filter of P.


The following terminology is not standard.








1.3.4 Definition. A subset S of the orthocomplemented

poset P is said to be a section of P in case S = IUF where

I is an order ideal of L, F is an order filter of L, and

xEI if and only if x'EF. If S = L(x,l)UL(0,x'), then S is

called a principal section of P and S is denoted by S x



1.3.5 Remark. Let P be an orthocomplemented poset.

Then every section of P is the union of principal sections

of P.

Proof. Let S be a section of P, then S = IUF for

some order ideal I and some order filter F. Then

S = U[Sx:xEI).



1.3.6 Lemma. Let P be an orthocomplemented poset, and

let S be a section of P. Then S is closed under the ortho-

complementation of P.

Proof. By Remark 1.3.5 we need only show that

any principal section S is closed under the orthocomple-

mentation of P. We may assume x $ 0. If yESx, then either

x s y or y x'. If x y, then y' : x' so that y'ES If

y s x', then x & y' so that y'ES Hence S is closed

under the orthocomplementation of P.



1.3.7 Remark. If Sx is a principal section of an

orthomodular lattice L, then S is a sub-orthomodular

lattice of L.





20


Proof. By Lemma 1.1.7 we need only prove that

the join (in L) of any two elements in Sx is again in Sx.

But this follows immediately from the definition of S
x












CHAPTER 2


THE PASTE JOB


1. New Lattices from Old

2.1.1 Convention. Throughout this chapter we assume
4t +
that (Ll,:, #) and (L2,2, ) are two disjoint orthomodular

lattices, and that S. c L. (i = 1, 2) are such that:

(1) [0,1) c Si 4 Li (i = 1, 2);

(2) S. is closed under the orthocomplementation of

Li (i = 1, 2);

(3) there exists e:S1 -> S2 such that 8 is an order J

ortho-isomorphism.



The ambiguity of notation generated by using the

symbol S1 for the principal section and for the subset of

L1 will be resolved by the context in which the symbol

appears. Unless otherwise stated S1 will denote the subset

of L1 given in the above convention.


2.1.2 Definition. (1) Let L0 = L UL2.

(2) Let P1 = ((x,y)EL0XLO:y = x)].

(3) Let A = {(x,x):xEL)}.
(4) Let P = APUP-1
(4) Let P = AUPlUPI
1 1*








2.1.3 Proposition. P is an equivalence relation on L0.

Proof. (i) P is reflexive since A c P.

(ii) P is symmetric by Remark 1.1.3.

(iii) To show that P is transitive, we must prove
that PP c P. We first note that PIP1 = since the range

of P1 is disjoint from the domain of Pl. For a similar

reason P 1 = (. We also note that PP11 = AIS and that

PI Pl = A S ; both identities are immediate from the
2
definition of PV. We now compute:
-1 -1
PP = (AUPLUP )(AUP1UP1

= A(AUUPP )UPP(AUPlUPI1 )u1 (PAUPlI1
1 1 1 -1 -1
= AAUAP U(API )UP1AUPP 1UP1P1 UP1 AUP 1PUP-P 1
1 -1 -1 -1 -1 -1
1 1 1 u1 1p1 UP1 1 1uplp1
= AUP UP-1UPUP P UPP 1 UPI UP 1pUP-P1
-1 1 -1 -1 -1
= AUP1UP1 UPIPUPIP1 UP1I P1UPp1 p1

= (AUPUP l)UVUAlS UAIS Ub
1 2
-1
= AUPIUP1 P.
Hence P is transitive.


2.1.4 Definition. (1) Let L = L0/P.

(2) Let R1 = (([x],[y])ELXL:there exist x1E[x] anc

ylE[y] such that xl 1l Y13}
(3) Let R2 = (([x],[y])ELXL:there exist x2E[x] anc

y2E[y] such that x2 2 Y2).
(4) Let R = {([x],[y])ELXL:there exists [z]EL such

that ([x],[z])ER1UR2 and ([z],[y])ER1UR21.


d
d








2.1.5 Remark. R = (R UR 2)

Proof. This is an immediate consequence of

Definition 2.1.4 part (4) and the definition of (RIUR2)


2.1.6 Lemma. (i) If y = x9, then xeS1 c L1 and

yES2 c L2.

(ii) If y = xG and z = x6, then y = z.

(iii) If y = x9 and y = z8, then x = z.

(iv) If [x]EL, then either [x] is a singleton set

whose only element is in L1-S1 or in L2-S2, or a

doubleton set consisting of an element xlES1 and

an element x2ES2 such that xl = x2.

(v) If xl,x2E[x], yly2E[y], x19 = x2, and yl1 = y2'

then the following are equivalent:

(a) xl 1 Y'

(b) x2 &2 Y2'

(c) y 1 x ,
+ +
(d) y2 2 x2'
Proof. (i) is clear because the domain of e is

contained in S1 which in turn is a subset of L1 and the

range of 0 is contained in S2 which in turn is a subset of

L2'

(ii) is clear because 0 is, among other things,

a function.


(iii) is clear because 0 is a monomorphism.








(iv) follows immediately from the fact that
-1
P = AUPUP-1
In (v), (a) is equivalent to (c) since is an
orthocomplementation for LI, and (b) is equivalent to (d)
+ .
since is an orthocomplementation for L2. Moreover (a) is
equivalent to (b) since 0 is an order isomorphism. Hence

all four statements are equivalent.


2.1.7 Notation. If [z]EL is such that ([x],[z])ER1UR2

and ([z],[y])ERlUR2, then we say that [z] implements

([x],[y])ER and we write [z]:[x]R[y].
For i,jEl1,2), if [z]EL is such that ([x],[z])ERi
and ([z],[y])ERj, then we write [z]:[x]R.iR[y] and we say

that [z] implements ([x],[y])ER.R..

In what follows [x] = {xl), [x] = (x2], and

[x] = lxl,x2) mean, respectively, that [x] = {xl c L -Sa ,

[x] = (x2] c L2-S2, and [x] = [(x,x2} where x1ES1, x2ES2,
and xl1 = x2. We will freely write [x1] for [x] whenever
xlE[x], and [x2] for [x] whenever x2E[x]. A subscript 1 or
2 will always denote the fact that the element in question

is in L1 or in L2, respectively. The home of unsubscripted

elements either will be made explicit or will be left

conveniently ambiguous if no confusion is generated by the
ambiguity.


2.1.8 Lemma. (1) If [z]:[x]R[y], [x] = (x1], and








[y] = [y2], then [z] = fzl,z2) and x 1 I zl'
z2 ;2 Y2'
(2) If [x]R[y], [x] = fxl,x2), and [y] = [ylY2],
then [x]:[x]R[y] and [y]:[x]R[y].
Proof. Ad (1). One of [z] = (zl], [z] = {z2'}
or [z] = (z1,z2] must obtain. If [z] = (zl), then
([z],[y])ER1UR2 which is impossible. If [z] = (z2], then
([x],[z])ERIUR2 which is impossible. Hence [z] = (zl,z2),
xl S1 Z1, and z2 2 Y2'
Ad (2). By the use of the order homomorphism 8,
it is easily shown that xl 1 y1 and x2 '2 Y2. The result
follows.


2.1.9 Proposition. R is a reflexive, anti-symmetric,
and transitive relation on L.
Proof. (1) R is reflexive since [x]:[x]R[x].
(2) To prove that R is anti-symmetric, assume
that [z]:[x]R[y] and [w]:[y]R[x]. We must show that
[x] = [y]. Note that if [x], [y], [z], and [w] all have
representations in L1, then the following obtains:

(I) 1 Z1 Z 1 y1 l w1 1 l
Hence x, = yl and [x] = [y]. Note also that if [x], [y],
[z], and [w] all have representations in L2, then the
following obtains:
(II) x2 s2 z 2 2 2 Y22 w22 x2
Hence x2 = y2 and [x] = [y].








One and only one of the following cases obtains:


Case (1) [x]

Case (2) [x]

Case (3) [x]

Case (4) [x]

Case (5) [x]

Case (6) [x]

Case (7) [x]

Case (8) [x]

Case (9) [x]

Cases (1) z
are Cases (3), (6),


= (x)l
= (xl)

= (xl)

= (x2)
= (x2)

= (x2)

= (xl,x23

= 1xl,x2)
= (Xl,x2]
nd (5), (2)

(7), and (8)


and

and

and

and

and

and

and

and

and

and (4)


[y] = IYl),

[y] = (y2),

[y] = (ylY2),
[y] = (y],

[y] = (y2),

[y] = (ylY21'
[y] = C(y)'

[y] = (y2),

[y] = [ylY2.
are symmetric, as


.Accordingly we only


consider Cases (1), (2), (3), and (9).

In Case (1), (I) obtains; hence [x] = [y].

In Case (2), by Lemma 2.1.8 part (1), we have
[z] = z11,22), [w] = [lW2)}, and x1 S1 Zl z2 2 2 Y2'

Y2 :2 w2' W 1 xl Hence wl 1 zl and z2 a2 w2. But,
since 0 is an order homomorphism, z2 S2 w2 = w 1 <2 z18 =22'

and so z2 = w2. Moreover, since 0 is a monomorphism,

z1 = w1. Hence xl 1 Zl = wl L xl and y2 92 w2 = z2 92 Y2

imply that xl = z and y2 = z2. Consequently

[x] = [xl] = [z1] = [z2] = [y2] = [y] which is the desired
result.

In Case (3), (I) also obtains which yields a con-
tradiction since no singleton equivalence class can equal a

doubleton equivalence class. Hence Case (3) cannot occur.








In Case (9) we may assume that [z] = [w] = [x]

(cf. Lemma 2.1.8 part (2)). Hence (I) occurs and [x]= [y].

Consequently R is anti-symmetric.

(3) To prove that R is transitive, we must show

that

(III) RR c R.
2
But R = (R UR22) by Remark 2.1.5, and hence we may compute

RR as follows:

RR = R(R UR2)2
2 2
= R(R2UR R2UR2R UR2)
2 2
= RR2URR1R2URR2R1URR2
21 2 2 2
= (RIUR2) 2R1U(R UR2) 2R1R2U( 2 RRRUR2 2R (R1UR2) R2
S2 2 2 2 2
= (R1UR1R2UR2RUR2) RU (R1UR1R2UR2R UR2)R1R2
112 21 21 11 2 1 22 2
U (R2UR1R2UR2RIUR2) R2RU (R2UR R2UR2RUR2)R2
4 2 2 22 3 2 2 2
= (RIUR R2R1UR2R UR2R1)U(R R2U(R R2) UR2RIR2UR2R1R2)
2 2 23 22 3 24
U (RR2RIUR1R2RIU (R2R) 2UR2R1)U(R1R2UR1R2UR2R R2UR) .

Hence (III) is equivalent to
4 2 3 2 2 3 2 2 2
(IV) R1UR R2R1UR2R1UR2R2UR1R2U (R R2) UR2R1R2UR2R1R2
2 2 2 3 2 2 3 2 4
UR1R2R1UR R2RU (R2R1) UR2R1UR2R2UR1R3UR2RR2U2R2
2 2
c R2UR1R2UR2R UR2.

To prove (IV) we consider 16 cases corresponding

to the 16 sets exhibited on the left hand side of (IV). In

each case we will show that the set in question is

contained in one of the sets on the right hand side of (IV).

Case (1). If ([x],[y]))ER, then there exist







[p],[q],[r]EL such that
([x],[p]),([p],[q]),([q],[r]),([r],[y])ER1. Hence there
exist xlE[x], plE[p], qlE[q], rlE[r], and ylE[y] such that
xi 1l pi 1 q1 1 r 1 Y which implies x 1 1q, Y1
Therefore ([x],[q]),([q],[y])ER1, [q]:[x]R2[y], and
([x],[y])ER2.
Case (2). If ([x],[y])ER1R2R, then there exist
[p],[q],[r]EL such that ([x],[p])(R1, ([p],[q])ER2, and
([q],[r]),([r],[y])ER.. By definition of R1 and R2,
[P] = (p',p2}' [q] ={qlq2], x1 '1 p1' P2 P 2 q2, and
q91 1 rl 1 YI Hence (via 0) p, 1 ,1 and therefore
xl1 Sp 1 1 1 q~i 1 rl1 ~ Yl which implies that [p]:[x]R2[y],
i.e., ([x],[y])R21.
Case (3). If ([x],[y])ER2R3, then there exist
[p],[q],[r]EL such that ([x],[p])ER2 and
([p1,[q]),([q],[r]),([r],[y])ER1. Hence [p] = fplP2]'
x2 &2 p2, and p, &l q1 11 r1 11 y1. Therefore
[p]:[x]R2Rl[y], i.e., ([x],[y])ER2R1.
22
Case (4). If ([x],[y])ER2R1, then there exist
[p],[q],[r]EL such that ([x],[p]),([p],[q])ER2 and
([q],[r]),([r],[y])ER Hence [q] = (ql,q2,
x2 2 P2 '2 q2, and q1 !1 rl 1 y which implies that
[ql:[x]R2R[y].
By arguments similar to those used in Cases (1) -
(4) we may conclude the following:








Case (5). If ([x],[y])ER R2, then there exists
[r]EL such that [r]:[x]R1R2[y].
Case (6). If ([x],[y])E(R1R2)2, then there
exists [r]EL such that [r]:[x]R1R2[y].

Case (7). If ([x],[y])ER2R2R2, then there exis
[r]EL such that [r]:[x]RRy].

Case (8). If ([x],[y])ER2R1R2, then there exis
[r]EL such that [r]:[x]R[y].
For n = 1, 2, .... 8. Case 16-(n-l) is obtained
from Case n by interchanging the subscripts 1 and 2.
Consequently R is transitive.


ts


ts


2.1.10 Notation. Since, by the preceding lemma, R is a
partial ordering on L, we write [x] A [y] whenever

([x],[y])ER. In particular, we will henceforth write
[z]:[x] S [y] for [z]:[x]R[y].
Moreover, in what follows, when considering a
relation such as [z]:[x] s [y], the statement "we may
assume that there exist xl6[x], zlE[z] such that xI sL1 Z"
will sometimes be abbreviated to the statement "we may

assume x 1 1 1 z"; also in case analyses the abbreviated
form "xI !1 z1" means "there exist x E[x], z E[z] such that

x1 1 zl*


2.1.11 Definition. Let 0. (resp., 1i) be the zero (resp.,
unit) element of Li (i = 1. 2). Define [0] to be [01]








and [1] to be [11]. Define ':L -> L by the following:

l[x= I[x4] if there exists x1EL1 such that x1E[x].
[x]=
[x2] if there exists x2EL2 such that x2E[x].


2.1.12 Notation. We henceforth write 0. and 1.
1 1
(i = 1, 2) as 0 and 1, respectively. When euphony demands,
the notation {0,1)i will be utilized to denote {Oi,li]. In
other occurrences, if a distinction is necessary it will be
clear from the context.


2.1.13 Lemma. [0] and [1] are the zero and unit
elements, respectively, of L.
Proof. If [x]EL, then [x] = [xl] or [x] = [x2].
If [x] = [xl], then [0] a [x] since 0 '1 xl; if [x] = [x2].
we may similarly conclude that [0] < [x]. Hence [0] is the
zero element of L. Dually, [1] is the unit element of L.


2.1.14 Proposition. ':L -> L is a well-defined ortho-

complementation for the poset L.
Proof. ':L -> L is well-defined. For, if

[x] = (x1}, or if [x] = (x2], then there is no ambiguity in
the definition of [x]' which is (x#] or fx +, respectively.
If [x] = {xlx2}, then, since x 9 = x, [x]' is unambiguous.
Now suppose that [x] S [a] and [x]' s [a].
Assume [z]:[x] [a] and [w]:[x]' a [a]. we may assume








that xl 1 zl and consider only the following eight

possibilities:


Case

Case

Case

.Case

Case

Case

Case

Case





x 1 r

[a] =



Xl 1

[a] =



xl 1l
hence


Xl 1 z1

[a] = [1]


(1) zl 1

(2) zl C

(3) z1 1~

(4) z1 ~i

(5) z2 i2

(6) z2 !2

(7) z2 '2

(8) z2 62

Ad (1).

Ad (2).

Z I al and

[1].

Ad (3).

z1 1 al and

[1].


al,
al,

al,

al,

a2,

a2,

a2,

a2,

Xl 1
[w] =

1


[x] =

x 4
1'1


x#1 ~1 Wl wl ~ al'

x I1 wl' w2 '2 a2'

x2 :2 W2' w1 ~l al'

x2 i2 2' w2 22 a2'

x S1 W1' W1 1 al'

X1 15 Wl w2 2 a2'
x2 22 w2 Wl S1 al'

x2 r2 w2' w2 2 a2'

al and x S1 al imply [a] =

(wl,w2] and [a] = (al,a2).

W1 1 al. Therefore al = 1



[x1,x2] and [w] = {w1,w2).

w1 1 al' Therefore al = 1


Ad (4). [x] = [xl,x2) and [a] = {al,a2).


[1].

Hence

and



Hence

and



Hence


1 al which implies x2 2 a2. But x2 2 a2 and
= 1. Consequently [a] = [1].

Ad (5). [a] a= (a,a2) and [z] = (z1,z2). Hen<

1 al and x#1 1 w al. Therefore al = 1 and

S.


Ad (6). [z] = [zl,z23 and [w] = {w1,w2}. Hence

xl i z1 which implies z1 ~1 x1 1 wl and therefore


ce







+
Z2 !2 w2 2 a2. But z2 2 a2 and hence a2 = i.

Consequently [a] = [1].

Ad (7). [x] = [(xX21}, [w] = (w1,w2},

[z] = [21,22], and [a] = fal,a2). Hence xL l Z1 '1 al and
x 1 W al. Therefore al = 1 and [a] = [1].

Ad (8). [x] = [Xl,x2) and [z] = (21,22]. Hence

x2 2 z2 a2, x2 2 w2 w 2 a2. Therefore a2 = 1 and

[a] = [1].

Hence [x]v[x]' exists and equals [1]. Dually,

[x]A[x]' exists and equals [0].

Clearly ':L -> L is an involution since
#:L--> L and +:L2-> L2 are involutions. The verifi-

cation that ':L -> L is anti-automorphic is immediate.

For if [z]:[x] s [y], then we may assume xl 1 zl and we

need only check the cases z2 1 y1 and z22 2 y2. In the

former case xl1 S y implies y# al xl; hence

[y]' s [x]'. In the latter case y 2 4z2 and z# X
imply that [y]' [x]'. Hence ':L -> L is an ortho-

complementation for the poset L.


We now assume that the S. are proper sub-
1
orthomodular lattices of L.. Then, by Lemma 1.3.2. the

order ortho-isomorphism 9 becomes a join-meet ortho-

isomorphism. This allows us to make statements such as

the following: if [x] = [xl,x2} and [y] = (y1.Y2}, then







(x1 V1 Yl)8 = x2 v2 Y2 and (x1 A1 Y1) = x2 A2 Y2' i.e.,

[x V1 y1] = [x2 V2 y2], and [xl A1 y1] = [x A2 Y2 ]


2.1.15 Lemma. Let 0O,1]i Si < Li (i = 1, 2). Let
[x]ES and [y]ES so that [x] = (x1.x2] and [y] = (ylY21}.
Then the following obtains:
(1) [x]v[y] exists in L. and
[x]v[y] = [xl V1 y1] = [x2 V2 y2];
(2) [x]A[y] exists in L. and
[x]A[y] = [x1 1 yl] = [x A2 Y2].
Proof. Let [x] = [x1',x2 and [y] = {y1,Y2}.
Then, since Si is a sublattice of Li (i = 1, 2), x ,ylES1
implies xl v1 YlES1; x2'Y2ES2 implies x2 v2 Y2ES2.
Moreover (xl V1 y1)8 = x1a v2 y18 = x2 v2 Y2' hence
x1,'Y ~1 x1 V1 Yl implies [x],[y] ( [xl V1 y1] = [x2 V2 y2].
Now suppose that [x],[y] < [z]. The following
statements are valid regardless of the composition of the
elements of L which implement [x] : [z] and [y] [z].
If [z] = (zl] or if [z] = [(1,22], then xl A1 21
and y1 '1 z1. In this case it follows that x, v1 y1 l zl'
and hence [xl v1 yl] [z].
If [z] = (z2], then x2 2 z2 and Y2 2 z2. In
this case it follows that x2 v2 Y2 2 z2' and hence

[x2 V2 y2] < [z].







We have shown that, for [x],[y]ES,
[x],[y] C [Ix V1 y1] = [x2 V2 y2], and if [x],[y] i [z]
then [xl VI y1] S [z]. Hence we conclude that [x]v[y]
exists and equals [xl v1 y1].


2.1.16 Lemma. Let [0,1]i S Si < Li (i = 1, 2). Let
[x]ES and [y]hL-S, then [x] = fx1,x2] and either [y] = [y1]


or [y] = (y2].
(1) If [y] = [yl), then [x]v[y]
[xv[y'] = [xI v1 y1];
(2) if [y] = [y21, then [x]v[y]
[x]v[y] = [x2 v2 Y2];
(3) if [y] = [y)1, then [x]A[y]
[x]A[y] = [x 1 A1 Y];
(4) if [y] = [y23, then [x]A[y]
[x]A[y] = [x2 A2 Y2].


exists in L and


exists in L and


exists in L and


exists in L and


Proof. Ad (1). [x] = [x,,x2) and
x1Y1 1 xl 1 V1 Y1 imply that [x],[y] C [xI v1 y1]. Now
assume [x],[y] a [z]. Let these inequalities be imple-
mented by [u] and [v], respectively. If [z] = [zl) or if
[z] = (zl,z2), then xl ~1 z1 and yl '1 z1; hence
xl V1 Y ~1 Z1 which implies that [xl v, yl] s [z]. If
[z] = (z2}, then [v] = (v1,V2], y1 Y Vl' v2 2 z2, and
x2 <2 z2; xl v11 1 x1 V1 V1 and
(x1 V1 V1)8 = x2 v2 v2 '2 z2; hence [xl V1 yl] [z].







Hence [x]v[y] exists and equals [xl vI y1].
A similar argument proves (2). (3) and (4)
are dual.


2.1.17 Lemma. Let [0,1)i Si < Li (i = 1, 2).
(1) If [x] = (xl] and [y] = (yl), then [x]v[y] exists
in L and [x]v[y] = [xl VI y1].
(2) If [x] = (x2) and [y] = (y2), then [x]v[y] exists
in L and [x]v[y] = [x2 v2 y2].
(3) If [x] = (x1) and [y] = (y1), then [x]A[y] exists
in L and [x]A[y] = [x1 A1 y1].
(4) If [x] = (x2] and [y] = (y2), then [x]A[y] exists
in L and [x]A[y] = [x2 A2 y2].
Proof. We need only prove (1) since (2) follows
by symmetry; (3) and (4) are dual.
Clearly [x],[y] s [xl vI y1]. Now assume that
[x],[y] a [z] where [u] and [v] are the elements of L which,
respectively, implement the inequalities. Then both [u]
and [v] have representatives in uI and vl, respectively, in
L1. x1 I1 u1 and Y1 1 v1 imply xl VI yl al ul Vl vl. If
[z] = (zl) or if [z] = {zl,z2), then clearly
[u1 V1 V1]:[x1 V1 y1] < [z]. If [z] = [z22, then
[u] = [ulU2], [v] = v1,v2)], and u2 V2 v2 S2 z2; hence
[u1 v1 V1]:[x1 vI y1] [z]. Consequently, in this case,
[x]vly] exists and equals [xl v1 y1].








2.1.18 Definition. Let S1 < L1, and S2 < L2. Then we
say that S1 and S2 are corresponding sections of L1 and L2

if and only if there exist Mi c Si (i = 1, 2) such that
(1) M1B = M2'
(2) S1 = S # and S2 = S and
mEM1 mEM2 m

(3) 91Sm :Sm#aS(m )+ is an ortho-isomorphism for each

mEM1.


Warning. Although a section need not, in general,

be a sublattice, in order for S1 and S2 to be corresponding

sections of L1 and L2, S1 and S2 must be sub-orthomodular
lattices of L1 and L2, respectively.


2.1.19 Lemma. Assume that S1 and S2 are corresponding

sections of L1 and L2. Under this assumption if [x],[y]EL

are such that [x] = (xl] and [y] = (y2), then [x] % [y].
Proof. Suppose the statement is false. Then

there exists [z] such that [z]:[x] < [y], [x] = {x)},

and [y] = (y23. It follows that [z] = [zl,z2), x1 L1 z'

z2 2 Y2, zlES1, and z2ES2. Hence z ESm# and z2ES(e)+

for some mEM1, so that either m# 5 zl or z1 !5 m. If
m# :1 zIl then (m) + 2 z 2 Y2. Therefore Y2ES(Me)+ S2

and hence [y]ES, which is a contradiction. If z1 ~1 m,








then xl 1 21 Z 1 m; and consequently x6ESm# C S i.e.,

[x]ES. This is a contradiction. Therefore no such [z]
exists and [x] [y].


2.1.20 Corollary. Assume that S1 and S2 are
corresponding sections of L1 and L2. Under this assumption
if [x],[y]EL are such that [x] = (xl1 and [y] = {y21, then

U([x],[y]) c S.
Proof. Suppose there exists [u]EU([x],[y]) such
that [u]OS. Then either [u] = {u1} or [u] = (u21. If

[u] = (ul), then [y] [u], contradicting Lemma 2.1.19. If
[u] = (u2), then [x] [u], again contradicting
Lemma 2.1.19. Hence no such [u] exists and U([x],[y]) c S.


2.1.21 Proposition. Let S1 and S2 be corresponding
sections of L1 and L2. Then L is an orthomodular poset.
Proof. By Proposition 2.1.14 L is an ortho-
complemented poset. Hence we need only show that
(1) if [e],[f]EL and [e] __ [f], then [e]v[f] exists,
and
(2) if [e],[f]EL and [e] s [f], then

[f] = [e]v([f]'v[e])'.
Ad (1). If [e] _L [f], then [e] < [f]'. If
there exists elE[e], then by Lemma 2.1.19 there exists
flE[f], and by one of Lemmata 2.1.15, 2.1.16, or 2.1.17,








[e]v[f] exists (and equals ([e vl f1]). If [e] = (e2'}

then by Lemma 2.1.19, there exists f2E[f] and hence by one

of Lemmata 2.1.15, 2.1.16, or 2.1.17, [e]v[f] exists (and

equals [e2 V2 f2]).

Ad (2). Assume that [x] s [y]. We will prove

that [y] = [x]v([x]v[y]')'. We need only consider two

cases:

Case (i) [x] and [y] both have a representative in L1.

Case (ii) [x] = (xl) and [y] = (y2).

Ad (i). xl 1 yl and hence, by Lemmata 2.1.15,

2.1.16, and 2.1.17, since L1 satisfies the OMI,

[x]v([x]v[y]')' = [xl]v([xl]v[y ])' = [xl]V[x1 V1 y ]
= [X ]V[x# A1 y1] = [y1] = [y].

Ad (ii). By Lemma 2.1.19 this case cannot obtain.

Since the OMI is satisfied in all consistent

cases, L is an orthomodular poset.


2.1.22 Theorem. Assume that S1 and S2 are corresponding

sections of L1 and L2. Moreover, assume Li is complete and

that Si is a subcomplete sub-orthomodular lattice of Li

(i = 1, 2). Then L is an orthomodular lattice.

Proof. Since L is an orthomodular poset, to show

that L is an orthomodular lattice we need only show that

[x]v[y] exists for all [x],[y]EL. Let [x],[y]EL. Exactly

one of the following must occur:








(1) [x],[y]ES,
(2) [x]ES, [y]EL-S (re-label [x] and [y]
if necessary),

(3) [x],[y]EL-S.
Ad (1). By Lemma 2.1.15 [x]v[y] exists.
Ad (2). By Lemma 2.1.16 [x]v[y] exists.
Ad (3). There are four cases:
Case (i) [x] = {xL) and [y] = (yl},
Case (ii) [x] = (x1) and [y] = [y21,
Case (iii) [x] = [x2} and [y] = {yl,'
Case (iv) [x] = {x2) and [y] = (y2].
By symmetry we need only consider Cases (i) and
(ii).
Ad (i). By Lemma 2.1.17 [x]v[y] exists.
Ad (ii). First note that U([x],[y]) is non-empty
since it contains [1], and that, by Corollary 2.1.20,
U([x],[y]) c S. Let M1 = {zl:there exists [z]EU([x],[y])
such that [z] = (zl,z2]}. Now inf, M1 (as computed in LI)
exists since L is complete. But since S is subcomplete,
inf M,, as computed in S1, exists and equals infl Mi. Let
z() = infI M1. Let z(2) = z(1), let [z]0 = [z(1)

= [z(2)], and let M2 = fz2:there exists [z]EU([x],[y]) such
that [z] = {z1,z2]}. Then by Lemma 1.3.2
z(2) = inf2 (zle:zlEM1] = inf2 M2 = inf M2, as computed in

S2 since S2 is subcomplete.








We claim that U([z]0) = U([x],[y]).

Xl 1 infl ML = z(1) and y2 r2 inf2 M2 = z(2)

[x],[y] [)]0. Hence U([z]0) c U([x],[y]).

[z]EU([x],[y]), then by Corollary 2.1.20 [z]

z(1) 1 z1, and z(2) 2 z2; hence [z]0 < [z],

[z]EU([z]0). Therefore U([z]0) = U([x],[y]).
L is an orthomodular lattice.


2.1.23 Example. If

need not be a lattice.

be the Boolean lattice


For,

imply that

Moreover, if

= {z1,z2],

i.e.,

Consequently


Si is not a section of Li, then L

For example, for i = 1, 2, let Li

given in Figure 2.


Li = Xi


Figure 2


Let Si = [O,l,xi,x). Then Si is not a section of Li.

Form L as in Definition 2.1.4. (The quotes about ai, bi,

mi, and ni indicate that ale 4 a2, bl0 / b2, m10 Y m2, and

n18 0 n2 in violation of Notation 2.1.7.) Then

["al"]v["b2"] does not exist since

(["m"],["n2"],[l] ) = U(["al" ["b2"]) has no smallest element.








2.1.24 Example. The following example illustrates the

fact that if the sub-orthomodular lattice Si is not a sub-

complete sub-orthomodular lattice of the complete ortho-

modular lattice L. (i= 1, 2), then L need not be a lattice.

Let L be the power set of an infinite set M, and

let Sl be the sub-orthomodular lattice of L1 consisting of

all finite or co-finite subsets of L1. (Recall that a

co-finite subset of M is a subset of M whose complement in

M is finite.) Let L2 be a disjoint "copy" of L1. Then

there exists a natural ortho-isomorphism c:L1"L2. Let

9 = cpS and let S2 = S A. Then e:SI S2 is an ortho-

isomorphism mapping Sl onto S2. Clearly S1 and S2 are

corresponding sections of L1 and L2. Moreover, Li is

complete, but S. is not a subcomplete sub-orthomodular

lattice of L. (i = 1, 2). Form L as ir Definition 2.1.4.

To show that L is not a lattice, consider two

subsets X and Y of M which are neither finite nor co-finite

and such that XUY is not co-finite. Let x be the element

of Ll which corresponds to X and let y be the element of L2

which corresponds to Y. Then xCL1-S1 and yEL2-S2. Suppose

there is a [z]eL such that [z] = [x]v[y]. Then by

Corollary 2.1.20, [z]cs, and hence [z] = (zl,z2}. Let Z be

the subset of M corresponding to zI (and z2). Since

XUY c Z, Z is a co-finite subset of M. But, since XUY is

not co-finite, there exists wEZ-(XUY) such that Z-(w} = XUY.









Let V = Z-([w and let v be the element of L1 which

corresponds to V. Then [v]EU([x],[y]) but [z] ; [v]. This

contradicts the fact that [z] = [x]v[y]. Hence [x]v[y]

does not exist and L is not a lattice.



The poset L given in the above example is of

interest for another reason. By Proposition 2.1.21 L is an

orthomodular poset. Hence L is an example of an ortho-

modular poset which is not a lattice. The only other known

example of such a poset is a finite poset given by

M. F. Janowitz in [5].


2.1.25 Example.

Hasse diagrams:


S





L

a b



0
Figure 3


Let L1 and L2 be given by the following




1



c' d' e'

SL2

Figure e


0
Figure 4


Let S1 = Sc, let S2 = S and let ce = x (thereby deter-
1 c' X x'
mining 9). Apply Theorem 2.1.22 to L1, L2, Sl, S2, and 8

(noting that all of the hypotheses are satisfied) to obtain








the following orthomodular lattice (for simplicity of

notation, we write z for [z]).


1


a' b'# c d# e.

Lo




0
Figure 5



This lattice will be used in the following proposition.



2.1.26 Proposition. Let the hypotheses of Theorem

2.1.22 be satisfied, then the lattice L of the conclusion

of Theorem 2.1.22 is complete. Moreover, if L1 and L2 (of

that theorem) are atomic, then L is atomic and the atoms of

L are of the form [m] where m is an atom of either L1 or

L2; however, not every element of L of the form [x], where

x is an atom of L1 or L2, need be an atom of L.

Proof. To show that L is complete, let M be any

subset of L. Let N = ([x]EM:there exists x E[x]], let

N1 = (x:xEL1 and [x]EN), let n = sup1 N1 (the supremum

exists since L1 is complete), let P = M-N, let







P2 = [x:xEL2, [x]EM, and [x] = [x2}), and let p = sup2 P2
(the supremum exists since L2 is complete).
We claim that sup N exists and equals [n]. It is
clear that [n] a [x] for all [x]EN. Assume that [z] a [x]
for all [x]EN. If there exists zlE[z], then z1 ~1 xl for
all xEN1; hence, in this case, z1 a1 n so that [z] a [n].
If [z] = (22), then for each xEN1 there exists
[w]x = (w,w2] such that z2 a2 w and wx a1 x. Hence
wx a1 n so that [w] 2 [n]. Consequently sup N exists and
equals [n].
We now claim that sup P exists and equals [p].
It is clear that [p] a [x] for all [x]EP. Assume that
[z] a [x] for all [x]EP. If there exists z2E[z], then
z2 !2 x for all xEP2; hence, in this case, z2 a2 p so that
[z] a [p]. If [z] = (zl), then, for each xGP2, there
xx x x
exists [w]x = [W1,W2 such that z1 a w1 and w2 a2 x.
Hence wx a2 p so that [z] a [p). Consequently sup P exists

and equals [p].
Since L is a lattice [n]v[p] exists in L. We
claim that [n]v[p] = sup M. Clearly [n]v[p] a [x] for all
xEM. If [z] a [x] for all [x]EM, then [z] a [n], [z] a [p],
and therefore [z] a [n]v[p]. Consequently sup M exists and
equals [n]v[p]. Since L is orthocomplemented, inf M also
exists. Therefore L is complete.
To show that L is atomic provided L1 and L2 are








atomic, let [x] be any non-zero element of L. Then we may

assume there exists a representative, xl, of [x] from L1.

Since L1 is atomic, there exists an atom a EL1 such that

al i x1. Either [al] is an atom of L or there exists

[y]EL such that [0] < [y] < [all. Assume the latter, then

[y] = [y2); and since L2 is atomic, it follows that y2
dominates some atom b2 of L2. Then

[0] < [b2] [y] < [al] [x]. Now [b2] is an atom of L,

for, if it were not then either al or b2 would fail to be

an atom in L1 or L2, respectively. Hence every non-zero

element of L dominates some atom [m] of L where m is either

an atom of L1 or an atom of L2.

To show that not every element of L of the form

[x], where x is an atom of L1 or L2, need be an atom of L
consider the following lattices:


Figure 6








Let L1 be the (Boolean) lattice given in Figure 6,

let S1 = (0,1,p,p',k,k',h,h'}, let L2 be the (orthomodular)

lattice given in Figure 5, let S2 = (0,l,a,a',b,b',c,c'),

and let

9 = {(0,0),(1,1), (a,p'),(b',k'), (c',h') ,(a',p),(b,k) ,(c,h)].

Apply Theorem 2.1.22 to obtain the lattice given in Figure

7. (Change the notation by dropping the brackets and

representing all two-element equivalence classes by the

representative in L1.)



1


Figure 7








Now [p'] = [a] is not an atom of L but a is an atom of L2.

Hence not every element of L of the form [x], where x is an

atom of L1 or L2, need be an atom of L.



2. A Partial Converse


2.2.1 Remark. We have shown that if an equivalence

relation is defined on the union of two disjoint ortho-

modular lattices in such a way that elements of isomorphic

sections (of a certain type) are equivalent to one another,

then, upon "dividing out" this equivalence relation, a new

orthomodular lattice is obtained. (We pictorially think of

the elements of one section as being "pasted" to the

corresponding elements of the other section.)

We now prove a partial converse of Theorem 2.1.22.

We maintain Convention 2.1.1 and make the further

assumption that L (cf. Definition 2.1.4) is an orthomodular

lattice. After obtaining some preliminary more general

information, we make the additional assumption that L1 and

L2 are complete Boolean lattices.

The preliminary information necessary (cf. Lemma

2.2.2, Corollary 2.2.3, and Lemma 2.2.4) is concerned with

comparabilities which may not obtain and restrictions on

those that may. These observations form the crux of our

main arguments.







2.2.2 Lemma. Assume that L is an orthomodular lattice
and that Si < Li. Let [z],[x],[y],[u]EL be such that
[z] s [x],[y] and [x],[y] a [u].
(1) If [x] = (x1) and [y] = [y2], then [u]Es.
(2) If [x] = [x2) and [y] = (yl}, then [u]ES.
(3) If [x] = fxli and [y] = {y2}, then [z]ES.
(4) If [x] = [x2) and [y] = [yl), then [z]ES.
Proof. (2) follows from (1) by symmetry; (3) and
(4) are dual to (1) and (2), respectively. Hence we need
only prove (1).

Suppose that (1) is false. Then either [u] = {(ul
or [u] = (u21 obtains. We may assume that [u] = 1ul}. Now,

u1 ! X1' and there exists [w]ES such that ul 1~ w1 and

w2 2 Y2. Let [a] = [u]A[y]', then [a]' = [y]v[u]'. Since
L is an orthomodular lattice and

(I) [y] 5 [w] < [u],
by the OMI and DOMI we have

(II) [u] = [y]v([u]A[y]') = [y]v[a]
and

(III) [y] = [u]A([y]v[u]') = [u]A[a]'.
Moreover, because of (I), by taking the infimum of both
sides of (III) with [w], we obtain

[y] = [w]A[y] = [w]A([u]A[a]')

= ([w]A[u])A[a]' = [w]A[a]',
i.e.,
(IV) [y] = [w]A[a]'.


,







If there exists alE[a], then, since Si < Li (cf. Lemmata
2.1.15 and 2.1.16), by (IV) we have

[Y2] = [l]A[al] = [w1 A1 al]. Hence [y]ES, contradicting
the fact that [y] = {y2). If there exists a2E[a], then,
since S. < Li (cf. Lemmata 2.1.15 and 2.1.16), by (II) we
have [u1] = [Y2]v[a2] = [Y2 V2 a2]. Hence [u]ES
contradicting the fact that [u] = f(ul.


2.2.3 Corollary. Assume that L is an orthomodular
lattice and that Si < Li. If x EL -S1 and Y2EL2-S2, then
[xl]v[y2]ES. If x2EL2-S2 and yfEL1-S1, then [x2]v[Y1]ES.
Proof. To prove the first statement let
[x1]v[y2] = [u] in Lemma 2.2.2. The second statement
follows similarly.


2.2.4 Lemma. Assume that L is an orthomodular lattice
and that Si < Li. Then there do not exist [a],[b]EL such
that both [a] s [b] and either
(1) [a] = (al), and [b] = {b2), or
(2) [a] = (a2], and [b] = (b1 .
Proof. By symmetry we need only prove (1).
Suppose the statement is false, then there exists [x]EL
such that [x]:[a] & [b], [a] = [al}, and [b] = fb2]. Then
[x]ES, and, since [a] t [x] s [b], by the DOMI we have
[a] = [b]A([a]v[b]'). Hence








[a] = [x]A[a] = [x]A([b]A([a]v[b]')) = ([x]A[b])A([a]v[b]')

= [x]A([a]v[b]'),

and therefore

[a] = [x]A([a]v[b]').

But by Corollary 2.2.3 [a]v[b]'ES and [x]ES. Hence [a]ES,

which is a contradiction. Therefore no such elements

exist.



Thus ends our preliminary results. We are now

prepared to deal with Boolean lattices. Henceforth, in

addition to Convention 2.1.1 and the assumption that L is

an orthomodular lattice, we assume that L1 and L2 are

complete Boolean lattices. In each of the following

results all assumptions augmenting Convention 2.1.1 will be

made explicit. With the exception of direct reference to

the lattices L1 and L2 of Convention 2.1.1, we will denote

a Boolean lattice by the symbol B.



2.2.5 Lemma. Let B be a Boolean lattice, and let T be

a proper sub-orthomodular lattice of B. If zEB is such

that B(0,z) c T, then there exists mEB(z,l) such that mT.

Proof. Suppose the statement is false. Then

there exists zEB such that B(O,z) c T and B(z,l) c T. Then

B(0,z') c T since T is a sub-orthomodular lattice of B.

Let xEB. Then x = (xAz)V(xAz') and consequently xET since








xAzET and xAz'ET. Therefore B c T, i.e., B = T, which

contradicts the fact that T is a proper sublattice of B.



2.2.6 Lemma. Let B be a complete Boolean lattice. Let

T be a subcomplete sub-orthomodular lattice of B. Let

M = (xEB:B(O,x) c T). Let c = sup M. Then cEM and, for

all xEB, B(O,x) c T if and only if x i c.

Proof. To prove that cEM, it suffices to show

that B(0,c) c T. Let a s c. Then

a = aAc = aA(sup M) = sup (aAm:mEM}. But, for all mEM,

(aAm)ET since aAm m and B(0,m) c T. Moreover, since T is

a subcomplete sub-orthomodular lattice of B, aET.

Consequently B(O,c) c T.

Now if xEB is such that B(O,x) c T, then x6M.

Hence x c. Moreover, if xEB is such that x 9 c, then

B(O,x) c B(O,c) c T.


2.2.7 Lemma. Let each Li of Convention 2.1.1 be a

complete Boolean lattice, let each Si be a subcomplete sub-

orthomodular lattice of L., let M. = {xEL.:L.(0,x) c Si),

and let c. = sup M. (i = 1, 2). Then

(1) Li(O,ci) c Si, i.e., ciEMi,

(2) Li(0,xi) c Si if and only if xi gi ci'

(3) xEM1 if and only if xe'M2, and

(4) c19 = c2.








Proof. (1) and (2) are immediate consequences of

Lemma 2.2.6.

To prove that if xEM1 then x6EM2, assume that the

statement is false. Then there exist xEM1 and mEL2(Ox9)-S2.

Also, by Lemma 2.2.5 there exists nL1I-S1 such that

x < n < 1. Then [m] < [x] < [n] which contradicts Lemma

2.2.4 since mEL2-S2, xES1, and nEL1-S1. Hence if xEM1,

then xE9M2. A similar argument utilizing 8-1 in place of 8

proves the converse. Hence (3) is proved.

Now by (3) and the fact that 9 is a complete join

homomorphism (cf. Lemma 1.3.2)

c1a = sup (m9:mEM1l = sup (m:mEM2) = c2. Hence (4) is

proved.



2.2.8 Lemma. Let each Li of Convention 2.1.1 be a

complete Boolean lattice, let each S. be a subcomplete

sub-orthomodular lattice of Li, let

Mi = (xELi:Li(0,x) c Si], and let ci = sup M (i = 1, 2).

Then the following statements are valid.

(1) If xES -S #, then there exists aEL1-S1 such
1
that a <1 x.

(2) If xESi-Sc#, then there exists bELL-S1 such

that x
(3) If xES2-S +, then there exists aEL2-S2 such
2
that a <2 x.








(4) If xES2-S +, then there exists bEL2-S2 such
2
that x <2 b.

Proof. Ad (1). Let xESI-Sc#, then x ; c. Hence
1
by Lemma 2.2.7 part (2) L(0,x) ; SL, i.e., there exists

aEL -S1 such that a <1 x.

Ad (2). Let xES1-S # and suppose that there does
1
not exist bEL1-S1 such that x <1 b. Then L(x,l) c S1 so

that L(0,x ) c S1. Hence by Lemma 2.2.7 part (2) x cl

and therefore xES #, which is a contradiction. The result
c1
follows.

Ad (3) and (4). These are proved by noting the

symmetry of the hypotheses.


2.2.9 Proposition. Let each Li of Convention 2.1.1 be

a complete Boolean lattice, let each Si be a subcomplete

sub-orthomodular lattice of Li, let Mi = xELi:Li(0,x) CSi],

and let ci =sup Mi (i = 1, 2). Then S1 = Sc # and S2 = S +.
1 2
Proof. By the symmetry of the hypotheses we

need only prove that S1 = Sc#. By definition of cl,


1 1
is false. Then there exists xES1-Sc#. By Lemma 2.2.8
1
part (1) there exists aEL1-S1 such that a <1 x. Moreover,

x8ES2-S +, and by Lemma 2.2.8 part (4) there exists bEL2-S2
2
such that x8 <2 b. Then [a] < [x] < [b], aEL -S1, xES1,

and bEL2-S2 which contradicts Lemma 2.2.4.









The following theorem is now immediate.

2.2.10 Theorem. Let (L1,c1, ) and (L2,2,+ ) be two

disjoint complete Boolean lattices each of cardinal number

strictly larger than two. Let S. < L. (i = 1, 2) be

subcomplete sub-orthomodular lattices of L such that there

exists an ortho-isomorphism 8:S -S2. Let L0 = L1UL2, let
-1
P1 = ((x,y)EL0XL0:y = x9), let P = AUPlUP. (By

Proposition 2.1.3 P is an equivalence relation on L0.)

Let L = LO/P. Then L is an orthomodular lattice if and

only if there exists clEL1 and c2EL2 such that S1 = Sc# and
1
S2 = Sc+
2
Proof. Assume that L is an orthomodular lattice.

Let M = (xELi:Li(0,x) c Si}, and let ci = sup Mi

(i = 1, 2). Then by Proposition 2.2.9 S1 = Sc# and
1
S2 = S +. Conversely, if S1 = S # and S2 = S +, then by
1 2
Theorem 2.1.22 L is an orthomodular lattice.



3. Variation on a Theme


In case S1 and S2 consist of only two elements

(0,1), L is said to be the horizontal sum (disjoint sum) of

L and L2, and L is written HS(L1,L2). The simplicity of

the operation HS allows us to make a more general

definition.


Definition. Let (L ,< ), aEI, be a family of


2.3.1








orthomodular lattices of cardinality larger than 2, indexed

by the set I, where I > 1, such that if a and B are

distinct elements of I, then LaniL = Let L0 = U L
cEI
let 1 and 0 denote the unit and zero, respectively, of

L Define a relation P1 on L0 by

P1 = ((x,y)EL XL0:there exist a,BEI such that x = 1a and

y = 1 or x = 0a and y = 0Og}U

where A = ((x,x):xELO). P, is clearly an equivalence

relation. Let L = LO/P. Then L is called the horizontal

sum (disjoint sum) of the lattices L written

L = HS(L :aEI).


2.3.2 Remark. If [x],[y]EL, then we write [x] a [y] in

case there exist representatives xE[x], yE[y], and aEI such

that x,yELa and x 9a y, where :a is the partial ordering in

L .
is easily seen to be a partial order for L;

moreover, for any aEI, [0 ] and [1 ] are the zero and unit,

respectively, for L. Since [0 ] and [1 ] are independent

of a, we write [0] for [0 ], [1] for [a ].

The orthocomplementations in La induce an ortho-

complementation in L defined as follows:

for [xa]EL, [x ]' = [x]

where x. is the orthocomplement of xa in La. Since, for

0 ,1 EL 0' = 1 and 1' = 0 and since all other
a a a a a a'








equivalence classes are singletons, ':L --> L is a well-

defined orthocomplementation for L.


2.3.3 Lemma. Let L = HS(L :aEI). Then L is an ortho-

modular lattice.

Proof. Let [x],[y]EL. Then an immediate conse-

quence of the definition of A is the fact that [x]v[y]

exists and

[x V y] if there exists aEI such that x,yEL .

[x]v[y] = [1] if there exists a,BEI such that a # 8,
xEL and yELB.

Since L is an orthocomplemented poset such that the

supremum of every pair of elements exists in L, it follows

that L is an orthocomplemented lattice. To prove that L is

an orthomodular lattice, let [x] s [y]. If [x] = [O],

[x] = [1], [y] = [0], or [y] = [1], then the computation of

the OMI is trivial. If none of these hold, then [x] and

[y] are singletons whose only representatives are x and y,

respectively, and there exists aEI such that x,yEL Hence

[x]V([y]A[x]') = [x]v([y]A[x']) = [x]v[y Aa X']
= [x V (y A x')] = [y]. Therefore L is an

orthomodular lattice.


2.3.4 Definition. Let P be an orthocomplemented poset.

Then a section S of P is said to be simple if and only if

S-(0,1l consists only of atoms and co-atoms of P.








The following theorem is a variation of Theorem

2.1.22. 1he function 6 is no longer required to be an

order homomorphism, and additional assumptions are made on

L1, L2, S1, and S2'

2.3.5 Theorem. Let Si be a simple section of the

orthomodular lattice Li (i = 1, 2), and let :SS1 2 be an

ortho-isomorphism. Assume that L1 = HS(L :aEI), for some

index set I, and that, for any aEI, S nL has cardinal

number 2 or 4. Moreover, assume that 06 = 0 and that xES1

and x an atom of L1 imply xG is an atom of L2. Define P,

R, L, and as in Definitions 2.1.2, 2.1.4, and 2.1.11.

Then L is an orthomodular lattice.

Proof. The proof that P is an equivalence

relation is exactly the same as in Proposition 2.1.3.

However, since 8 is not an order isomorphism, the proof

that R is a partial order is not the same as Proposition

2.1.9, but it follows the latter proof closely enough

that we shall, in general, only point out those parts which

are different. (In this proof the symbols (I), (II),

(III), and (IV) refer to the comparabilities bearing these

labels and appearing in the proof of Proposition 2.1.9;

moreover, all unidentified elements are defined in the

corresponding parts of the proof of Proposition 2.1.9.)

(1) R is reflexive since [x]:[x]R[x].

(2) To prove that R is anti-symmetric, assume







that [z]:[x]R[y] and [w]:[y]R[x]. We must show that
[x] = [y]. There are nine cases as listed in Proposition
2.1.9. We may assume [[x],[y]]n{[0],[l]} = b.
Case (1). [x] = (xl) and [y] = (yl). Since S1

is a simple section (I) obtains.
Case (2). [x] = (xl1 and [y] = (y2]. [z] must
be an atom or a co-atom of L. Each possibility is contra-
dictory.


Case (3). [x] = {xl) and [y] = (y1,Y2).
implies [y] is a co-atom and [y]R[x] implies [y] is

atom. Hence [x] = [y] (which in this case is a
contradiction).
Case (4). [x] = (x2) and [y] = tyll. As
Case (2) this is contradictory.

Case (5). [x] = (x2] and [y] = (y2). Sin
is a simple section (II) obtains.

Cases (6), (7), and (8) are resolved as in


[x]R[y]
an




in


ce S2


Case (3).
Case (9). [x] = (Xl,x21 and [y] = {ylY2l.

Suppose [x] # [y]. Then [x]R[y] implies xl is an atom but
not a co-atom of L1 and yl is a co-atom but not an atom of

L1; [y]R[x] implies y, is an atom but not a co-atom of L1.
This is a contradiction.
(3) To prove that R is transitive we must again
consider the sixteen cases provided by (IV). Cases (1),







(3), (4), and (5) are proved exactly as before. Case (2)
is resolved as follows: If [x] = [0], then[r]:[x]R2[y]. If
[x] = [y], then [x]:[x]R [y]. If [y] = [1], then
[p]:[x]R2[y]. Hence we may assume [0] t [x], [x] # [y],
and [y] # [1]. Then [x] = [p] or [p] = [y]. In the former
case [p] = [q] implies [q]:[x]R2[y], and [p] # [q] implies
[q] = [r] = [y] and [p]:[x]R R2[y]. In the latter case
[p] = [q] = [r] = [y] and [p]:[x]RR2[y].
In Case (6), if [x] = [0], [x] = [y], or
[y] = [1], then [r]:[x]R1R2[y]. Hence we may assume
[x] # [0], [x] X [y], and [y] ? [1]. Since [p] = {plP2,
[q] = (qlq2), [r] = (rl,r2), xl1 pi' P2 "2 q2' q1 '1 rl
and r2 !2 y2, it follows that [p], [q], and [r] are atoms
or co-atoms. Hence either [p] = [q] or [q] = [r]. If
[p3 = [q] and [p] is an atom, then [p] = [x] and
[r]:[x]R1R21[]; if [p] = [q] and [p] is a co-atom, then
[p] = [y] and [p]:[x]R1R2[y]. Hence we may assume
[p)] [q]. It follows that [x] = [p] and [q] = [r] = [y]
and therefore [q]:[x]R R2[y]. Consequently, in all cases,
([x],[y])ER1R2.
In Case (7) we may assume [x] X [0], [x] A [y],
and [y] A [1]. If [p] = [r], then [p]:[x]R2[y]. If
[p] [r], then [x] = [p], [r] = [y], and [q]:[x]R2[y].
Hence ([x],[y])ER2UR2 C R.
In Case (8) we may assume [x] 4 [0], [x] $ [y],








and [y] 4 [1]. If [p] = [r], then [p]:[x]R2[y]. If
[p] [r], then [x] = [p], [r] = [y], and [q]:[x]R2R1[y].
Since we have not made use of the hypotheses that
L = HS(L:a~EI) and that, for any aEI, (S nLa) = 2 or 4,
we may obtain Cases (9) (16) by symmetry of hypotheses.
Consequently L is a poset.
We must now prove that ':L -> L is a well-
defined orthocomplementation. That it is well-defined is
proved exactly as in Proposition 2.1.14.
Assume that [x],[x]' 9 [a]. We must prove that
[a] = [1]. Suppose that [a] # [1]. Then [x] # [0],[1].
If [a] = (al), then, since the only two-element equivalence
classes are [0], [1], and [m] where [m] is either an atom
or a co-atom of L, either [x] = (xl or [x] = (x1,x2). If

[x] = (x1], then we have xl,x 9l al and hence
[a ] = [a] = [1]. If [x] = {xlx2}, then x, is either zero
or an atom of LL; it cannot be zero since [x]' & [a], hence
xI is an atom of L1 and therefore x1 is a co-atom of L1
which contradicts the fact that [x]' < [a] = (a1}. If
[a] = (a2), a similar argument yields a contradiction. If
[a] = {al,a2], then al is either an atom or a co-atom of
L1. It cannot be an atom since [x],[x]' [a]. Hence al
is a co-atom of L1. If [x] = (xl), then xl,x' 1 al and
hence [a] = [1]. If [x] = (x2}, then there exist [w],[z]
such that x2 12 w2 wl al, x2 2 2', and z 1I a Now








[w] and [z] must be co-atoms of L, hence w1 = zI = al and

w2 = z2. Therefore x2,x+ 2 z2 and hence [z] = [a] = [1].
If [x] = (xl,x2], then [x] is an atom of L, [a] is a
co-atom of L, and [x]' s [a] yields a contradiction. Hence
[a] = [11, and consequently [x]v[x]' exists and equals [1].
Dually [x]A[x]' exists and equals [0]. Therefore ':L -> L
is an orthocomplementation.
To show that L is a lattice we need only show
that [x]v[y] exists for every [x],[y]EL. Hence assume
[x],[y]EL. We need only consider the following six cases:
Case (1) [x] = [(xl and [y] = {y1),
Case (2) [x] = (xl) and [y] = {y2],
Case (3) [x] = (xl] and [y] = (ylY2,
Case (4) [x] = (x2] and [y] = [y2],
Case (5) [x] = (x2) and [y] = (yl'Y2]'
Case (6) [x] = (xl,x2) and [y] = (yly2l.
Ad (1). [x1 V1 y1] a [x],[y]. If [z] z [x],[y],
then there exists z E[z] such that z 1 1 xI v, y1 and hence
[z] [x1 V1 y]. For, if [z] = (z2], then there exists
[w] = {wl,w2) such that z2 "2 w2 and w1 1 x1; in this case
[w] is a co-atom of L or [1] which yields a contradiction
since [z] = {z2).
Ad (2). Let [z]EU([x],[y]). It follows that
[z] = [1] or [z] = (zl,z2) is a co-atom of L. We may
assume [z] is a co-atom of L. Since, by hypothesis,








L1 = HS(L :aEI), xiEL$ for some BEI; it follows that
#(S1nL) = 4, zlESInL8 and [z] is unique with these

properties. Consequently [z] = [x]v[y].
Ad (3). Note that [xl v1 y1] a [x],[y]. Let
[z] a [x],[y]. If [z] = (zl, then [y] is an atom of L,
z1 a1 Y1, and z1 l xl hence zl ,1 x1 V1 y1 and
[z] a [x1 v1 y ]. The case [z] = {z2] is contradictory
since [z] 2 [x] and Si is a simple section of Li
(i = 1, 2). If [z] = (z1,z21, then [z] is a co-atom of L,
z1 2 X, and z1 21 y1; hence [z] a [xl v1 yl].
Ad (4). The proof, in this case, is similar to
that of (1).
Ad (5). The proof, in this case, is similar to
that of, (3).
Ad (6). We may assume that
([x],[y]]n([[],[1]j = 0 and that [x] / [y]. We claim that
[x]v[y] = [x2 V2 y]. Clearly [x V2 y2 ] a [x],[y]. Let
[z] a [x],[y]. We may assume that neither [x] nor [y] is a
co-atom in L, and hence that both [x] and [y] are atoms in
L. [z] $ ([zl, since [z] = (zl] contradicts the fact that

#(LanS1) 1 4, for every aEI, since xl,Y1 are (in this case)
atoms of the same L If [z] = [z2], then z2 22 x2'Y2 so
that z2 a2 x2 v2 y2 and hence [z] a [x2 V2 y2]. If
[z] = [zL,z2], then [z] = 1 or [z] is a co-atom of L. We
may assume that [z] / 1; hence [z] is a co-atom of L. The








time has come to note that x v1 yl = 1 since #(S1L ) a 4
for each aEI. Hence z2 22 x2 and z2 2 Y2; therefore
z2 2 x2 v2 2 and [z] a [x2 v2 y2]. Hence in this case

[x]v[y] = [x2 v2 Y2]. Consequently L is a lattice.
To show that L satisfies the OMI, assume that
[x] s [y]. We must show that [y] = [x]v([y]A[x]'). We may
assume that [0] < [x] < [y] < [1]. We consider nine cases.
Case (1). [x] = xl), and [y] = {yl). Since S1
is simple, all computations occur in L1 which is ortho-
modular. Hence the OMI is satisfied.
Case (2). [x] = (xlj, and [y] = [y2). Then
there exists [z] = fzl,z2) such that xl ~1 z1 and z2 S2 Y2-
But this contradicts the fact that S. is a simple section
(i = 1, 2).
Case (3). [x] = (xl), and [y] = {yl1Y2]. Then

[y] is a co-atom of L. It follows that xl 1 y1 and hence
that the OMI is satisfied since all computations are made
in L1.
Case (4). [x] = (x2), and [y] = (yl). The proof
is similar to that of Case (2).
Case (5). [x] = {x2}, and [y] = (y2l. The proof
is similar to that of Case (1).
Case (6). [x] = (x2), and [y] = (yl,Y2]. The
proof is similar to that of Case (3).
Case (7). [x] = {x1X2]), and [y] = (yl). The








proof is similar to that of Case (3).

Case (8). [x] = (x ,x2), and [y] = [y2]. The

proof is similar to that of Case (3).

Case (9). [x] = fxl,x2), and [y] = (yL,y2}. Then

[x] is an atom of L, [y] is a co-atom of L, and [y] / [x]'.

It follows that x2 -2 y2 since L1 = HS(L :aEI) and
(SlnL ) 4 for each aEI. Therefore all computations are

made in L2 and hence the OMI is satisfied.

Consequently L is an orthomodular lattice.



2.3.6 Remark. Let L = HS(L :aEI). If at least one La

admits a chain of length 3, then L is non-modular.

Proof. Assume that L admits a chain of length 3.

Then there exist a,bEL -(0,1l such that a < b. Let

cELO-(0,1l for any 8 # a. Then M(c,b) does not hold.

Hence L is not modular.


The only (previously) known finite orthomodular

non-modular lattices are those horizontal sums admitting

at least one chain of length 3 and the following lattice

(see Figure 10) given by R. P. Dilworth in [1]. Rather

than exhibit this lattice "out of the blue" we construct it

by applying Theorem 2.3.5.

2.3.7 Example. Let L1 and L2 be the following

orthomodular lattices:








1



a' b' cg e' 'f g'

L =

a b c e f g




0
Figure 8

1


x' d' y

L2

x d y

Figure 9




Let S1 = [0,l,c,c',e,e'5 and S2 = (0,l,x,x',y,y']. Define

e:S I S2 by 9 = ((0,0),(1,1),(c,x),(c',x'),(e,y),(e',y')}.

Note that the hypotheses of Theorem 2.3.5 are satisfied.

Hence the lattice given in Figure 10, L, is an orthomodular

lattice. Re-labeling the equivalence classes {a), (b},

(c,x), (d], (e,y), {f}, and (g) by a, b, c, d, e, f, and g,
respectively, we obtain the following diagram for L:
























Figure 10


L is frequently denoted by the symbol D16.

non-modular since M(g,a') fails to hold.

Note that D16 may be constructed

applications of Theorem 2.1.22. The first

Theorem 2.1.22 yields the lattice given in


Note that L is



from 23 by two

application of

Figure 5. Form


Figure 11


Let L1 be the lattice given in Figure 5, let









S1 = (0,1,e,e'}, let L2 be the lattice given in Figure 11,

let S2 = (0,l,z,z'], and let

8 = ((0,0),(1,1),(e,z),(e',z')]. Apply Theorem 2.1.22 and

obtain D16 (given in Figure 10) by dropping the brackets

and representing two-element equivalence classes by the

representative in L1.



In [3], the process of applying Theorem 2.1.22

with L2 = 23 and with simple four-element sections was

baptized adiunction of a crown to the lattice L We now

prove a theorem which generalizes the method of adjoining

crowns and which allows us to exhibit a countable family

of non-isomorphic finite orthomodular non-modular lattices

The construction is a generalization of the second method,

exhibited above, of constructing D16.

2.3.8 Theorem. For each i = 1, 2, 3, let L. be a

complete orthomodular lattice admitting a chain of length

3, let xiELi be such that the principal section Sx is a
1

proper sub-orthomodular lattice of Li, and let ylEL, be

such that S ylnn = (0,1. Let there exist a,bEL -Sxi
y1 x1 i xi
i = 2 or 3, such that a < b. Let S be ortho-isomorphic

to Sx and let Sy be ortho-isomorphic to Sx. Let L be
x 2 1 x3
the orthomodular lattice obtained by applying Theorem

2.1.22 to L1, L2, S l, and Sx2; then the principal section
1 2 q ^







S = [[z]EL:zES y is a proper sub-orthomodular lattice

(which is ortho-isomorphic to S ) of L. Let L be the

orthomodular lattice obtained by applying Theorem 2.1.22
to L, L3, Sy1, and S3. Then L is non-modular.

Proof. S is ortho-isomorphic to S For, the

function x -> [x] is clearly an ortho-isomorphism. To
show that it is an epimorphism, assume [z]ES ; then we may

assume that [yl] a [z]. Since S ylnS = [0,1),

[yl] = fyl). We claim that there exists zlE[z]. For, if
[z] = (z2), then there exists [w] = fwl,w2} such that

Yl '1 Wl and w2 s2 z2. Then w ESy lSxl. Hence w1 = 1 and

therefore z2 = 1 which contradicts the fact that

[z] = [z2]. Consequently there exists zlE[z] and y al zl;
therefore zlESyI and z1 -> [z]. Hence x -> [x] is an

epimorphism.
There exist a,bEL2-Sx2 such that a <2 b, and, in

L, [a] = (a) and [b] = [b}; moreover, in L, [[a]] = ((a)]
and [[b]] = [[b}3. Now there exists cEL3-S3
[[c]] = ({c)). Let [[m]] = [[b]]A[[c]]. By the dual of
Corollary 2.1.20, [[m]]ES. Hence [[m]] has a
representative [m]S y. It follows that [m] e [yl]'

Hence if [m] 4 0, then since S nS = (0,1), [m] is a
y1 xl




69


singleton. But [c] is a singleton. Hence by Lemma 2.1.19,

[m] ; [c]. This is a contradiction. Hence [m] = [0].
Similarly, [[a]]v[[c]] = 1. Now [[a]] < [[b]] and

[[a]]v([[c]]A[[b]]) = [[a]]VO < 1A[[b]]

= ([[a]]v[[c]])A[[b]].
Hence M([[c]],[[b]]) does not obtain and consequently L is

non-modular.













CHAPTER 3


THE REPRESENTATION OF ORTHOCOMPLEMENTED
POSETS BY SETS


1. F-sequences


In [7] Zierler and Schlessinger prove that every

orthocomplemented poset P may be (in some sense) "embedded"

in a Boolean lattice of sets. The set utilized is the set

of all order ortho-homomorphisms of P into the two-element

Boolean lattice. The proof is made by noting that the

kernel of such homomorphisms is a maximal ideal (of a

special type) and by developing some of the properties of

these ideals.

In the sequel we give a proof for this theorem

which is valid in orthocomplemented posets which are, in

some sense, "not too wide." In particular, our proof

holds for all finite orthocomplemented posets. The

advantage of our method is that it prescribes an algorithm

for generating the order ortho-homomorphisms. Because of

this we are able to associate a certain matrix, called a

deterministic Travis matrix, of zeroes and ones with such

posets. The columns of this matrix, partially ordered

vector-wise, form an orthocomplemented poset which is

ortho-isomorphic to the given poset.

70









3.1.1 Convention. Let P be an orthocomplemented poset.



3.1.2 Notation. (1) Let N denote the set of natural

numbers, let R denote the set of real numbers,

and let [0,1] denote the closed unit interval

in R.

(2) Let 2P denote the power set of P.

(3) If X and Y are two sets and if (x,y)EXXY, then by

nl((x,y)) we mean x, and by n2((x,y)) we mean y.
(4) For M c P, let XM denote the characteristic

function of the set M, i.e., yM:P -> R is

defined as follows: for xEP,

f1 if xEM.
XMX) O0 if xM.



3.1.3 Definition. Let xEP. Define f :S -> (0,1] c
x x
as follows: for zESx, let fx(z) = XL(x,l)().



3.1.4 Remark. Let xEL.

(1) If x = 0, then fx(z) = 1 for all zESx

1 if x z.
(2) If x Z 0, then, for zESx, f (z) = if x
0 if x & z'.

(3) The range of fx is a subset of [0,1].

Proof. These are immediate consequences of the

fact that Sx = L(x,l)UL(0,x'), the definition of fx, and








the fact that P is an orthocomplemented poset.


3.1.5 Definition. A r-sequence on P is a function

y:N -> PX2 such that the following conditions hold:

(1) If y(i) = (b,B), then b = 0 if and only if B = 0.

(2) If y(i) = (b,B), then B 0 0 implies bEB.

(3) n2(Y(1)) = P.
(4) If y(i) = (0,0), then y(i+l) = (0,0).

(5) If y(i) = (b,B), b # 0, then y(i+l) = (c,C)

where C = B-Sb.

If y is a r-sequence, then define

B = [7l(Y(i)):iE6N-([)

and

D = U S
Y xEB x
Y


3.1.6 Notation. In the sequel y always denotes a

r-sequence. If only one y is under discussion, then we

sometimes write bi for Trl(Y(i)) and Bi for r2(y(i)); then

y(i) = (bi,Bi).


3.1.7 Lemma. Let y be a r-sequence. Then the

following statements are valid.

n
(1) Bn+1 = P-( Sb ).
j=l j
(2) If i < j and Bi 0 0, then Bi Z Bj.









(3) If P < m, then there exists mEN such that

n : m implies B = 0.

(4) If P < -, then D = P.
Y
(5) If a,bEB a / b, then aESb or bOSa.

(6) If a,bEB xESanSb, then fa(x) = f (x).

Proof. Ad (1). If n = 1, then

1
B2 = Bl-Sb = L-( U Sb ). Assume that the formula is valid
for j=l j

for n = k-1. If bk 0, then


k-i k
B+l= Bk-Sb = (P-(U) Sb ))-S = P-(U Sb)
k j=l j k j=l

then Bk = 4 and therefore Bk+1 = 0. But

k
P-( Sb ) c P-S = P-P = 6. Hence Bk+l = 0
j=l j bk


If bk = 0,




k
= P-(U Sb
j=l j


Ad (2). Since biEB ilSb.
i


B i B-Sb = B+ B .
1


Ad (3). Suppose the statement is fali

there exists a r-sequence y such that Bi ? d foi

S1
and by part (2) P = B1 2 B2 B3 .... Hence

P > #B1 > B2 > 3 > ..., and #B > 0 for a

this is a contradiction since there does not exj

infinite strictly decreasing sequence of natural

less than the natural number (P)+l.


se. Then

r all iEN


ii iEN. I

ist an

i numbers








Ad (4). Suppose the statement is false. Then

there exists xEP-D Let m = inf (i:b. = 0). Now m > 1

since B1 = P # 0 (cf. Definition 3.1.5), Bm = and

m-1
bm_ i 0. But, by part (1), xEL-D = L-( U S = B = 0
m1y i1 b m
which is a contradiction.

Ad (5). Let a = nl(y(i)) and b = T2(y(j)); then

i 7 j. If i < j, then bEBj c Bi+1 = Bi-Sa; hence b9Sa. If

j < i, then aEBi c Bj+1 = Bj-Sb; hence aOSb.

Ad (6). Suppose the statement is false. Then we

may assume that f (x) = 1 and fb(x) = 0. Hence a s x and

b 9 x'; therefore a 6 x 4 b'. Consequently bESa and aESb

which contradicts part (3).


3.1.8 Definition. (1) For each P-sequence y, define

g :D -> (0,1) by g (x) = fb(x) whenever bEB

and xESb.

(2) If y,6 are F-sequences, we define y a 6 in case

D = D6 and g (x) = g6(x) for all xED Clearly

a is an equivalence relation. Denote the

equivalence class containing y by [y).

(3) Let Z = ([y]:D = L). Each [y]EZ is said to be a

F-state; Z is called the P-state space for P.


3.1.9 Lemma. Let y be a r-sequence. Then








(1) g :D -> (0,1) is a well-defined function from
D into [0,1].

(2) If xED then x'ED g (x') = 0 if g (x) = 1,

and g (x') = 1 if g (x) = 0.

(3) g (0) = 0 and g (1) = 1.
(4) If xED g (x) = 1, yEP, and x : y, then yEDy
and g (y) = 1.

Proof. Ad (1). By Lemma 3.1.7 part (6), g is

well-defined. The domain is D by definition; the range is

a subset of [0,1] by Remark 3.1.4.

Ad (2). If xED then xESb for some bEB Hence
by Lemma 1.3.6 x'ESb c D Moreover, since b a x if and

only if x' I b', and x a b' if and only if b e x', it
follows that fb(x') = 1 if and only if fb(x) = 0 and

fb(x') = 0 if and only if fb(x) = 1. The result follows.

Ad (3). Since 1ESb for all bEP, it follows that

fb(1) = 1 for all bEP and hence that g (1) = 1. Therefore
by part (2) g (0) = 0.
Ad (4). Since g (x) = 1, there exists bEB such
that 0 < b & x. Hence b i y, and yESb; therefore yED and

g (y) = 1.


Condition I. For every r-sequence y, D = P.


Condition II. (1) If aEP-(0), then there exists








a T-sequence y such that l1(Y(1)) = a and D = P.

(2) If a,bEP-(0}, b'Sa, then there exists a F-sequence

y such that nr(y(l)) = a, nl(y(2)) = b, and

D = P.
Y


3.1.10 Lemma. Condition I implies Condition II.

Proof. Assume that Condition I holds.

(1) Let aEP. Define y(l) = (a,P). For n > 1,

let Fn = T2(Y(n-l))-Sl(y(n-l)) and define


(n) (0,) if F = ~
y(n) n
(f,Fn) for any fEFn if Fn .

Then y is a F-sequence, Tl(Y(1)) = a, and, by Condition I,

D = P.
Y
(2) Let a,bEP be such that bOSa. Define

y(l) = (a,P) and y(2) = (b,P-Sa). For n > 2, let

Fn = 2(Y(n-l))-Snl(y(n-1)) and define

) (0,0) if F = ~.
y(n) =
(f,Fn) for any fEFn if Fn 0.
Then y is a P-sequence, Tnl((l)) = a, nl(y(2)) = b, and, by

Condition I, D = P.


The following examples of horizontal sums show

that the converse of Lemma 3.1.10 is not valid and that

there are posets which satisfy neither.









3.1.11 Example. Let L = HS(24:iEF). A r-sequence on L

satisfying (1) (or (2)) of Condition II may be easily

constructed by noting the following fact: for every set of

atoms A of L consisting of exactly one atom of each 2k,

there exists a r-sequence y such that B = A (and hence

D = L). But the r-sequence defined by

n-1
y(n) = (a2n,L-U Sa ) where a2n is an atom of 22n has the
i=l 2i
property that

D y 2 2 L.
Y n=1 2n

Hence L satisfies Condition II but not Condition I.



3.1.12 Example. Let L = HS(2a:aEI) where I is any

uncountable index set. Then L satisfies neither Condition

I nor Condition II since, for any r-sequence y, D is a

countable union of finite sets and hence countable, whereas

L is uncountable. It follows that for every r-sequence y,
C
D p L.
Y


3.1.13 Remark. (1) Condition I implies that [y]EZ for

every r-sequence y. Condition II implies that, for every

aEP-(O}, there exists [y]EZ such that g (a) = 1; and, for

every a,bEP-{0) such that b)Sa, there exists [y]EZ such

that g (a) = g (b) = 1.

(2) By Lemma 3.1.7 part (4), if #P < m, then








Condition I holds. Hence, by Lemma 3.1.10, if P < ,

then Condition II holds.


3.1.14 Lemma. Let P satisfy Condition II, and let

e,fEP. Then e f if and only if g (e) e g (f) for all

[y]EY.

Proof. Assume that e 9 f. We must prove that

g (e) s g (f) for all [y]EZ. Let [y] be any fixed element
of Z, and let yE[y]. We may assume that (e,f}~n{,l1= 0 and

that g (e) = 1. We must show that g (f) = 1. But
1 = g (e) implies that there exists bEB such that fb c g

and fb(e) = 1. Hence b s e so that b 9 f; therefore

fb(f) = 1 and consequently g (f) = 1.
Conversely, assume that g (e) s g (f) for all

[y]EZ. We must prove that e 9 f. Suppose the statement is

false. Then e e f, e ? 0, and f # 1. If f = 0, then by

Lemma 3.1.9, g (f) = 0 for all [y]EZ; hence by hypothesis

g (e) = 0 for all [y]EZ and consequently, by Condition II,
e = 0 which contradicts the fact that e 3 0. Therefore

f f 0. If f r e', then e f' and, by Condition II, there

exists [y]EZ such that g (e) = 1; hence by Lemma 3.1.9

g (f') = 1 and g (f) = 0; consequently 1 = g (e) which is a contradiction. Therefore f $ e' and

consequently ffs Hence f'kS and by Condition II there

exists [y]EZ such that g (e) = g (f') = 1; therefore by









Lemma 3.1.9 1 = g (e) : g (f) = 0 which is a contradiction.

Consequently e E f.


3.1.15 Theorem. Let P be an orthocomplemented poset

satisfying Condition II, and let e,fEP. Then there exists

a set X and a function I:P -> 2X such that

(1) O0 = 0 and 1i = X,

(2) e'5 = X-(e$),

(3) e f if and only if eS c f$.

Proof. Let X = Z. Define S as follows:

for eEP, eS = [[y]EZ:g (e) = 11.

(1) follows from Lemma 3.1.9. (2) follows from the fact

that [y]Ee'S if and only if g (e') = 1, and g (e) = 0 if

and only if [y]9et. To prove (3), note that if e a f, then

g (e) = 1 implies g (f) = I for all [y]EZ. Therefore
eS c fS. Conversely, assume that eS c f$ and that [y]ES.

Then g (e) = 1 implies [ly]ce; hence [y]EfS, so that

g (f) = 1. Therefore g (e) = 1 implies g (f) = 1.
Consequently, g (e) < g (f) for all [y]ES, and hence by

Lemma 3.1.14, e t f.


2. Order Ortho-homomorphisms


3.2.1 Definition. Let P be an orthocomplemented poset.

Let (0,1) c A, and define 0' = 1 and l' = 0; then (0,L) may

be regarded as an orthocomplemented poset.








(1) Let H = (a:P -> [0,l]:(i) e'a = (ea)',

(ii) e < f implies ea s fa, (iii) if

[xi] is a (not necessarily countable)

chain, if inf (xi) exists, and if

x = inf (x.i, then x a = 1 for every

element x. of the chain (xi. implies

xa = 1].

(2) For each aEH, let

E = [xEP:xa = 1 and y < x imply ya = 0}.

(3) For each aEH, let

F = (xEP:there exists a chain ?xi), maximal with

respect to x.a = 1, such that x is the

smallest element of (xi)].


3.2.2 Lemma. If aEH, then Oa = 0 and la = 1.

Proof. It is sufficient to prove Oa = 0.

Suppose the statement is false. Then Oa = 1. Since 0 < 1

implies Oa < la, it follows that la = 1 and hence that

1 = la = 0'a = (Oa)' = 1' = 0 which is a contradiction.



3.2.3 Lemma. Let P be an orthocomplemented poset such

that if (xi) is a chain in P, then inf ([xi exists in P.

Let aEH. Then the following statements obtain.

(1) E j 0 and E = F .

(2) For yEP, ya = 1 if and only if there exists xEEa

such that x : y.









Proof. Ad (1). Let aEH. Then la = 1. (1) is a

chain in P; extend it to a chain (x.i, maximal with respect

to xia = 1. Let x = inf [xi). Since xia = 1, it follows

that xa = 1. By maximality, xE[xi}, therefore x is the

smallest element of (xi.. Since xa = 1, if y < x, it

follows that y = 0; hence xEEa and therefore E f 0.

Clearly Ea 3 F .

Suppose that E a. F Then there exists xEEC

such that for every chain (xi), maximal with respect to

xia = 1, x is not the smallest element of fxi3. Now xEE,

implies xa = 1. Since x is not the smallest element of any

such maximal chain, there exists y < x such that ya = 1.

But xEE and y < x imply ya = 0 which is a contradiction.

Hence E = F

Ad (2). The necessity of the condition is clear.

To prove the sufficiency, note that, since ya = 1, (y] may

be extended to a chain tyi], maximal with respect to

ya = 1. As above, define x = inf {yi). Then xE{yi] and

xEE



3.2.4 Corollary. Let P be an orthocomplemented poset

such that if (xij is a chain in P, then inf [xi. exists in

P. Then for any aEH, P = U S .
xE x

Proof. Suppose the statement is false. Let

yEP-(U x), then y'EP-( U S ). We may assume ya = 1.
xEE xEE x
ci ci








By Lemma 3.2.3 there exists xEE such that x 6 y. Then

yESx which is a contradiction.



For each P-state [y], the corresponding function

g :P -> (0,1) is an order ortho-homomorphism (provided we

regard g as mapping onto (0,1), and define 0' = 1 and

1' = 0). In what follows we prove that every order ortho-

homomorphism from a finite orthocomplemented poset into

[0,1) arises in this way.

3.2.5 Proposition. Let P be a finite orthocomplemented

poset, let a be any order ortho-homomorphism mapping P onto

(0,1) (regarded as a subset of [0,1] with O' = 1 and

1' = 0). Then there exists a unique F-state [y] such that

g = a.

Proof. We first note that the third condition on

the elements of H (cf. Definition 3.2.1 part (1)) is

redundant since P is finite; hence H is, in fact, the set

of all order ortho-homomorphisms mapping P onto (0,1).

Well-order Ea; since P is finite, E may be

written as [bl,b2,...,b n for some nER, where bi # bj if

i F j. Define y:R -> PX2P by y(i) = (bi,B ) where biEE

if 1 i S n, bi = 0 if i > n, B1 = P, and Bi+1 = Bi-Sb

for all iEN. By the definition of E and the fact that

Oa = 0 for all aEH, it follows that b. # 0 for
1









i = 1, ..., n; hence y is a P-sequence. By Corollary 3.2.4,

[y] is a r-state. Finally, by Definition 3.1.8 parts (1)

and (3) and Definition 3.2.1 part (2), it follows that

gy = a and that [y] is unique.


3. Travis Matrices


The following definition is motivated by certain

tables which appear in the 1962 Wayne State University

Master's Thesis of R. D. Travis [6].

3.3.1 Definition. Let 2* and P* be two non-empty sets.

Let ':P* -> P* be a mapping of period 2. By a Travis

matrix over (Z*,P*,') we mean a function T:Z*XP* -> [0,1]

which satisfies the following postulates. (For simplicity

we write a(e) for T(a,e).)

(1) 0 K a(e) : 1 for all aE.*, eEP*.

(2) a(e') = 1 a(e) for all aE6*, eEP*.

(3) If e,fEP* are such that a(e) = a(f) for all aEZ*,

then e = f.

(4) If a,5E6* are such that a(e) = B(e) for all eEP*,

then a = 8.

(5) There exists OEP such that a(0) = 0 for all aE.*.

(6) k = inf (sup (a(e)) > .
eEP* a.E*
e O

We define the "weak" partial order induced on P* by _* as

follows: Let e,fEP*, then e S* f if and only if








a(e) < a(f) for all aES*.

If the range of T is (0,1), then T is said to be

a deterministic Travis matrix over (Z*,P*,').



3.3.2 Lemma. (P*,s*,') is an orthocomplemented poset.

Proof. Let e,f,gEP*. Since a(e) = a(e) for all

aSE*, e s* e. If e s* f and f <* e, then a(e) s a(f) and

a(f) i a(e) for all acE*; hence a(e) = a(f) for all aEZ*

and therefore e = f by Postulate (3) of Definition 3.3.1.

Moreover, if e S* f and f -* g, then a(e) S a(f) S a(g) for

all aE2*; hence e t* g. Consequently 5* partially orders

P*. Since a((e')') = 1 [1 a(e)] = a(e) for all aEZ*,

e' = e. If e S* f, then a(e) 9 a(f) for all aES* and

1 a(f) i 1 a(e) for all aE6*, i.e., a(f') s l(e') for

all aEZ*; hence f' s* e'.

By Postulate (5) there exists E0P. Define 1 to

be 0'; 0 S* e' for all eEP implies that e S* 1 for all eEP.

We claim that eve' exists and equals 1. Suppose e,e' S f;

we must prove that f = 1. Suppose that f # 1. Then f' # 0

and 1 < k a sup (a(f'):aE2*}. Therefore there exists aEZ*

such that < a(f') = 1 a(f); hence a(f) < a. Since

(I) a(e) s a(f) < a,

it follows that

(II) 1 a(e) a(f) < .

By adding the extremes of the inequalities (I) and (II) we








obtain 1 < 1 which is a contradiction. Therefore f = 1.
Similarly, eAe' exists and eAe' = 0. Conse-

quently (P*,t*,') is an orthocomplemented poset.


3.3.3 Definition. Let T denote the matrix

T:ZXP --> 0,1) defined by T([y],e) = g (e).


3.3.4 Lemma. T is well-defined.

Proof. If [y],[6]EZ and [y] = [6], then D = D6
and g (e) = g8(e) for all eEP.


3.3.5 ,Theorem. If P satisfies Condition II, then T is

Travis matrix over (Z,P,').
Proof. We must show that the following state-

ments are valid.

(1) 0 1 g (e) 1 for all [y]EZ, eEP.

(2) g (e') = 1 g (e) for all [y]EZ, eEP.
(3) If e,fEP are such that g (e) = g (f) for all

[y]EZ, then e = f.
(4) If [y],[6]EZ are such that g (e) = g6(e) for all

eEP, then [y] = [6].

(5) There exists 0EP such that g (0) = 0 for all
[y]EZ.

(6) k = inf (sup g (e)) > .
eEP
eEo [y]EY








Ad (1). Since [y]EZ, the domain of g is P. The

inequalities follow immediately from the fact that, for any

bEB eEP, fb(e) equals either 0 or 1.

Ad (2). Let eEP and [y]EZ. Then the result

follows immediately from Lemma 3.1.9.

Ad (3). By Lemma 3.1.14 g (e) g (f) for all

[y]EZ implies that e z f, and g (f) s 9 (e) for all [y]EY

implies that f & e. Hence e = f.

Ad (4). If [y],[6]EE, then D = D = P. There-

fore, since g (e) = g,(e) for all eED [y] = [6] by

Definition 3.1.8.

Ad (5). Since P is an orthocomplemented poset,

there exists OEP such that 0 9 b for all bEP. Hence, for

any P-sequence y, 0 b' for all bEB Therefore

fb(0) = g (0) = 0 for all [y]EE and for all bEB .
Ad (6). For any eEP such that e X 0, there

exists a F-sequence y such that g (e) = 1 by Condition II.

Hence sup g (e) = 1. Since this holds for all non-zero


eEP, it follows that inf (sup (g (e))) = 1 > .
eEP [y]EZ Y
e#O


3.3.6 Corollary. Let P satisfy Condition II, let <* be

the "weak" partial order induced on P by E (cf. Definition

3.3.1), and let e,fEP. Then e : f if and only if e d* f.

Proof. By Lemma 3.1.14 e s f if and only if





87



g (e) : g (f) for all [y]EZ. But by Definition 3.3.1,
since the T of Definition 3.3.3 is a Travis matrix for

(2,P,'), g (e) g (f) for all [y]EZ if and only if e :* f.

Therefore e f if and only if e !* f.



4. Examples


We now give deterministic Travis matrices which

correspond to some of the posets previously mentioned.

Figures 12, 13, 14, 15, and 16 are deterministic Travis

matrices for the posets given in Figures 1. 3. 5, 8, and

10, respectively. Figure 17 is the Hasse diagram of the

orthomodular post which is not a lattice, given by

Janowitz [5] and mentioned in Example 2.1.24; Figure 18 is

a deterministic Travis matrix for it.







0 e f 1 e' f'

S 0 1 1 1 0 0

S2 0 0 1 1 1 0

S3 0 0 0 1 1 1


Figure 12
























Figure 13


a' b' c' d' e'

0 1 1 0 1

0 1 1 1 0

0 1 1 1 1

1 0 1 0 1

1 0 1 1 0

1 0 1 1 1

1 1 0 1 1

1 1 1 0 1

1 1 1 1 0

1 1 1 1 1


Figure 14


0

S1 0

S2 0

S3 0

S4 0

S5 0

S6 '0

S7 0

Ss
Sg 0
S9 0
S10 0


a b

1 0

1 0

1 0

0 1

0 1

0 1

0 0

0 0

0 0

0 0


c d e

0 1 0

0 0 1

0 0 0

0 1 0

0 0 1

0 0 0

1 0 0

0 1 0

0 0 1

0 0 0


I - -


_
















0 a b c e f q 1 a' b' c' e' f' q'

S 0 1 0 0 1 0 0 1 0 1 1 0 1 1

S2 0 1 0 0 0 1 0 1 0 1 1 1 0 1

S3 0 1 0 0 0 0 1 1 0 1 1 1 1 0

S4 0 1 0 0 0 0 0 1 0 1 1 1 1 1

S5 0 0 1 0 1 0 0 1 1 0 1 0 1 1

S6 0 0 1 0 0 1 0 1 1 0 1 1 0 1

S7 0 0 1 0 0 0 1 1 1 0 1 1 1 0

S8 0 0 1 0 0 0 0 1 1 0 1 1 1 1

S9 0 0 0 1 1 0 0 1 1 1 0 0 1 1

S10 0 0 0 1 0 1 0 1 1 1 0 1 0 1

S11 0 0 0 1 0 0 1 1 1 1 0 1 1 0

S12 0 0 0 1 0 0 0 1 1 1 0 1 1 1

S13 0 0 0 0 1 0 0 1 1 1 1 0 1 1

S14 0 0 0 0 0 1 0 1 1 1 1 1 0 1

S15 0 0 0 0 0 0 1 1 1 1 1 1 1 0

S16 0 0 0 0 0 0 0 1 1 1 1 1 1 1


Figure 15








Figure 16



0 a b c d e f g 1 a' b' c' d' e' f' q'

S1 0 1 0 0 1 0 1 0 1 0 1 1 0 1 0 1

S2 0 1 0 0 1 0 0 1 1 0 1 1 0 1 1 0

S3 0 1 0 0 1 0 0 0 1 0 1 1 0 1 1 1

S4 0 1 0 0 0 1 0 0 1 0 1 1 1 0 1 1

S5 0 1 0 0 0 0 1 0 1 0 1 1 1 1 0 1

S6 0 1 0 0 0 0 0 1 1 0 1 1 1 1 1 0

S7 0 1 0 0 0 0 0 0 1 0 1 1 1 1 1 1

S8 0 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1

S9 0 0 1 0 1 0 0 1 1 1 0 1 0 1 1 0

S1 0 0 1 0 1 0 1 0 0 1 1 0 1 0 1 1 1
S11 0 0 1 0 0 1 0 0 1 1 0 1 1 0 1 1

S12 0 0 1 0 0 0 1 0 1 1 0 1 1 1 0 1

S13 0 0 1 0 0 0 0 1 1 1 0 1 1 1 1 0

S14 0 0 1 0 0 0 0 0 1 1 0 1 1 1 1 1

S15 0 0 0 1 0 0 1 0 1 1 1 0 1 1 0 1
S6 0 0 0 1 0 0 0 1 1 1 1 1 1

S17 0 0 0 1 0 0 0 0 1 1 1 0 1 1 1 1

S18 0 0 0 0 1 0 1 0 1 1 1 1 0 1 0 1

S19 0 0 0 0 1 0 0 1 1 1 1 1 0 1 1 0

S20 0 0 0 0 1 0 0 0 1 1 1 1 0 1 1 1

S21 0 0 0 0 0 1 0 0 1 1 1 1 1 0 1 1

S22 0 0 0 0 0 0 1 0 1 1 1 1 1 1 0 1

S23 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0

S24 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1

















































Figure 17






Figure 18

0 a b c d e f q h a' b' c' d' e' f' q' h'
S1 0 1 1 0 0 0 0 0 0 1 0 0 1 1 1 1 1 1
S2 0 1 0 0 1 0 1 0 0 1 0 1 1 0 1 0 1 1
S3 0 1 0 0 1 0 0 0 0 1 0 1 1 0 1 1 1 1
S4 0 1 0 0 0 0 1 0 0 1 0 1 1 1 1 0 1 1
S5 0 1 0 0 0 0 0 0 0 1 0 1 1 1 1 1 1 1
S6 0 0 1 1 0 1 0 0 0 1 1 0 0 1 0 1 1 1
S7 0 0 1 1 0 0 0 0 0 1 1 0 0 1 1 1 1
Sg 0 0 1 0 0 1 0 0 0 1 1 0 1 1 0 1 1 1
S9 0 0 1 0 0 0 0 0 0 1 1 0 1 1 1 1 1 1
S10 0 0 0 1 1 1 1 0 0 1 1 1 0 0 0 0 1 1
S11 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0 1 1 1
S12 0 0 0 1 1 0 0 0 0 1 1 1 0 0 1 1 1 1
13 0 0 0 1 1 0 1 0 0 1 1 1 0 0 1 0 1 1
S14 0 0 0 1 1 0 0 1 0 1 1 1 0 0 1 1 0 1
S15 0 0 0 1 0 1 0 0 0 1 1 1 0 1 0 1 1 1
S16 0 0 0 1 0 1 1 0 0 1 1 1 0 1 0 0 1 1
S17 0 0 0 1 0 0 1 0 0 1 1 1 0 1 1 0 1 1
S18 0 0 0 1 0 0 0 1 0 1 1 1 0 1 1 1 0 1
S19 0 0 0 1 0 0 0 0 0 1 1 1 0 1 1 11 1
S20 0 0 0 0 1 1 1 0 0 1 1 1 1 0 0 0 1 1
S21 0 0 0 0 1 1 0 0 0 1 1 1 1 0 0 1 1 1
s22 0 0 0 0 1 0 1 0 0 1 1 1 1 0 1 0 1 1
S23 0 0 0 0 1 0 0 1 0 1 1 1 1 0 1 1 0 1
524 0 0 0 0 1 0 0 0 0 1 1 1 1 0 1 1 1 1
s25 0 0 0 0 0 1 1 0 1 1 1 1 1 1 0 0 1 0
S26 0 0 0 0 0 1 1 0 0 1 1 1 1 1 0 0 1 1
27 0 0 0 0 0 1 0 0 1 1 1 1 1 1 0 1 1 0
S28 0 0 0 0 0 1 0 0 0 1 1 1 1 1 0 1 1 1
S29 0 0 0 0 0 0 1 0 1 1 1 1 1 1 1 0 1 0
S30 0 0 0 0 0 0 1 0 0 1 1 1 1 1 1 0 1 1
31 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0
S32 0 0 0 0 0 0 0 1 0 1 1 1 1 1 1 1 0 1
S33 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 0
S34 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1




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