Title: Implicativity and irreducibility in orthomodular lattices
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Title: Implicativity and irreducibility in orthomodular lattices
Physical Description: 39 leaves : illus. ; 28 cm.
Language: English
Creator: Catlin, Donald Edward, 1936-
Publication Date: 1965
Copyright Date: 1965
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Subject: Lattice theory   ( lcsh )
Mathematics thesis Ph. D
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Thesis: Thesis - University of Florida.
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IMPLICATIVITY AND IRREDUCIBILITY

IN ORTHOMODULAR LATTICES





















By
DONALD EDWARD CATLIN










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


August, 1965













ACKNOWLEDGEMENTS


I wish to thank Professor D. J. Foulis, not

only for his invaluable aid in preparing this work,

but for many, many inspiring discussions with him

about various other mathematical concepts.














TABLE OF CONTENTS



Page
ACKNOWLEDGEMENTS . . . . . . . ... ii

Chapter
I. PRELIMINARIES ............. .. 1

II. MAIN RESULTS ............... 13

III. ADDITIONAL RESULTS ............ 31

BIBLIOGRAPHY . . . . . . . . . . 38

BIOGRAPHICAL SKETCH . . . . . . . . 39



































iii












CHAPTER I


PRELIMINARIES


1. Introduction

The current model for the "logic" of (non-

relativistic) quantum mechanics is the lattice of closed

subspaces of a separable, infinite dimensional, Hilbert

space. However, as Mackey (5) points out, this assump-

tion is rather ad hoc in character. One would like to be

able to deduce such an assumption from a list of physically

plausible assumptions. Now one can exhibit some quite con-

vincing arguments to show that the "logic" of any experi-

mental theory ought to be an orthocomplete, orthomodular,

poset: perhaps even an orthomodular lattice (7). From this

point, however, it is not clear how to transfer various

physical assumptions into suitable lattice theoretic state-

ments and thereby deduce the "correct logic" for quantum

mechanics. What appears to be needed is a list of physi-

cally interpretable lattice theoretic definitions. One

possible candidate for such a list is the topic of the

present work, namely, the notion of an implicative pair.

This and some of its consequences will be treated in Chapter

2.








Let us make it clear at the outset that we do not

intend to engage in polemics regarding the physical inter-

pretations (if any) of our work. Such discussions must a-

wait subsequent research on the connection between lattice

theory and physics. Hence, our approach will be from a for-

mal mathematical point of view.


2. Basic definitions and theorems.

In this section we give some of the basic, well-

known, theorems and definitions found in the field of ortho-

modular lattice theory. We will not present proofs since

they are readily available in (2).

Definition 1.2.1. An orthomodular lattice is a

lattice L having a 0 and a 1, and equipped with an ortho-

complementation ':L->L such that the orthomodular identity

is satisfied:

x y implies y = x V (y x').

Convention 1.2.2. Throughout the rest of this work,

L will always represent an orthomodular lattice.

Definition 1.2.3. Let e,f E L with e s f. We then

define the interval L(e,f) by: L(e,f) = (x E L : e < x < f).

Lemma 1.2.4. If e,f E L with e < f, then L(e,f) is

itself an orthomodular lattice with orthocomplementation

x-->x# = e V (f A x') = (e v x') A f.

Definition 1.2.5. Let e E L. We then define a

mapping 0e:L--->L, called the Sasaki projection determined

by e, by the rule: x0e = e A (x V e ').
e








Lemma 1.2.6. Let e E L. Then the following hold.

(i) 0 =0 2
e e
(ii) (x V y)0e = X0e V YOe

(iii) x0e = x if and only if x < e.

(iv) x0e = 0 if and only if x < e'.

Definition 1.2.7. Let e,f E L. Then f is said to

commute with e, in symbols fCe, if and only if

f0 = f Ae.
e
We will write f d e providing f does not commute with e.

Lemma 1.2.8. Let e,f,g E L. Then the following

hold.

(i) f e implies fCe.

(ii) fCe implies eCf.

(iii) fCe implies fCe'.

(iv) If any two of the three conditions eCf,

fCg, or eCg hold, then (e v f) Ag =

(e A g) V (f A g) and (e A f) v g =

(e V g) A (f V g).

(v) If fCea for each a E A, and if V EAea

exists, then fC V ee
aEA a
Definition 1.2.9. Let A c L. Then we define

C(A) = (x E L : xCy for every y E A).

If A = L then the set C(L) is called the center of L. If

A is a singleton set, say (a), we will use the abusive no-

tation C(a) for C((a)). L is said to be irreducible if

and only if C(L) = [0,11. If e E C(L), e / 0, and e / 1,








then one can prove that L can be "factored" into the Car-

tesian product: L = L(0,e) x L(0,e'), hence the reason for

the word "irreducible".

Lemma 1.2.10. Let a E L. Then the following hold.

(i) C(a) = ((x V a)A(x V a') : x E L).

(ii) C(a) = ((x A a)v(x A a') : x E L).

(iii) C(a) = Nx0a V x0a, : x E L}.


3. The atomic bisection property.

Throughout this section we will follow the treat-

ment given by Janowitz (3).

Definition 1.3.1. L is said to be atomic if and

only if x E L and x / 0 imply there exists an atom e such

that e < x.

Definition 1.3.2. L is said to have the atomic

bisection property if and only if L is atomic and for every

pair of atoms b,c E L, there exists an atom a E L such that

a + b, a c, and a < b V c.

Remark. 1.3.3. In definition 1.3.2, it suffices

to suppose that b and c are orthogonal atoms, i.e. b c'.

One can easily see this by noting the identity

b V c = b v[(b V c) A b'].

Lemma 1.3.4. If L has the atomic bisection pro-

perty, then L is irreducible.

The proof of this lemma is omitted since it can be

obtained as a consequence of theorems 2.3.2 and 2.3.3 of

Chapter II.









Lemma 1.3.5. If L has the atomic bisection pro-

perty, then so does any interval L(e,f).

Proof. Note that the mapping a->a A e' is an

orthoisomorphism of L(e,f) onto L(0,f A e'). Thus, since

L(0,f A e') has the atomic bisection property, so does

L(e,f).

Lemma 1.3.6. If L is atomic, complete, and modu-

lar, then the irreducibility of L is both a necessary and

sufficient condition for L to have the atomic bisection

property.

Proof. According to Kaplansky (4), every com-

plete, modular, orthocomplemented lattice is a continuous

geometry. The result now follows from (6, p. 80, Th. 2.4).

Theorem 1.3.7. Let L be atomic. Then the follow-

ing conditions are mutually equivalent.

(i) L has the atomic bisection property.

(ii) Every interval L(e,f) is irreducible.

(iii) Every interval L(0,a) is irreducible.

If in addition, L is complete and modular, we can add

(iv) L is irreducible.


4. The relations V and S.

In this section we introduce the relations V and

S on an orthomodular lattice. The relation V has been

studied by Maeda (6) for arbitrary lattices and by S.

Holland, Jr. and M. F. Janowitz for orthomodular lattices.

The relation S was apparently first defined by S. Holland,


Jr.







Definition 1.4.1. Let e,f E L.

(i) eNCf if and only if e A f = e A f'

= 0.

(ii) We say that e and f are detached, in

symbols e V f, if and only if e and

f are orthogonal and whenever g E L

and gNCe, then gCf.

(iii) We say that e and f are separated,

in symbols eSf, if and only if e and

f are detached in the interval

L(0,e v f).

Lemma 1.4.2. Let e,f E L. Then eNCf if and only

if every non-zero subelement of e fails to commute with f.

Proof. Let e,f E L and suppose that eNCf. Then

by definition, e Af = e A f' = 0. Let 0 / el s e. Then,

clearly, el A f = e1 A f' = 0. If elCf, then e =

(el A f) v (el A f') = 0. But el 1 0, so this is a con-

tradiction. Hence, el does not commute with f.

Conversely, suppose every non-zero subelement of

e fails to commute with f. e A f and e A f' are both sub-

elements of e and both commute with f. Hence, e A f =

e A f' = 0. By definition, then, eNCf.

Theorem 1.4.3. (Holland). Let e,f E L. Then

there are uniquely determined elements el and e2 in L

such that el is orthogonal to e2, elCf, e2NCf, and e

eI V e2. In fact e = (e A f) v (e A f') and 02 =

(f'0 ) A (f0W).







Proof. Note that (e A f) v (e A f') :

((f A e) V e') v ((f' A e) v e') so that el is orthogonal

to e2. Clearly elCf. To show that e2NCf we compute:

e2 A f = [(f' e') A e] A [(f V e') A e] Af =
[(f' v e') A e] A [f A e] = [f' v e'] A [f A e] = 0 ;

e2 A f' = [(f' V e') A e] A f' A [(f V e')Ae] =

[f' A e] A [(f v e') Ae] = [f' A e] A [f V e'] = 0.

Hence e2NCf. Since e2 = e A ei and eI < e, we have

e = el v (e A e') = (e, V e2). It remains to show

uniqueness. Suppose gl and g2 have the same respective

properties as eI and e2. Then el = (e A f) v (e A f') =

[(gl V g2) A f] V [(gl V g2) A f'] = (g1 A f) V (g1 A f')

g1, i.e., gl = el. Hence, e2 = e A e' = e A g

gl V g2)A g. = 92 A g{ = g2'
Definition 1.4.4. Let e,f E L and let el, e2

be as in theorem 4.3. We shall refer to e = el v e2 as

the C-NC decomposition of e with respect to f.

Theorem 1.4.5. (Holland-Janowitz) Let e,f E L.

Then the following are equivalent.

(i) e V f.

(ii) For all g E L, g e' = 0 implies

that f is orthogonal to g.

(iii) For all g E L, g v e = 1 implies

f g.

(iv) If g is any complement of e in L,

then f < g.

(v) For all g E L, f is orthogonal to e0 .








(vi) For all g E L, g = (e v g) A (f V g).

(vii) For all g E L, f e V g implies

f < g.
Proof. (i) implies (ii). Suppose that e V f and

that g E L with g A e' = 0. Let g = gl V 92 be the C-NC

decomposition of g with respect to e. Then gl = (g A e) v

(g A e') = g A e and g2 = g A gl = g A (g' V e'). Since

e V f and g2NCe, we have by definition that g2Cf. Also,

since e is orthogonal to f we have gl orthogonal to f.

Hence, f A g = f A (gl V g2) = (f A g) V (f A g2) =

f A g2 = f A g A (g' V e') = (f A g A g') V (f A g A e')
= 0. Since fCg1 and fCg2, then fCg. But fCg and f A g

= 0 implies that f and g are orthogonal.

(ii) implies (iii) and (iii) implies (iv)

are trivial.

(iv) implies (v). Assume (iv) and let h

(e' v g) A [g' v (e' A g)]. Then h A e = (e' V g) A

[g' V (e' A g)] A e = (e' V g) A [(g' A e) v (e' A g A e)]
= (e' v g) A (g' A e) = 0 and h V e = [e v (e' v g)]

[g' v (e' A g) v e] = 1. Hence h is a complement of e.

Therefore f < h < g' V (e' A g) = (e0 )'.

(v) implies (vi). Put h = (e V g) A (f V g).

Since g < h, it suffices to prove h A g' = 0. But this

follows easily: h A g' = (e0 ,) A (f V g) < f' A g' A

(f v g) = 0.









(vi) implies (vii). g = (e V g) A (f v g)

implies g A f = (e V g) A (f v g) A f which in turn im-

plies g A f = (e V g) A f.

(vii) implies viii). f < e V g implies

(e v g) A f = f. By (vii), g A f = (e V g) A f = f, so

that f 9 g.

(viii) implies (i). First put g = e' in

(viii) and obtain f orthogonal to e. Now suppose hNCe.

We must prove that hCf. Since hNCe, then hA e' = 0

so that e v h' = 1. By (viii), f c h', hence fCh.

Corollary 1.4.6.

(i) The relation v on any orthomodular

lattice is symmetric.

(ii) e E C(L) if and only if e V e'.

(iii) Let (e a be a family of elements

of L with e = V (e ). If f E L

and if e V f for all a, then e V f.

Theorem 1.4.7. Let e,f E L. Then the follow-

ing statements are equivalent.

(i) eSf.

(ii) For all g < e v f, g A e' = 0 im-

plies f orthogonal to g.

(iii) For all g < e V f, g V e = f v e

implies f < g.

(iv) If g < e V f and g is a complement

of e in L(O,e V f), then f < g.









(v) If g < e v f, then f and e0 are

orthogonal.

(vi) If g t e Vf, then g = (e V g) A

(f v g).

(vii) If g s e vf, then (e v g) A f =

gA f.

(viii) If f a e v g 9 e v f, then f < g.

Proof. Follows at once from 4.6 and definition

4.1.

Corollary 1.4.8.

(i) The relation S on any orthomodular

lattice is symmetric.

(ii) Let e,f E L with e < f. Then e is

central in L(O,f) if and only if

eSe' A f.

Theorem 1.4.9. Let e,f E L with e A f = 0.

Then the following statements are equivalent.

(i) esf.

(ii) For all g E L, g < e v f, g A e = 0,

g A f = 0 imply g = 0.

(iii) For all g E L, g < e V f implies

g = (e A g) V (f A g).

(iv) e central in L(0,e v f).

(v) There is no g + f such that e A g = O

and e v g = e v f.









(vi) For all g E L, g A (e V f) = (g A e) V

(g A f).
(vii) For all g E L, gC(e v f) implies gCe

and gCf.

Proof. (i) implies (ii). From theorem 1.4.7,

part (v), setting g = f, we obtain f orthogonal to e0f.

Therefore, e0f = 0 = e A f, i.e., eCf. Now let g E L(O,e

v f) satisfy g A e = g A f = 0. By (vi) of theorem 1.4.7,

g = (e v g) A (f v g). Now 0 = (f A g) V (e A g) =

[f A (e V g) A (f V g)] V [e A (e V g) A (f V g)] =

[(e v g) A f] V [(f v g) A e]. Now (e V g) A f f v g

and [(evg) A f]Ce, so that 0 = [f V g] A [((e V g) A f) v e]

= (f v g) A (e V g) A (f V e) = g A (f V e) = g

(ii) implies (iii). Let g e v f. Since

(e A g) v (f A g) < g, it will suffice to show that h = g

A (e A g)' A (f A g)' = 0. But h < g < e v f and h ^ e =

h f = 0. Hence, h = 0 by (ii).

(iii) implies (iv). The orthocomplement of
e in L(0,e V f) is given by e# = (e v f) A e'. Condition

(iii) with g = e# yields e# < f. Thus condition (iii)

gives g = (e A g) V (e# A g) for every g EL(0,e V f).

Hence, by lemma 1.2.10, e is central in L(0,e v f).

(iv) equivalent to (v) is clear.

(iv) rhplies (vi). Since e is central in
L(0,e V f) and since e A f = 0, then f = e = e' A (e v f).








Given g E L, set h = g A (e V f). Since h commutes with

e in L(0,e V f), then g = (h A e) v (h A e#) A (h A e) v

(h A f). In other words, g A (e V f) = [g A (e V f) A e] V

[g A (e v f) A f] = (g A e) v (g A f).
(vi) implies (vii). By (vi) we have g A

(e v f) = (g A e) v (g A f) and g' A (e V f) = (g' A e) V

(g' A f). Since gC(e v f), we have e v f = [(e v f) A g] v
[(e V f) A g'] = (e A g) V (f A g) v (e A g') V (f A g') =

[(e A g) V (e A g')] V [(f A g) V (f A g')]. Now in (vi),

let g = e', and obtain eCf. Thus, since e A f = 0, we

have that e and f are orthogonal. Hence (e A g) V (e A g')

is orthogonal to (f A g) v (f A g'). Therefore, e =

(e v f) A e = ([(e A g) V (e A g')] v [(f A g) V (f A g')]

A e = (e A g) V (e A g'), i.e., eCg. Similarly, fCg.
(vii) implies (i). Condition (vi) of theorem

1.4.7 follows from (vii) at once.

Corollary 1.4.10. If L is a modular orthomodular

lattice, then in L we have V = S.













CHAPTER II


MAIN RESULTS


1. Implicative Pairs

In this section we shall study the notion of an

implicative pair in an orthomodular lattice. This notion

is motivated by the study of classical logic as follows.

Let B be a Boolean lattice, e,f E B. We then define an

element of B called "e implies f", written eDf, by the

equation eDf = e' v f. Two of the theorems which then

hold are

(1) e A (e=f) < f, (modus ponens), and

(2) e A h < f if and only if h < (e)f),

(exportation).

Now in B these theorems hold for any pair of

elements e and f. A reasonable question is: Can we de-

fine an operation = in an arbitrary orthomodular lattice

such that (1) and (2) hold? The answer is no, in general,

since Skolem (1) has shown that any lattice L in which (1)

and (2) hold for every e and f in L is necessarily dis-

tributive. This suggests that the following definition

will be non-trivial for orthomodular lattices.

Definition 2.1.1. The elements e and f in the

orthomodular lattice L are said to form an implicative








pair, written I(e,f), if and only if there exists g E L

such that

(1) e A g f and,

(2) if h E L, then e A h < f implies h < g,

both hold.

Remark 2.1.2. If h 9 g and if (1) holds, then

e A h < e Ag < f, i.e., (1) implies the converse of (2).

Lemma 2.1.3. g in definition 2.1.1 is unique

and in fact g = e' V f.

Proof. Let gl and g2 both satisfy definition

2.1.1. Since e A gl < f, we have by (2) that gl $ g2.

Similarly, e A g2 f implies that g2 g1. Hence g =

g2 and so g is unique. Since e A f < f, we have by (2)
that f < g. Also, e A e' < f so that e' g. Hence,

e' v f < g. Since e' v f < g, it suffices to show that

e A f' A g = 0. But e A g < f by (1), and so e A g f'

< f A f' = 0. Thus g = e' V f.

Convention 2.1.4. Henceforth we shall write

eDf = e' v f if and only if I(e,f).

Lemma 2.1.5. I(e,f) implies eCf.

Proof. By lemma 2.1.3 we know that e A (e' V f)

< f, i.e., f0e f. Hence, f0e f A e. But f0e f A e

always holds so that f0e = f A e. By definition, eCf.

Theorem 2.1.6. The following hold in L.

(i) I(a,c) and I(b,c) imply I(a v b,c).

Moreover, (a>c) A (b:c) = (a v b)Dc.








(ii) I(a,b) and I(a,c) imply I(a,bA c).

Moreover, (amb) A (a=c) = aD(b A c).

(iii) I(a,b) and I(b,a) imply I(a V b,

a A b). Moreover, (aDb) A (bDa) =

(a v b)D(a A b).

(iv) I(a,b) implies a A (aDb) = a A b.

(v) I(a,b) implies (a v b) A (aDb) = b.

Proof. (i) Note that if I(a,c) and I(b,c) hold,

then by 2.1.5 we have aCc and bCc. It follows that

(a v b)Cc. We will show that (1) and (2) of definition

2.1.1 hold for the pair (a v b,c). Since (a v b)Cc,

(a v b) A [(a V b)' V c] = (a V b) A c c so that (1)

holds. To show that (2) holds, suppose that (a v b) A

h < c. Then a A h & c and b A h < c. Since I(a,c) and

I(b,c), we obtain that h < a' v c and h < b' v c. Hence,

h < (a' v c) A (b' V c) = (a' A b') v c = (a V b) v c.

Finally, (amc) A (bDc) = (a' v c) A (b' v c) = (a v b)' v c

=(a v b)Dc.

(ii) Suppose that I(a,b) and I(a,c). Then

aCb and aCc so that aC(b A c). As in part (i), we check

(1) and (2) of definition 2.1.1. a A [a' v (b A c)] =

a A b A c < b A c so that (1) holds. For (2), suppose

a A h < b A c. Then a A h < b and a A h < c. Since I(a,b)

and I(a,c), we conclude that h < (a' v b) A (a' V c) =

a' v (b A c). Finally, (aDb) A (aoc) = (a' v b) A (a' V c)

= a' v (b A c) = aD(b A c).









(iii) Note that I(a,a) and I(b,b) hold.

Thus, by (i), I(a v b,b) and I(a v b,a). Now apply (ii)

to these and obtain I(a v b,a A b). The rest is simple

computation.

(iv) and (v) are direct computations.

Remark 2.1.7. (i) It is reasonable to ask

whether or not the conclusions to parts (i) and (ii)

of theorem 2.1.6 can be replaced by I(a A b,c) and

I(a,b v c) respectively. The answer is yes and the

proof depends upon consequences of our next theorem.

(ii) A reasonable conjecture is that

I(e,f) is equivalent to I(f',e'). This conjecture is

false, however, as the following example shows.


e1



a b c d e

0
The above lattice is orthomodular, I(b',e) holds, and

I(e',b) does not. In 2.1.18 we will give a necessary

and sufficient condition for both I(e,f) and I(f',e')

to hold.

Theorem 2.1.8. Let e,f E L. Then I(e,f) if

and only if eCf and e' V (e A f').

Proof. Suppose I(e,f). Then by lemma 2.1.5,

eCf. To show that e' V (e A f'), we use part (viii) of









theorem 1.4.5. Suppose g E L and g has the property

e A f' e' V g. Then e A g' & e' v f. Hence, e A g'

& e A (e' v f) = e A f 9 f. But e A g' 9 f and I(e,f)

imply g' 9 e' V f. Therefore, e A f' 9 g.

Conversely, suppose eCf and e' V (e A f').

Since eCf we have e A (e' V f) = e A f < e; hence (1)

of 2.1.1 holds. Now suppose that e A h < f. Then

e' v h' t f'. Thus f' A e < e' v h'. By part (viii)

of theorem 1.4.5, we obtain f' A e < h'. Thus

h < e' v f and so (2) of 2.1.1 holds.

Corollary 2.1.9. a V b if and only if a and

b are orthogonal and I(a',b'). Hence, a orthogonal to

b implies that I(a',b') if and only if I(b',a').

Proof. If a and b are orthogonal, then a' 2 b.

By 2.1.8, I(a',b') implies that a V (a' A b). Therefore

a V b.

Conversely, if a V b, then a and b are ortho-

gonal. Therefore a V (a' A b). By 2.1.8, I(a',b').

Corollary 2.1.10. a,b E L, a b imply I(a,b).

Proof. a b implies a b' = 0. Since xV 0

for any x E L, in particular a' V 0. Thus a' V (a A b').

By 2.1.8, I(a,b).

Corollary 2.1.11. In L(H) I(a,b) if and only

if a = 1 or a < b.

1-
L(H) = the lattice of closed subspaces of a
Hilbert space H.








Proof. In L(H), x V y if and only if x = 0

or y = 0. By 2.1.8 then, I(a,b) is equivalent to aCb

and either a' = 0 or a A b' = 0. Thus, I(a,b) is equi-

valent to a = 1 or aCb and a A b' = 0. But aCb and

a A b' = 0 is equivalent to a b.

Lemma 2.1.12. Let e,f E L. Then the following

are equivalent.

(i) I(e,f).

(ii) I(e' v f,e) and eCf.

(iii) I(e,e' v f) and eCf.

(iv) I(e,e A f) and eCf.

Proof. We use theorem 2.1.8 and the fact that

V is symmetric. I(a,b) is equivalent to a' V (a A b')

and aCb. This is then equivalent to (a A b') V a' and

aCb. But applying 2.1.8 we see that this last statement

is equivalent to I(a' v b,a) and aCb. This gives (i)

equivalent to (ii). To obtain (ii) equivalent to (iii),

let a = e' v f and let b = e. Finally, (ii) equivalent

to (iv) follows by setting a = e and b = e A f.

Theorem 2.1.13. Let a,b E L.

(i) I(a,b) and I(a,c) imply I(a,b V c).

Moreover, (a)b) v (anc) = aD(b V c).

(ii) I(a,c) and I(b,c) imply I(a A b,c).

Moreover, (a=c) v (bnc) = (a A b)nc.

Proof. (i) By 2.1.12, I(a,b) and I(a,c) imply

that I(a' V b,a) and I(a' v c,a). Applying theorem 2.1.6,









part (i), we obtain I((a' v b) v (a' v c),a), i.e.,

I(a' v (b v c),a). Since aC(b v c), we again apply

2.1.12 and obtain I(a,b v c). The rest is straight

forward computation.

(ii) I(a,c) and I(b,c) imply I(a' v c,a)

and I(b' v c,b). Also, I(a,c) and I(b,c) imply aCc and

bCc, hence (a A b)Cc. We claim that I(a' V b' v c,

a A b). Since (a A b)Cc, it is clear that (1) of defi-

nition 2.1.1 holds. To check (2) suppose that

(a' v b' V c) A x < a A b. It follows that (a' v c) A

x < a and (b' v c) A x < b. But since I(a' v c,a) and

I(b' v c,b), we obtain x < (a' v c)' v a = (a A c') V a

(a A c') v a = a and x < (b' V c)' v b = (b A c') v b = b.

Thus x < a A b and so (2) of definition 2.1.1 holds.

Hence, I(a' V b' v c,a A b), i.e., I((a A b)' V c,a A b).

This and the fact that (a A b)Cc imply by 2.1.12 that

I(a A b,c). The rest is straight forward computation.

Definition 2.1.14.

(i) A(a) = (x : x E L and I(a,x)).

(ii) P(a) = (x : x E L and I(x,a)).

Corollary 2.1.15. For every a E L, A(a) and

P(a) are sublattices of L.

Theorem 2.1.16. Let a L. Then the following

statements are equivalent.

(i) a E C(L).

(ii) I(a,x) for every x E L.

(iii) I(a,x) and I(a,x') for some x E L.








(iv) I(a,a').

(v) I(a,0).

(vi) A(a) = L.

(vii) A(a) is a sub-orthomodular lattice

of L.

Proof. (i) implies (ii). Let x E L. Then for

any g E L we have (a' v g) A ((a A x') V g) = [a' A

((a A x') V g)] v g = (a' A g) v g = g. Hence, by part

(vi) of 1.4.5 we have a' V (a A x'). Since a E C(L),

aCx. Therefore by 2.1.8 we have I(a,x).

(ii) implies (iii) is clear.

(iii) implies (i). I(a,x) implies a' V

(a A x'); I(a,x') implies a' V (a A x). By (iii) of

1.4.6 we then have a' V (a A x') V (a A x), i.e., a' V a.

Thus I(a,x) and I(a,x') imply that a E C(L).

(ii) implies (iv) is clear.

(iv) implies (v). This follows from lemma
2.1.12, parts (iv) and (i).

(v) implies (i). I(a,0) implies a' V (aA 0),
i.e., a' V a.

(i) implies (vi). This follows from the fact

that (i) implies (ii) and (ii) implies (vi).

(vi) implies (vii) is clear.

(vii) implies (i). This follows from the fact
that (vii) implies (iii) and (iii) implies (i).

Corollary 2.1.17. C(L) = nfP(x): x E L).









Corollary 2.1.18. I(e,f) and I(f',e') both

hold if and only if eCf and e' v f E C(L).

Proof. If I(e,f) and I(f',e'), then we have

by 2.1.8 that eCf, e' V (e A f'), and f V (f' A e). By

(iii) of 1.4.6 we obtain (e' v f) V (e A f'), i.e.,

e' V f E C(L).

Conversely, suppose that (e' v f) E C(L) and

eCf. By parts (i) and (ii) of 2.1.16 we obtain I(e' v f,e)

and I(e' v f,f'). By parts (i) and (ii) of 2.1.12 we ob-

tain I(e,f) and I(f',e').


2. Weakly Implicative Pairs.

By strengthening the hypothesis of (2) of defi-

nition 2.1.1 and assuming a priori that g = e' V f, we

can define a weaker form of implicativity than I(e,f).

Definition 2.2.1. We say that e and f form a

weakly implicative pair, written W(e,f), if and only if

(1) e A (e' v f)< f

(2) e A f < e A h < f imply h < e' v f,

both hold.

Remark 2.2.2. (i) I(e,f) implies W(e,f).

(ii) Definition 2.2.1 is equivalent to the

following definition: W(e,f) if and only if

(1) e A (e' v f) f,

(2) e A f = e A h implies h < e' v f,

both hold.


(iii) Many of the theorems which are true








for implicative pairs do not hold for weakly implicative

pairs, hence we will not study weakly implicative pairs in

any detail. One theorem of consequence which does hold,

however, is the analog of theorem 2.1.8. We state this

without proof, noting that the proof makes use of part

(viii) of theorem 1.4.7.

Theorem 2.2.3. Let e,f E L. Then W(e,f) if and

only if eCf and e'S(e A f').


3. Irreducibility Conditions.

As we pointed out in chapter I, a lattice L is

irreducible if and only if C(L) = f0,1). In section 3 of

that same chapter we saw that in the case of an atomic

orthomodular lattice, one can give a condition stronger

than irreducibility, namely, the atomic bisection pro-

perty. In this section we define and investigate condi-

tions which are stronger than irreducibility and which

make sense in any orthomodular lattice.

Definition 2.3.1. (i) L is said to be hyper-

irreducible, abbreviated H,I,, if and only if the fol-

lowing hold.
2 2
(1) L 22. 2

(2) 0 < e < f < 1 implies that there

exists g E L such that eCg and f(g.


222 is used to denote the lattice .








(ii) L is said to be weakly hyper-irreduci-

ble, abbreviated W.H.I., if and only if the following hold.

(1) L / 22

(2) If g E L, g / 0, g / 1, and g

not an atom, then there exists

f E L such that f A g / 0 and f(g.

(iii) L is said to satisfy condition (I) if

and only if there does not exist e,f E L such that

0 < e < f < 1 and I(f,e).

(iv) L is said to satisfy condition (W) if

and only if there does not exist e,f E L such that

0 < e < f < 1 and W(f,e).

Lemma 2.3.2. H.I. implies W.H.I. and W.HI.

implies irreducibility.

Proof. H.I. implies WoH.I. Let f E L be such

that f X O, f / 1, and f is not an atom. Then there

exists an h different from 0 and 1 such that eCh and fWh.

Let g = e V h. g A f A (e V h) A f < e A f = e so that

g A f / 0. It remains to show that ggf. Suppose gCf.

Then g0f = g A f = (e V h) A f A (e A f) v (h A f) =

e v (h A f). Also g0f = (e V h)0f = eOf V h0f A e v h0f.

Therefore, e v (h A f) = e V h0 It follows that

[e v (h A f)] A h0f = h0f. Now eCh and eCf imply that

eC(h0f). Thus h0f = (e A h0 ) v (h A f A h0f) = (e A h0f)

V (h A f) = [e A f A (h V f')] v (h A f) = (e A h) v

(h A f) = h A f. Hence h0f = h A f, i.e., hCf. This is








a contradiction, hence g(f.

W.H.I. implies irreducibility. Suppose L is

reducible and W.H.I. We claim that there exists f E L

such that f is different from 0 and 1, f E C(L), and f

is not an atom. For, suppose not. Then x E C(L), x / 0

or 1 implies that x is an atom. Now, by reducibility

there exists a g such that 0 < g < 1 and g E C(L). Also,

g' E C(L). But by assumption, g and g' are both atoms and

so we have L L(Og) x L(0,g') = 22, a contradiction.

Hence, there exists f E C(L) such that 0 < f < 1 and f is

not an atom. But W.H.I. implies that there exists an h

such that fgh. But since f E C(L), this is a contradiction.

Hence, W.H.I. implies irreducibility.

Theorem 2.3.3. Let L be atomic. Then L is H.I.

if and only if L has the atomic bisection property.

Proof. Let a and b be orthogonal atoms, a / b.

Then 0 < a < a v b. If a v b = 1, then since 1 / 22, there

exists an atom c such that c 4 a and c / b. (Clearly c < 1 =

a v b.) Hence we may suppose that 0 < a < a v b < 1. This

is equivalent to 0 < a' V b' < a' < i. By H.I. there exists

an element g E L such that gC(a' A b') and g(a'. Now g

a' A b' or else gCa'. Therefore we have g9a v b / 0. Let

c < ga v b be an atom. If c = b, then since gC(a v b) we

have b = c < g A (a V b). Thus b < g and so bCg. But bCg,

a' A b'Cg imply a'Cg, a contradiction. Thus c / b. Similar-

ly, c 4 a.









Conversely, suppose that 0 < e < f < 1. Let a

and b be atoms such that a f' and b < e' A f. (Note

e' A f # 0 or else e = f.) By the atomic bisection prop-

erty, there exists an atom c such that c < a V b, c / a,

and c 7 b. Suppose c < e' A f. Then, noting that a and

b are orthogonal and hence commute, we have c < (e' A f)

A (a v b) = (e' A f A a) V (e' A f A b) = b. Thus c = b,

a contradiction. Therefore, c e' A f. Suppose now that

c f'. Since aCb we have c < f' A (a V b) = (f' A a) V

(f' A b) = a. Hence c = a, a contradiction. Therefore,

c 4 f'. We claim that eCc and fec. It is clear that eCc

since c 4 e'. Suppose that fCc. Then cOf = c A f = 0 or

c. If c0f = 0, then c f', a contradiction. If c A f = c,

then c < f. But c < e' so this would imply c < e' A f,

a contradiction. Thus cPf.
2
Lemma 2.3.4. Suppose L / 22. Then W.H.I. if

and only if it has the following property: If h E L with

h / 0, h / 1, and h' not an atom, then there exists a com-

plement k of h such that k / h' and k A h' / 0.

Proof. Suppose that L is W.H.I. Since h' 3 0,

h' / 1, and h' is not an atom, there exists f E L such

that h' A f # 0, h(f. Put k = [(h' V f') A f] V (h' A f').

Note that k A h' = (f A h') v (h' A f') z h' A f so that

k A h' / 0. If k = h', then h' = (f A h') V (h' A f')

which implies that hCf, a contradiction. Thus k / h'.

Finally we note: k V h = [(h' v f') A f] V [(h' A f') V hb

= [(h' v f') v (h' A f') V h] A [f V (h' A f') v h] = 1 A








[(f V h) v (f V h)'] = 1 and k A h = (h' V f) A (f A h)

= 0.

Conversely, suppose that for h E L with h # 0,

h / 1, and h' not an atom, there exists k 6 L such that

g' V k = 1, g' A k = 0, g / k, and k A g / 0. If gCk

we would have g' A k' = (g' v k) A k' = k', and so g < k.

But then, since g' A k = 0, we would have g = k. This

is a contradiction, thus g(k.
2
Lemma 2.3.5. Let L / 22. Then condition (W)

implies condition (I) and condition (I) implies irreduci-

bility.

Proof. Condition (W) implies condition (I) by

remark 2.2.2. To show condition (I) implies irreduci-

bility we will show the contrapositive. Suppose L is

reducible. Then this together with the fact that L / 22

imply that there exists f E C(L) such that f / 0, f / 1,

and f is not an atom. Let e be chosen so that 0 < e < f

< 1. By theorem 2.1.16, I(f,e).

Theorem 2.3.6. L is hyper-irreducible if and

only if condition (W) holds.

Proof. We first claim that if 0 < e < f < 1,

then C(e)CC(f) is equivalent to W(f,e). For W(f,e) if

and only if eCf and f'S(f A e'). But f'S(f A e') is

equivalent to the condition gC(f' V (f A e')) implies

gCf' and gC(f A e') by Theorem 1.4.9, part (vii), Since

0 < e < f < 1, this last condition is equivalent to gCe'








implies gCf', i.e. C(e) e C(f). The theorem now follows

by noting that L is hyper-irreducible if and only if

0 < e < f < 1 implies C(e) 7 C(f).

Corollary 2.3.7. If V = S, then L is hyper-

irreducible if and only if condition (I) holds. In

general, hyper-irreducibility implies condition (I).

Theorem 2.3.8. If L is irreducible and modular,

then L is W.H.I.

Proof. Suppose g E L, g / 0, g 1 i, and g is

not an atom. Since L is irreducible, then there exists

f E L such that f / g and f is a complement of g'. If

fCg then f = g. Hence we must have fCg. If f A g y 0

we are done. Hence suppose f A g = 0. Since g is not'

an atom, there exists c E L such that 0 < c < g. Then

(c v f) A g c v (f A g) = c so that (c v f) A g / 0.

If (c v f)Cg, then c = (c V f) A g = (c V f V g') A g = g.

This is a contradiction, hence (c V f)Cg. Thus c V f

will work.

Example 2.3.9. The following is an example of

an orthomodular lattice in which property (I) holds (and

hence is an irreducible lattice) but which is not W.H.I.
1









It is a simple matter to see that (2) of defi-

nition 2.1.1 fails for all pairs (f,e) where 0 < e < f < 1.

Thus condition (I) holds. To show that L is not W.H.I. we

can use lemma 2.3.4. For, a and a' are the only comple-

ments of b other than b', but a A b' = a' A b' = 0.

Example 2.3.10. The following is an example of

an orthomodular lattice that is WoH.I. but in which con-

dition (I) fails. This lattice was first given by Dil-

worth and thus is usually denoted by the symbol D16.

1




g9 f' el d' c' b a







a c e g




0



We can use lemma 2.3.4 to show that D16 is W.HI.

For, the complements given in the following table satisfy

that lemma.









element complement
a c
b d'
c a
d b
e b'
f d'
g c'

On the other hand, I(e',d) holds so that con-

dition (I) fails. Note that since condition (I) fails,

D16 is not hyper-irreducible.

Example 2.3.11. L(H), H a Hilbert space, is

hyper-irreducible by corollary 2.1.11 and the fact that

V= S.

Remark 2.3.12. (i) The consequences of the

above theorems and examples can be depicted in diagram

of implication as follows:

W.H.I.

cond.(W)'= H.I. irreducible

cond. (I)

(ii) We can now extend theorem 1.3.7 to

the following theorem.

Theorem 2.1.13. Let L be atomic. Then the

following are mutually equivalent.

(i) L has the atomic bisection property.

(ii) L is hyper-irreducible.

(iii) Every interval L(e,f) is irreducible.

(iv) Every interval L(0,a) is irreducible.

If in addition L is complete and modular, we can add:





30


(v) L is irreducible.

(vi) L is weakly hyper-irreducible.

(vii) Condition (I) holds for L.

(viii) Condition (W) holds for L.













CHAPTER III


ADDITIONAL RESULTS


In the last chapter it was clear that our study

of implicativity was intimately associated with V and S.

Thus it is reasonable to have a brief look at some theorems

which are involved with the general problem of finding ex-

actly when V and S are equal.

1. The relative center property.

Definition 3.1.1. (i) L is said to have the

relative center property providing that for any a E L,

e is central in L(0,a) if and only if e = a A z for some

z E C(L).

(ii) The central cover of e (when it

exists) is defined to be the infimum of all of the

central elements which are greater than e. The central

cover of e is denoted by e y.

(iii) e = V ( x : x V e ), whenever this

supremum exists.

Lemma 3.1.2. Let e,a E L. Then, e = z A a for

some z E C(L) if and only if e < a < e v z', e < z, and

z E C(L).

Proof. Let z E C(L) and suppose e < a < e v z'

and e 4 z. Then e = e A z < a A z (e V z') A z = e A z

= e.









Conversely, suppose that e = z A a for some

z E C(L). Then e' = a' V z'. Thus e' A a = (a' v z')

A a = z' A a. a = e V (a A e') = e v (z' A a) = (e v a)

A (z' v e) = a A (z' V e). Therefore e a a e v z'.

e 9 z is clear.

Corollary 3.1.3. If L has the relative center

property and y exists, then e E C(L(0,a)) if and only if

e < a 9 e V (e y)'.

Proof. If e E C(L(0,a)), then e = z A a for

some z E C(L). For any such z, e < z. Therefore e y z.

Hence, e < a and e < e y imply that e < a A e y % a A z = e.

Therefore e E C(L(0,a)) if and only if e = (e y) A a. The

result now follows from the lemma.

Theorem 3.1.4. If L has the relative center pro-

perty, then the following are equivalent.

(i) a V b.

(ii) a A b = 0 and there exists no x / b

such that a ; x = a 9 b.

(iii) a A b = 0 and a and b are central in

L(O,a v b).

(iv) aSb.

(v) There exists central elements za and A

Zb such that a z A z' and b < z'
b a b a
zb'

(vi) (If y exists) a y A b y = 0.









Proof. (i) implies (11). If a V b and a x =

a i b, then x A b = (a V x) A b = (a v b) A b = b. There-

fore b < x. Note that x E L(0,a v b). Now a V b implies

a is orthogonal to b, whence a = b in L(0,a v b). Now

b 9 x implies that aCx in L(0,a v b). Therefore,

b = bA x = aA x = a x = xA (a v X) = x (a x)#

= XA 0 = x.

(ii) implies (iii) is clear.

(iii) implies (iv). Let x E L such that

x < a v b. Then x = (a A x) V (b A x). Hence aSb.

(iv) implies (v). aSb implies a and b are

central in L(0,a V b). Thus there exists central elements

z and Zb such that a = z A (a v b) and b = zb A (a b).

Then zb A a = zb A z A (a v b) and z A b = zb z
T b a a b a
A (a v b). But aSb implies that a A b = 0, so 0 = a A b

= z A zb A (a v b). Thus zb A a = 0 and z A b = 0.

Hence zb a and z' > b. It follows that a < z A z and

b < z' A .
a b
(v) implies (i). If b < za A zb and
a b
a 9 za A zb, then a A b = 0. We will make use of theorem

1.4.5, part (vii). Let x A L. Then (za v x) A b =

(za A b) V (x A b) = x A b. Thus x A b < (a v x) A b =

(za V x) A b & x A b. Therefore we have (a v x) A b

x A b. Thus a V b.

(v) implies (vi). By (v), a = a y A (a V b),

b = b y A (a v b). 0 = a A b = a y A b y A (a v b) implies

that a y A b v = 0.









(vi) implies (v). Merely take za = a y

Zb = b y.

Lemma 3.1.5. Let L have the relative center

property. If either e or e y exists for some e, then

they both exist and e = (e y)'.

Proof. Suppose e y exists. e A (e y)' = 0

and for every x E L, (e V x) A (e y)' = x A (e y)'. Thus

by (vii) of 1.4.5 we have e V (e y)'. If e V f, then by

theorem 3.1.3, e and f are central in L(O, e V f). Thus

e = (e y) A (e V f) = e v (e y A f). Hence, e y A f 9 e

implies e y A f = e y A f Ae = 0.. Therefore, f r (e y)'.

Hence e = (e y)'.
V V V
Suppose e exists. Then e V e so e, e are

central in L(0,e V e ). Hence there exists z E C(L) such

that e = z A (e V e) = (z A e) (z A e ) = (z A e) V e.

Hence z A e < e and so z A e = 0. This result together

with the fact that for all x E L, (e V x) A z = x A z,

implies that e V z. Hence z < e Thus we have shown

that z = e Since e A e = 0 and e E C(L), we have that

e < (e V'. If z1 E C(L), zI e, then e V z1. (Use part

(v) of theorem 1.4.5.) Thus zi eV and so (e )' zl.

Hence (e) = e y.

Lemma 3.1.6. (Holland). Suppose eV and e y

exist for e E L and moreover that eV = (e y)'. Then if

eSf implies e V f, then L has the relative center property.

Proof. Let e be central in L(0,a). Then









a = e I e# and e# = a A e'. Now e central in L(0,e v e )

implies eSe. By hypothesis then, e V e Since e = (e y)'.

Thus e A (e y) = 0. Now e < e y A a implies e y A a =

e v (e y A a A e') = e v (e y A e ) = e.

Lemma. 3.1.7. If b is central in L(0,a), z E C(L),

then b A z is central in L(0,a A z).

Proof. Let x E L(0,a A z). Then x E L(0,a).

(b A z)0x = x A ((b A z) v x ) = x A (b V x ) A (z V X) =

x A (b A (x' A a A z)) A (z V x ) = x A (b V (x' A a))

A (b V z) A (z V x#) = (x A b) A (b V z) A (z V x ) =

(X A b) A (z V X ) (x A b A z) V (x A b A X#) = X A b A z.

Therefore, b A z commutes with x for every x E L(0,a A z).

Theorem 3.1.8. If y exists for the lattice L,

then the following statements are equivalent.

(i) L has the relative center property.

(ii) e is central in L(0,a) if and only

if e < a 9 e V (e y)'.

(iii) eV exists for all e, eV = (e y)',

and eSf implies e V f.

(iv) a A b = 0, a y A b y p 0 imply that

there exists x / b such that a 9 x =

a b.

Proof. (i) equivalent to (ii) was corollary

3.1.3.

(i) implies (iii) follows from lemma 3.1.5

and theorem 3.1.4.









(ii) implies (i) was lemma 3.1.6.

(i) implies (iv) was theorem 3.1.4.

(iv) implies (i). Let e be central in

L(0,a). Suppose e < e y A a. Then e is central in

L(0,e y A a). Also, e y A a = e I e# where e# -
e y A a A e' / 0. Then e# A e y / 0. Hence there

exists x r e such that e 9 x = e 9 e Therefore e is

not central in L(0,e y A a), a contradiction. So e

e yA a.

Definition 3.1.9. L is said to be relatively

irreducible if and only if every interval L(0,a) is ir-

reducible.

Theorem 3.1.10. If C(L) is atomic and y exists,

then L has the relative center property if and only if

L(0,z) is relatively irreducible for every atom z E C(L).

Proof. Suppose L has the relative center prop-

erty. Let z be an atom in C(L) and let a d z. If

b E C(L(0,a)), then b = a A x for some x E C(L). Now

b = b A Z = a A x A z. But z an atom in C(L) and

x E C(L) imply that x A z = 0 or z. Therefore, b=

a A x z = 0 or a A z. But a A z = a so that b = 0 or

b = a. Hence L(0,a) is irreducible for every a E L(0,z).

Conversely, suppose e is central in L(0,a).

Then by lemma 3.1.7, e is central in L(0,e y A a). By

the same lemma, e A z is central in L(0,e y A a A z) for

every atom z E C(L). By hypothesis, e A z = 0 or





37



e A z = e y A a A z. If e A z = 0 we have (e A z) y =

e yA z = 0 and so e y Aa Az = 0. Thus in any case we

have e A z = e y A a A z for every atom z in C(L). Hence,

e = e A 1 = e A V(z:z an atom in C(L)] = V(e A z: z an

atom in C(L)j = V(e y A a A z: z an atom in C(L)) =

e y A a.













BIBLIOGRAPHY


(1) Curry, H. B., Foundations of Mathematical Logic,
McGraw-Hill, 1963, 143-144.

(2) Foulis, D. J., A Note on Orthomodular Lattices,
Port. Math., 21, Fasc 1 (1962), 65-72.

(3) Janowitz, M. F., Quantifiers and Orthomodular
Lattices, Pac. Journ. Math., Vol. 13,
No. 4 (1963), 1241-1249.

(4) Kaplansky, I., Any Orthocomplemented Complete
Modular Lattice is a Continuous Geometry,
Ann. of Math., 61 (1955), 524-541.

(5) Mackey, G. W., The Mathematical Foundations of
Quantum Mechanics, W. A. Benjamin, Inc.,
1963, 71-72.

(6) Maeda, F., Kontinuierliche Geometrien, Berlin
(1958).

(7) Travis, Raymond D., The Logic of a Physical Theory,
Masters Thesis, Wayne State Univ., 1962.













BIOGRAPHICAL SKETCH


Donald Edward Catlin was born the son of Fay H.

and Marion L. Catlin on April 29, 1936 in Erie, Pennsyl-

vania.

In June of 1954, he was graduated from Erie

Academy High School. In June of 1958, he received the

degree of Bachelor of Science from Pennsylvania State

University, and in June of 1961, he received his Master

of Arts degree from the same institution. In September

of 1961, he enrolled in the Graduate School of the Uni-

versity of Florida and since that time has been working

toward the degree of Doctor of Philosophy.

Donald Edward Catlin is married to the former

Mary Edith Higgins and is the father of one child,

Jeffrey Donald Catlin. He is a member of the Mathe-

matical Association of America, the American Mathe-

matical Society, Pi Mu Epsilon, Tau Beta Pi, and Pi

Tau Sigma.














This dissertation was prepared under the direction

of the chairman of the candidate's supervisory committee

and has been approved by all members of that committee. It

was submitted to the Dean of the College of Arts and Sciences

and to the Graduate Council, and was approved as partial ful-

fillment of the requirements for the degree of Doctor of

Philosophy.


August, 1965







Dean, College of Arts and
Sciences




Dean, Graduate School


Supervisory Committee:



Chairman








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