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Implicativity and irreducibility in orthomodular lattices
Creator:
Catlin, Donald Edward, 1936-
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1965
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English
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39 leaves : illus. ; 28 cm.

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Implicative logic ( jstor )
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Mathematical lattices ( jstor )
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Dissertations, Academic -- Mathematics -- UF
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Mathematics thesis Ph. D
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Full Text

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

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

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

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

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.

(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

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,

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

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.

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

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

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

Supervisory Committee:

Chairman

3-3 7----L-
I 7 --

Full Text

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IMPLICATIVITY AND IRREDUCIBILITY IN ORTHOMODULAR LATTICES By DONALD EDWARD CATLIN A DISSERTATION PRESENTED TO THE GRADUATE COUNQL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY UNIVERSITY OF FLORIDA August, 1965

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UNIVERSITY OF FLORIDA 3 1262 08552 2604

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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. ii

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TABLE OF CONTENTS Page ACKNOWLEDGEMENTS ii Chapter I. PRELIMINARIES 1 II. MAIN RESULTS 13 III. ADDITIONAL RESULTS 31 BIBLIOGRAPHY 38 BIOGRAPHICAL SKETCH 39 111

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CHAPTER I PRELIMINARIES 1 . Introduction The current model for the "logic" of (nonrelativistic) quantum mechanics is the lattice of closed subspaces of a separable, infinite dimensional, Hilbert space. However, as Mackey (5) points out, this assumption 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 convincing arguments to show that the "logic" of any experimental 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 statements and thereby deduce the "correct logic" for quantum mechanics. What appears to be needed is a list of physically 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.

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Let us make it clear at the outset that we do not intend to engage in polemics regarding the physical interpretations (if any) of our work. Such discussions must await subsequent research on the connection between lattice theory and physics. Hence, our approach will be from a formal mathematical point of view. 2 . Basic definitions and theorems . In this section we give some of the basic, wellknown, theorems and definitions found in the field of orthomodular 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 and a 1, and equipped with an orthocomplementation ' :L Â— >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 Â€ L with e < f. We then define the interval L(e,f) by: L(e,f) ={x^L:ex# =eV(fAx')=(evx') Af. Definition 1.2.5. Lete ^ L. We then define a mapping :L Â— >L, called the Sasaki projection determined by e_, by the rule : x0 = e a (x V e ' ) .

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Lemma 1.2.6. Let e ^ L. Then the following hold. (i) 0e = ^e Â• (ii) (x V y)0^ X0^ V y0g . (iii) x0 = X if and only if x < e. (iv) x0 =0 if and only if x < e'. Definition 1.2.7. Let e,f g L. Then f is said to commute with e, in symbols f Ce , if and only if f = f A e . e We will write f d e providing f does not commute with e. Lemma 1.2.8. Let e,f,g Â€ L. Then the following hold. (i) f s e implies f Ce . (ii) fCe implies eCf. (iii) fCe implies f Ce ' . (iv) If any two of the three conditions eCf, fCg, or eCg hold, then (e v f) a g = (e A g) V (f A g) and (e a f) V g = (e V g) A (f V g). ( v) If fCe for eachaÂ€A, andif V^.e a af A a exists, then fC V ^.e ' aÂ€A a Definition 1.2.9. LetACL. Then we define C(A) = {x Â€ L : xCy for every y Â€ AlIf 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 notation C(a) for C([a]). L is said to be irreducible if and only if C(L) fO.ll. If e Â€ C(L), e ^ 0, and e i>^ 1,

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then one can prove that L can be "factored" into the Cartesian product: L = L(0,e) x L(0,e')> hence the reason for the word "irreducible". Lemma 1.2.10. Let a 6 LThen the following hold. (i) C(a) = {(x V a)A(x V a') : x Â€ L}. (ii) C(a) = {(x A a)v(x A a') : x Â€ L}. (iii) C(a) = |;x0, V x0^, : x Â€ L]. 3 . The atomic bisection property . Throughout this section we will follow the treatment given by Janowitz (3). Definition 1.3.1. L is said to be atomic if and only if X Â€ L and x ;^ 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 Â€ L, there exists an atom a 6 L such that a =1= b, a f 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 i 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 property, 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 .

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Lemma 1.3.5. If L has the atomic bisection property, then so does any interval L(e,f). Proof . Note that the mapping a Â— >a A e ' is an orthoisomorphisra 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 modular, 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 complete, 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 following 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 JÂ£ and S^. In this section we introduce the relations 7 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.

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Definition 1.4.1. Let e,f 6 L. (i) eNCf if and only ifeA f=e Af' = 0. (ii) We say that e and f are detached , in symbols e ^ f , if and only if e and f are orthogonal and whenever g Â€ 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 Â€ L. Then eNCf if and only if every non-zero subelement of e fails to commute with f. Proof . Let e,f Â€ L and suppose that eNCf. Then by definition, e Af=e Af =0. Let 7^ e. i e. Then, clearly, e, A f = e, a f ' = 0. If e,Cf, then e, = (e, A f ) V (e, A f ') = 0. But e, i^ 0, so this is a contradiction. Hence, e, 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 ' ^^e both subelements 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 Â€ L. Then there are uniquely determined elements e, and ein L such that e, is orthogonal to q^, e.Cf, e-NCf, and e = e, V Q^. In fact e, = (e a f) v (e A f) and e^ = (f'0g) A (Â£0^).

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Proof . Note that (e a f ) v (e a f ) Â« ((fAe) ve') V ((f'A e) ve') so that e, is orthogonal to eÂ• Clearly e-jCf. To show that e-NCf we compute: e2^f = [(f' Ve') Ae] A[(fve')Ae] Af = [(f ve') Ae] A [fAe] = [f ve'] a [fAe] =0 ; 62 ^ f = [(f* ve') Ae] A f A [(f ve')Ae] = [f Ae] A [(fve')Ae] = [f Ae] a [fve'] =0. Hence e2NCf. Since e2 = e a e' and e, < e, we have e = e, V (e A e|) = (e, v 62). It remains to show uniqueness. Suppose g, and g2 have the same respective properties as e, and e^Then e, = (e a f) v (e a f ') = [(g^ V g2) A f] V [(g^ V g2 ) A f] = (g^ A f) V (g^ A f) g. , i.e . , g. = e, . Hence , eÂ„ = e A e ' = e a g ' = gj^ V g2) A g{ = g2 /\ <3[ = <32' Definition 1.4.4. Let e,f Â€ L and let e,, e2 be as in theorem 4.3. We shall refer to e = e, v 62 as the C-NC de compo s i t i on of e_ with respect to f_. Theorem 1.4.5. (Holland -Janowitz ) Let e,f Â€ L. Then the following are equivalent. (i) e 7 f . (ii) For all gCL, g Ae' =0 implies that f is orthogonal to g. (iii) For all g
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(vi) For all g Â€ L, g = (e v g) A (f V g) . (vii) For all gÂ€L, fievg implies f < g. Proof . (i) implies (ii) . Suppose that e V f and that g Â€ L with g A e' =0. Let g = g, V gbe the C-NC decomposition of g with respect to e. Then g, = (g a e) v (g A e') = g A e and g2 = g A g, = g a (g' v e'). Since e V f and g-NCe , we have by definition that gjCf. Also, since e is orthogonal to f we have g, orthogonal to f . Hence, fAg=fA (g^vg2) = (fAg^) v (fAg2) = f A g2 = f A g A (g' V e') (f A g A g') V (f A g A e') = 0. Since fCg, and fCg, then fCg. But fCg and f A g = 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) = and h V e = [e v (e ' v g) ] [g'v(e' Ag)ve] =1. Hence h is a complement of e. Therefore f
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(vi) implies (vii) g=(evg)A (fvg) implies gAf=(evg)/Â»^(fvg)Af which in turn implies 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 < 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 h A e' =0 so that e V h' = 1. By (viii), f i h', hence fCh. Corollary 1.4.6. (i) The relation v on any orthomodular lattice is symmetric. (ii) e Â€ C(L) if and only if e v e'. (iii) Let {e ] be a family of elements of L with e = V [e }. If f Â€ L and if e V f for all a, then e V f. a Theorem 1.4.7. Let e,f Â€ L. Then the following statements are equivalent. (i) eSf. (ii) For allgievf, g/ve'^O implies f orthogonal to g. (iii) For all g
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10 4.1 (v) If g < e V f, then f and e0 are orthogonal Â• (vi) If g < e Vf, then g = (e V g) A (f V g). (vii) If g < e vf, then (e v g) A f = g A f . (viii) Iff
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11 (vi) For all g e L, g a (e v f ) = (g a e ) V (g A f). (vii) For all g Â€ 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 e0^. Therefore, e0^ = = e A f, i.e., eCf. Now let g Â€ L(0,e V f) satisfy gAe=gAf=0. By (vi) of theorem 1.4.7, g = (e V g) A (f v g). Now = ( 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 (evg)Af
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12 Given g ? L, set h = g A (e v f ) . Since h commutes with it e in L(0,e v f ) , then g= (h Ae) v (h Ae^) a (h Ae) V (h A f ) Â• I"^ other words, gA(eVf) = [gA(evf) Ae]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 evf=[(evf) Aglv [(e V f) ^g'] = (e A g) V (f A g) V (e Ag') v (f Ag') = [(e Ag) V (e Ag')] v [(f Ag) v (f Ag')]. 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 vf) Ae = {[(e Ag) v (g Ag')] v [(f Ag) v (f Ag')]} 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 y = S.

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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 Â€ B. We then define an element of B called "e implies f", written eof, by the equation eof = e' v f. Two of the theorems which then hold are (1) e A (e=>f) * f, (modus ponens), and (2) e A h Â« 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 define an operation r> 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 distributive. 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 13

PAGE 18

14 pair , written I(e,f), if and only if there exists g Â€ L such that (1) e A g Â« f and, (2) if h c.

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15 (ii) I(a,b) and I(a,c) imply I(a,bA c). Moreover, (arsb) a (ar>c) = az)(b A c) . (iii) I(a,b) and I(b,a) imply l(a v b, a A b) . Moreover, (a::Db) A (bra) = (a V b)3(a A b) . (iv) I(a,b) implies a A (a3b) = a a b. (v) I(a,b) implies (a v b) a (ar>b) = 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 Â« a' v c and h < b' v c. Hence, h< (a' V c) A (b' V c) = (a* Ab') v c = (a vb)' v c. Finally, {az5c) A (boc) = (a ' v c) a (b' v c) = (a v b) ' v c = (a V b)oc. (ii) Suppose that I(a,b) and I(a,c). Then aCb and aCc so that aC(b Ac). As in part (i), we check (1) and (2) of definition 2.1.1. a a [a' v (b a c) ] = a AbAcCbAcso that (1) holds. For (2), suppose a A h < b A c. Then a a 1^ < t) and a ^ h < c. Since I(a,b) and I(a,c), we conclude that h < (a' v b) a (a' ^ c) = a' V (b A c). Finally, (aob) a (aoc) = (a' v b) a (a* v c) = a' V (b A c) = ar>(b a c) .

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16 (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. 1 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

PAGE 21

17 theorem 1.4.5. Suppose g Â€ L and g has the property e A f* < e' V g. Then e a g' Â« e' v f. Hence, e ^ g' Â« e A (e* V f) = e A f Â« f . But e a g' < f and I(e,f) imply g' Â« e' V f. Therefore, e A f' * q. Conversely, suppose eCf and e' V (e A f ). Since eCf we have eA(e'Vf)=eAfÂ«e; hence (1) of 2.1.1 holds. Now suppose that e A h < f. Then e' V h' s f. Thus f A e Â€ 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' a b. By 2.1.8, I(a',b') implies that a V (a' A b) . Therefore a y b. Conversely, if a v b, then a and b are orthogonal. Therefore a V (a' A b) . By 2.1.8, I(a',b*). Corollary 2.1.10. a,b 6 L, a < b imply I(a,b). Proof , a < b implies a a b' = 0. Since x V for any x Â€ 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. L(H) = the lattice of closed subspaces of a Hilbert space H.

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18 Proof . In L(H) , x V y if and only if x = or y = 0. By 2.1.8 then, I(a,b) is equivalent to aCb and either a' =OoraAb' =0. Thus, I(a,b) is equivalent to a = 1 or aCb and a a b' =0. But aCb and a A b' = is equivalent to a < b. Lemma 2.1.12. Let e,f ^ L. Then the following are equivalent. (i) I(e,f). (ii) I(e Â• 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 Â€ L. (i) I(a,b) and I(a,c) imply l(a,b V c) . Moreover, (ar^b) v (arDc) = a3(b v c). (ii) I(a,c) and I(b,c) imply I (a A b,c). Moreover, (auc) v (boc) = (a A b)3c. 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,

PAGE 23

19 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 Kb* 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 definition 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 Â« 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 Ac') V a = a and x< (b' vc)' vb= (b ac') vb=b. Thus X < a A b and so (2) of definition 2.1.1 holds. Hence, I(a' Vb' v c,a Ab), i.e., I((a Ab)' v c,a Ab). 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 Â€ L and I(a,x)]. (ii) P(a) = {x : X Â€ L and I(x,a)}. Corollary 2.1.15. For every a Â€ L, A(a) and P(a) are sublattices of L. Theorem 2.1.16. Let a c l. Then the following statements are equivalent. (i) a f C(L) . (ii) I(a,x) for every x Â€ L. (iii) I(a,x) and I(a,x') for some x Â€ L.

PAGE 24

20 (iv) I(a,a'). (v) l(a,0). (vi) A(a) = L. (vii) A(a) is a sub-orthomodular lattice of L. Proof . (i) implies (ii) . Let x Â€ L. Then for any g Â€ L we have (a* Vg) a ((a ax') Vg) = [a' A ((a ax') V g) ] V g = (a* a g) v g = g. Hence, by part (vi) of 1.4.5 we have a' V (a Ax'). Since a 6 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 ax') v (a Ax), i.e., a' V a. Thus I(a,x) and I(a,x') imply that a Â€ 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 (a a 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) = nrP(x): x f L] .

PAGE 25

21 Corollary 2.1.18. I(e,f) and I(f ,e') both hold if and only if eCf and e' v f Â€ 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) 7 (e a f), i.e., e' V f Â€ C(L) . Conversely, suppose that (e ' v f) Â€ 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 obtain I(e,f) and I(f',e'). 2 . Weakly Implicative Pairs . By strengthening the hypothesis of (2) of definition 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) eAf
PAGE 26

22 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 Â€ L. Then W(e,f) if and only if eCf and e'S(e a f). 3. Irreducibilitv Conditions . As we pointed out in chapter I, a lattice L is irreducible if and only if C(L) = [0,11. 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 property . In this section we define and investigate conditions which are stronger than irreducibility and which make sense in any orthomodular lattice. Definition 2.3.1. (i) L is said to be hyperirreducible, abbreviated H,I., if and only if the following hold . (1) h / 2^. ^ (2) 0
PAGE 27

23 (ii) L is said to be weakly hyper-irreducible, abbreviated W.H.I. , if and only if the following hold. (1) L 7^ 2^. (2) Ifg6L, g?^0, gj^l, and g not an atom, then there exists f Â€ L such that f A g ?^ and f0g. (iii) L is said to satisfy condition (I) if and only if there does not exist e,f Â€ L such that < 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 Â€ L such that < e < f < 1 and W(f ,e) . Lemma 2.3.2. H.I. implies W.H.I, and W.H.I, implies irreducibility . Proof . H.I, implies W.H.I . Let f Â€ L be such that f 7^ 0, f 7^ 1, and f is not an atom. Then there exists an h different from and 1 such that eCh and f(?h. Let g=e^h. gAfA(eVh)Af
PAGE 28

24 a contradiction, hence gCff. W.H.I, implies irreducibility . Suppose L is reducible and W.H.I. We claim that there exists f 6 L such that f is different from and 1, f Â€ C(L), and f is not an atom. For, suppose not. Then x Â€ C(L), x 7^ or 1 implies that x is an atom. Now, by reducibility there exists a g such that < g < 1 and g 6 C(L). Also, g' Â€ C(L). But by assumption, g and g* are both atoms and so we have L ~ L(Opg) x L(0,g') = 2^, a contradiction. Hence, there exists f 6 C(L) such that < f < 1 and f is not an atom. But W.H.I, implies that there exists an h such that f^h. But since f Â€ 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 atcxnic bisection property. Proof . Let a and b be orthogonal atoms, a / b. Then 0
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25 Conversely, suppose that < e < f < 1. Let a and b be atoms such that a < f and b < e' A f. (Note e* A f 7^ or else e = f.) By the atomic bisection property, there exists an atom c such that c^avb, cT^a, and c 5^ 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
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26 [(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 Â€ L with h ?^ 0, h 7^ 1, and h' not an atom, there exists k Â€ L such that g' Vk = l, g'Ak = 0, g^^k, and k A g 7^ 0. If gCk we would have g' a^' = (g' vk) a^^' = ^' t ^nd so g < k. But then, since g' /^ "k = 0, we would have g = k. This is a contradiction, thus g(^k. 2 Lenuna 2.3.5. Let L ?^ 2 . Then condition (W) implies condition (I) and condition (I) implies irreducibility. Proof . Condition (W) implies condition (I) by remark 2.2.2. To show condition (I) implies irreducibility we will show the contrapositive . Suppose L is 2 reducible. Then this together with the fact that L ;^ 2 imply that there exists f Â€ C(L) such that f 7^ 0, f ^ 1, and f is not an atom. Let e be chosen so that < e < f < 1. By theorem 2.1.16, I(f,e). Theorem 2.3.5. L is hyper -irreducible if and only if condition (W) holds. Proof . We first claim that ifO
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27 implies gCf ' , i.e. C(e) (^ C(f). The theorem now follows by noting that L is hyper-irreducible if and only if < e < f < 1 implies C(e) 7 C(f). Corollary 2.3.7. If V = s, then L is hyperirreducible 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 gÂ€L, g?^0, gj^l, and g is not an atom. Since L is irreducible, then there exists f Â€ L such that f 7^ g and f is a complement of g'. If fCg then f = g. Hence we must have f^g. If f A g ?^ we are done. Hence suppose f A g = 0. Since g is not' an atom, there exists c Â€ L such that < c < g. Then (cvf) Ag = cv (fAg) =cso that (c v f) a g ?^ 0. If (c V f)Cg, then c= (cVf) Ag= (cVfVg') Ag=g. This is a contradiction, hence (c ^ t)(ig. Thus cvf 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 a'

PAGE 32

28 It is a simple matter to see that (2) of definition 2.1.1 fails for all pairs (f,e) where < 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 complements of b other than b', but aAb'=a'Ab'=0. Example 2.3.10. The following is an example of an orthomodular lattice that is W^H.I. but in which condition (I) fails. This lattice was first given by Dilworth and thus is usually denoted by the symbol D,^. 1 We can use lemma 2.3.4 to show that D,, is W.H.I, For, the complements given in the following table satisfy that lemma .

PAGE 33

29 element

PAGE 34

30 (v) L is irreducible. (vi) L is weakly hyper-irreducible (vii) Condition (I) holds for L. (viii) Condition (W) holds for L.

PAGE 35

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 exactly 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 Â€ L, e is central in L(0,a) if and only if e = a A z for some 2 Â€ 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=Vfx:xVe}, whenever this supremum exists. Lemma 3.1.2. Let e,a Â€ L. Then, e = z a a for some z Â€ C(L) if and only ifeÂ«a
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32 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') Aa=z'Aa. a=eV(aA e') = e v (z ' A a) = (e v a) A (z' ve) =a A (z* ve). Therefore e * a < e v z*. e < z is clear. Corollary 3.1.3. If L has the relative center property and y exists, then e Â€ C(L(0,a)) if and only if e < a < e V (e y) ' Â• Proof . If e Â€ c(L(0,a)), then e = z a a for some 2 Â€ C(L). For any such z, e < z. Therefore e \ < z. Hence, e < a and e < e y imply that e^af\ey*aAZ=e. Therefore e Â€ 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 property, then the following are equivalent. (i) a V b. (ii) a A b = and there exists no x 7^ b such that a ^ x = a ^ b. (iii) a A b = and a and b are central in L(0,a V b) . (iv) aSb. (v) There exists central elements z and a z, such that a * z a z ' and b < z ' b aba ^b(vi) (If Y exists) a y a b y = 0.

PAGE 37

33 Proof . (i) implies (11) . If a V b and a C' x = a C' b, then x A b = (a v x) A b = (a v b) a b = b. Therefore b < X. Note that x Â€ 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 Â« X implies that aCx in L(0,a v b) . Therefore, b=bA x=a*A x= a^0^ = x a (a* v x*) = x a (a a ^) = X a 0^ = X. (i i) implies (iii) is clear. * (iii) implies (iv) . Let x Â€ 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 z^ such that a = z A (a v b) and b = z, A (a . b) . a b a D Then z a a = Zj^ A z A (a v b) and Zg ^ b = z^ a z^ A (a V b) . But aSb implies that aAb=0, soO=aAb = z A Zt^ A (a V b) . Thus z a a = and z^ A b = . Hence z ' a a and z' 2 b. It follows that a < z a z ' and b < z ' A z, . a b (v) implies (i) . If b < z' a z and a < z ' A z, , then a a b = 0. We will make use of theorem a b' 1.4.5, part (vii) . Let x A L. Then (z^ v x) A b = (z A t)) V (x A b) = X A b. Thus xAb< (avx) Ab = (z Vx)Ab
PAGE 38

34 (vi) implies (v) . Merely take z = a y 2^ = b Y. Lemma 3.1.5, Let L have the relative center property. If either e or e y exists for scane e, then they both exist and e = (e y) ' Â• Proof . Suppose e y exists, e A (e y) * =0 and for every x Â€ 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(0, e v f ) . Thus e = (e y) A (e V f ) = e v (e y A f ) . Hence, e y A f < e implies eyAf=eyAfAe=0. Therefore, f Â« (e y) ' . Hence e = (e y) ' . V V 7 Suppose e exists. Then e V e , so e, e are central in L(0,e v e ). Hence there exists z Â€ C(L) such that e^= z A (e V e^) = (z ^ e) V (z A e^) = (z A e) V e^. Hence z A e Â« e and so z A e = 0. This result together wi th the fact that for all xÂ€L, (eVx) Az=xAz, implies that e V z. Hence z Â« e . Thus we have shown that z = e . Since e A e =0 and e Â€ C(L) , we have that e < (e^)*. If z^ 6 C(L), z^ a e, then e V z^. (Use part (v) of theorem 1.4.5.) Thus z^ < e and so (e ) ' Â« z^. Hence (e^) ' = e y. Lemma 3.1.6. (Holland). Suppose e'^ and e y exist for e Â€ L and moreover that e^ = (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

PAGE 39

35 # # # a = e ^ e and e = a as'. Now e central in L(0,e v e ) # V implies eSe. By hypothesis then, e V e . Since e = (e y)'Â» it Thus e A (e y) = 0. Now e < e y ^ ^ implies e y a a = it ev (eY^^Ae') = ev {e y A\ e ) =e. Lemma . 3.1.7. If b is central in L(0,a), z Â€ C(L), then b A z is central in L(0,a a z). Proof . Let X Â€ L(0,a A z). Then x Â€ L(0,a). (b A 2)0^ = X A ((b A Z) V X*) = X A (b V X^) A (Z V X*) = X A (b V^ (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*) = (XAbAZ) V (XAbA X*) = X A b A Z. Therefore, b a z commutes with x for every x Â€ 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 iS e V (e y) ' Â• (iii) e exists for all e, e^ = (e y) ' . and eSf implies e y f. (iv) aAb = 0, aYAbYJ^O imply that there exists x j^ b such that a 'J' x = a ^3b. Proof . (i) equivalent to (ii) was corollary 3.1.3. (i) implies (iii) follows from lemma 3.1.5 and theorem 3.1.4.

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36 (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 YAa=e^e where e it e YA aA e' j^O. Then e y A e Y ?^ 0. Hence there exists X ?^ e such that e ^ x = e '^ e . Therefore e is not central in L(0,e y ''^ a) Â» ^ contradiction. So e = e Y A a. Definition 3.1.9, L is said to be relatively irreducible if and only if every interval L{0,a) is irreducible . 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 Â€ C(L). Proof . Suppose L has the relative center property. Let z be an atom in C(L) and let a < z. If b Â€ C(L(0,a)), then b = a A x for some x Â€ C(L). Now b=bAZ=aAXAz. But z an atom in C(L) and X Â€ C(L) imply that xAz=Oorz. Therefore, b = aAxAZ=OoraAz. But a a z = a so that b = or b = a. Hence L(0,a) is irreducible for every a Â€ 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 Â€ C(L). By hypothesis, e a z = or

PAGE 41

37 eAz=eY^S''^z. ifeA2 = 0we have (e A z) y = e Y ^ 2 = and soe y Aa az =0. Thus in any case we have eAz=eY''^3^2 for every atom z in C(L). Hence, e=eA l=eA v(z:z an atom in C(L)} = v{e A z: z an atom in C(L)3 = V(e y ^ a a z: z an atom in C(L)} = e y ^ a.

PAGE 42

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 Orthocoraplemented 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, 38

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BIOGRAPHICAL SKETCH Donald Edward Catlin was born the son of Fay H, and Marion L. Catlin on April 29, 1936 in Erie, Pennsylvania. 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 University 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 Mathematical Association of America, the American Mathematical Society, Pi Mu Epsilon, Tau Beta Pi, and Pi Tau Sigma. 39

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This dissertation was prepared under the direction of the chairman of the candidate's supervisory coinmittee 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 fulfillment of the requirements for the degree of Doctor of Philosophy. August, 1965 Dean, College of Arts and Sciences Supervisory Committee: ^ Chairman 7T \A

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