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 nonzero 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 nonzero 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 CNC decomposition of e with respect to f.
Theorem 1.4.5. (HollandJanowitz) 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 CNC
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 nontrivial 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 suborthomodular 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 hyperirreduci
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 hyperirreducible 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 hyperirreducible 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, hyperirreducibility 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 hyperirreducible.
Example 2.3.11. L(H), H a Hilbert space, is
hyperirreducible 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 hyperirreducible.
(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 hyperirreducible.
(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,
McGrawHill, 1963, 143144.
(2) Foulis, D. J., A Note on Orthomodular Lattices,
Port. Math., 21, Fasc 1 (1962), 6572.
(3) Janowitz, M. F., Quantifiers and Orthomodular
Lattices, Pac. Journ. Math., Vol. 13,
No. 4 (1963), 12411249.
(4) Kaplansky, I., Any Orthocomplemented Complete
Modular Lattice is a Continuous Geometry,
Ann. of Math., 61 (1955), 524541.
(5) Mackey, G. W., The Mathematical Foundations of
Quantum Mechanics, W. A. Benjamin, Inc.,
1963, 7172.
(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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