AN INTRODUCTION TO BOOLEAN ALGEBRA

by

R. J. SQUIRES

A T the Annual General Meeting of the Students' Society in 1961,a motion was passed that articles of a less serious nature than thosenormally published in the past be accepted by the Editors of J.S.S.It is in the spirit of this motion that this article is offered. It doesnot pretend to advance actuarial science in any way, and its onlypurpose is to introduce members of the profession to a techniquewhich is interesting in that it is different from our normal tech-niques and to show that it has possibilities in the actuarial field bydemonstrating its applications in other fields.

This is not the first time that the subject has been written aboutin an actuarial journal. Edmund C. Berkeley (1937) wrote a paperfor the American Institute of Actuaries going into the subjectin some detail, and it is from this paper that I have taken myexamples of applications to life office practice.

At this point it would seem meet to record my thanks to him forpermission to use this material, and also to F. M. Goldner ofAlleyn's School who first introduced me to this subject, and towhom I owe my examples of logical problems.

The algebra derives its name from the English mathematician,George Boole, who lived between 1815 and 1864, and is thought tobe the first person to demonstrate the possibility of paraphrasinglogical argument by algebraic manipulation. The following quota-tions sum up the power that is thereby added:

' The more complex forms of inference cannot be studied untila specially devised symbolism is introduced.. .which enables usto keep different meanings distinct, to concentrate upon what isessential, to show the form of propositions and to save labourand thought.' M. R. Cohen and E. R. Nagel.

'Arithmetic at first lacked any more appropriate medium thanthat of ordinary language. Ancient Greek mathematicians had

BOOLEAN ALGEBRA 315

no symbol tor zero and used letters 0f the alphabet for othernumbers. As a result it was impossible to state any general rulefor division—to give only one example. Operations which anyfourth grade child can accomplish in the modern notation, taxedthe finest mathematical minds of the age of Pericles. Had it notbeen for the adoption of new and more versatile ideographicsymbols, many branches of mathematics could never have beendeveloped, because no human mind could grasp the essence oftheir operations in terms of the phonograms of ordinarylanguage.' C. I. Lewis.In some fields advances have been made with this technique up

to the expectations one might have of it—in others, and mainly itsfirst field, that of logical deduction, there has been no advance, asfar as I know, for many years. There is, however, no reason to thinkthat further advances cannot be made, and what more can a field ofresearch offer?

Turning to the details of the subject, I present it here as anabstract algebra where the rules are to be learnt as they stand, andthere is no question of understanding the relationships. However,most people will find it easier to skim through these rules and passrapidly to the applications and examples. For those brave soulsintending to try to master the conditions of manipulation first, itmay be some help to mention that for most applications the vari-ables have two values only, o and 1, and these are not to be thoughtof as integers, but rather symbols of 'nothingness' and 'every-thineness'.

RULES OF THE ALGEBRA

Some of the rules of our normal algebra we take so for granted thatwe do not even realize their existence. The first group of laws,given below, hold both for our normal algebra and for Booleanalgebra:

Addition MultiplicationCommutative Law

x+y - y+xAssociative Law

(x+y)+z = x+(y + z)Distributive Law

(3)

xy = yx ( 2 )

(xy)z = x(yz)x(y+z) = xy+xz

(4)(5)

(z)

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The Algebra of Classes, which is a Boolean Algebra, has thefollowing additional laws:

Addition Multiplication

Law of Absorptionx+x = x

Distributive Law(6)

(8)

X.X = X (7)

x+yz = ( x + y ) ( x + z )At this point we introduce the conception of the complement,

and define x as all of the class with which we are dealing (the'reference class') which is not x. We use 1 to represent the 'uni-versal class', that is the total of the class we are referring to, and oto represent the 'null class', or empty class.

We now have the further laws:

x+x = 1 (9); X.X = O (1o); (x) = x (II).

Also:1+x = 1 (12);o+x = x (14);o = 1 (16);

o.x = o (13);1.x = x (15);1 = 0 (17).

If x+y = o,then x = o and y = o (18).

then x = 1 and y = 1 (19).

The last two laws, (20) and (21), are sometimes referred to as DeMorgan's Laws.

It will be noticed that the laws go in pairs; for each law of addi-tion there is a corresponding law of multiplication (Law (5) pairswith Law (8)). Further inspection yields the Principle of Duality.This says that having proved a certain relationship, if we sub-stitute multiplication for each addition, addition for each multipli-cation, and replace each variable or constant by its complement,the resulting relationship will be true also. For example, Law (5) isx.{y+z) = xy+xz. Following the above procedure we arrive at

{x+y) x.y (20); (xy) x+y (21).

lix.y = i,

BOOLEAN ALGEBRA 317

the relationship x +y. z = (x +y)(x+z) which is Law (8) written interms of the variables x, y and z.

The expression may be paraphrased in terms of x, y and z ifrequired. The question may then be asked, what is the need to takethe complements of the variables? The answer lies in the relation-ship between laws involving o and 1; only by this device do theytranslate correctly. (To make this position quite clear, perhaps itshould be pointed out that in the transformation, o becomes 1 and1 becomes o.)

We are now ready for the important step that since this trans-formation holds for Laws (1)-(21) and since any other functions orrelationships we form will be based on these laws, these functionsand relationships may also be transformed according to the prin-ciple of duality. This is evidently a powerful technique.

THE ALGEBRA OF CLASSES—VENN DIAGRAMSo far, we have developed rules for an algebra on a completelyabstract basis except for the definitions of complement, o and 1.Even here it was not necessary to introduce terms of the algebra ofclasses, but it was convenient to do so to avoid difficulties in defin-ing abstract values for these. We now proceed to define the vari-ables and operators in terms of this algebra.

' x' represents the class of objects having the property px, out of agiven reference class.

'y' represents the class of objects having the property py, out ofthe same reference class.

'xy' represents the class of objects having both the property px

and the property py.'x+y' represents the class of objects having either the property

px or the property pv or both.'x' represents the class of all objects in the universal class not

having the property px.' 1' represents the universal class, so that 'x = 1' means 'every

object in the reference class has the property px,.

' o ' represents the null class, so that 'x = o' means 'no object inthe reference class has the property px

,.A further operator sometimes used is that of inclusion, usually

318 R. J. SQUIRES

designated < or c , a c b is to be read as 'the class a is whollycontained by the class b\ This immediately leads us to the equa-tion ab = a, so that the operator is not indispensable although use-ful in the translation of statements into algebraic form.

To see how the algebra operates on classes, let us take as anexample the Distributive Law of Addition—Law (8). This states:x+yz = (x+y)(x+z), and to the mind trained to normal algebrais probably the most strange-looking of all the laws. Translatedinto terms of classes it states:' The class of objects having either thepropertypx or both the propertypy and the propertypz is equal tothe class of objects having (i) either the property px or the propertypy, and (ii) either the property px or the property pz.' To see thatthis is true consider first objects having the property px. These areincluded by either statement of the rule. Next, consider thoseobjects which do not have this property. To be included accordingto either statement of the rule, these must have both the propertypy and the property pz. Hence the two statements are equivalent.

These conceptions can be represented diagrammatically, and formany people the principles will be easier to grasp if presented inthis way. The diagrams below, known as Venn diagrams, illustratesome of the points above. The rectangle in each case is supposed torepresent the universe or reference class with which we are dealing,and the classes which we are considering are represented by irregu-lar shapes. It will be realized that although, in the diagram below,each class is represented by a single area contained by a con-tinuous line, this is not necessary, and a class could consist of oneobject represented by a point in the top right-hand corner andanother represented by a point in the bottom left-hand corner, orindeed anywhere else within the rectangle. Perhaps at this point itwould be helpful to point out that the components of the classesneed not necessarily be objects. I have worked in terms of objectsbecause I feel the ideas are easier to grasp if put this way; it is notnecessary to confine the discussion to objects.

In Fig. 1 the shaded area represents x+y, that is the class ofobjects having either the propertypx or the property py. In Fig. 2,the shaded area represents x.y, that is the class of objects havingboth the property px and the property py. Thus the white area in

BOOLEAN ALGEBRA 319

Fig. 1 represents {x+y). Alternatively, we may say that the whitearea is 'not x' and 'not y' or x.y, thus demonstrating Law (20).Law (21) may similarly be demonstrated from Fig. 2. Law (8)which was explained verbally above in terms of classes may beverified by reference to Fig. 3.

Fig. 1 Fig. 2

Case I.Fig. 3.

Case II.

In each case the double-hatched area represents the function.Other cases exist but are mostly trivial.

As a simple example of the working of this algebra, let us considerthe classic question posed by Lewis Carroll:' In a very hotly foughtbattle, at least 70 % of the combatants lost an eye, at least 75 % anear, at least 80 % an arm and at least 85 % a leg. How many lost allfour members?'

For convenience, let us suppose the total number involved was100: we may revert to percentage form at the end.

represent the class of combatantthat lost an

eye; /mln. = 70.ear; Emin = 75.arm;Amln. =80.leg; Lmln_ = 85.

Let (/)]Let (E)Let (A)Let (L),

320 R. J. SQUIRES

Then (/+E) represents the class of combatant that lost either aneye or an ear, etc.

Similarly, (^L)mm.

= (I) + (E)-(IE),= (I)min. + (E)min.-(I+E)max.'

= 70+75-100 = 45.= (A)min. + (L)mln.-(A+ L)max. = 65,= (IE)mln.+(AL)min.-(IE+ AL)max.= 45 + 65 — IOO = IO.

Thus 10% at least lost all four members.In the above example it has been implicitly assumed that (IE)

can take its minimum value when (I+E) takes its maximum value.No proof is offered here, but reference to the Venn diagram shouldconvince most readers of the validity of this.

As a further example, try working Example 3.1 of Probability(Bizley, 1957) using this notation, and relationships such as:EMF = EMF(G+G). This I felt certain had been done in someother text-book, but was unable to trace it.

THE ALGEBRA OF PROPOSITIONS

In this algebra our variables represent propositions. Thus xmight represent the proposition ' I am now writing with a pen' andy, ' I am now writing with a pencil'. Addition represents proposi-tions taken in disjunction; that is one or other or both are asserted.Multiplication represents propositions taken in conjunction, that isboth asserted simultaneously, o represents falsity and 1 verity,rather as in probability o represents impossibility and 1 certainty.Thus, with the above example x+y represents the proposition'either I am now writing with a pen or I am now writing with apencil, or both', and xy represents the proposition ' I am nowwriting with a pen and with a pencil'. If pens and pencils were theonly writing implements we could write x +y = 1: for most normalpeople we could write xy — o. In this algebra x represents theopposite assertion; that is for our example, ' I am not now writingwith a pen'.

The algebra follows Laws (1)-(21) above: most solutions comefrom Laws (18) and (19). Very often the most difficult part of the

••• (/£)«!«.Now,(/+£)

and (IE. AL)mln.

BOOLEAN ALGEBRA 321

problem is to derive a satisfactory notation. Once this has beendone, and all the information made use of, the solution followswithout difficulty. Two examples, one simple, one more difficult,are now given.

First example

Three counters A, B and C are coloured red, white and blue, butnot necessarily respectively. One only of the following statementsis true:

(a) A is red, (b) B is not red, (c) C is not blue.What colour is each counter?

Let ar represent the proposition that A is red,ab represent the propositon that A is blue,etc.

Since one statement is true, we may assert them in disjunctioncorrectly. Since only one is true any conjunctive assertion of twostatements will be false.

In algebra:

Now a..br = o, since if A is red, B cannot be red, and vice versa.(Law 5).

(Law 9).(Law 15).

But

(Since if c is blue, one of A or B must be red.)Also 1But since (Law 16).

(Law 15).(Laws 5 and 9).

But

(Law 15).

(Law 14).(Law 14).

(Law 14).(Law 16).

Translation: A is blue, B is red, and C is white.

322 R. J. SQUIRES

Second example

Out of 6 boys, 2 were known to have been stealing apples. Butwho? Harry said 'Charles and George'. James said, 'Donald andTom'. Donald said, 'Tom and Charles'. George said, 'Harry andCharles'. Charles said, 'Donald and James'. Four of the boysinterrogated named just one miscreant correctly. The fifth had liedoutright. Who stole the apples?

Here the notation suggests itself immediately: Let C denote theproposition that Charles stole the apples, etc.

Since in each case there was at least one lie, each of the productsCG, DT, CT, CH and DJ equals zero.

Since in exactly one case there was a double lie, all the sums equalone, save one. Thus the product of all five sums is zero, and theproduct of one set of four sums is one, so that the sum of all fivepossible sets of four must be one.

In symbols:CG = DT=CT= CH = DJ=o (i)...(v),(C+G)(C+H)(C+T){D+T)(D+J) = o (vi),(C+G)(C+H)(C+ T)(D+T) + (C+G)(C+H)(C+ T)(D+J)

+ (C+G){C+H)(D+T)(D+J) + (C+G)(C+T)(D + T)(D+J)+ (C+H)(C+T)(D+T)(D+J) = 1 (vii).

Expression (vi) immediately simplifies to (C+GHT)(D + TJ) = 0by continued application of Law 8. Now since only two boys stoleapples, any product of three terms is automatically zero, so that theabove equation becomes:

and ultimatelyC{D+TJ) = 0,

CD = 0.Expression (vii) simplifies, as above, to

(C+GTH)(D + T) + (C+GTH)(D+J) + (C+GH)(D+Tf)+(C+GT)(D+TJ) + {C+HT)(D+TJ) = 1,

C(D+T+J) + (C+GH+GT+HT)(D+TJ) = 1or(remembering X+X = X, and products of three terms are zero).Multiplied out this yields: CD+CT + CJ+CD = 1. ButCD = CT= 0. Hence CJ = 1.

Translation: Charles and James stole apples.

BOOLEAN ALGEBRA 323

There is a further conception and operator that is sometimesuseful, but like the operator 'inclusion' in the algebra of classes isnot indispensable. It is that of implication. We say 'A implies B,

,

and write 'A B', meaning ' If A is true, then B is also true.' Thismay alternatively be phrased 'Either A is not true, or else B istrue' which leads us to our alternative form A+B = 1. The typeof case where this operator is most useful is that where we wish toexpress 'The fact that A implies B implies that C implies D;where we can write {A B) (C D), and proceed by easystages to the algebraic form. Let us consider another simpleexample:

In a certain strange community, everyone is either an Actuary ora Statistician. Actuaries always tell the truth. Statisticians alwayslie (in this strange community!). Three men meet and X introduceshimself, mentioning his profession. Y then remarks to Z, 'He sayshe is an Actuary'. Z replies, 'No, he is a Statistician'. How manywere there of each profession?

Solution: Let XA be the proposition that X is an Actuary, andsimilarly for Y and Z.

Now XA YA, (since X will have made a true statement, and Ycorrectly reported it).

Also XA YA (since X will have made an untrue statement andY correctly reported it), i.e.

Also

Therefore either ZA.XA = 1 or XA.ZA = 1. (Not both in thiscase.)

Hence one of X and Z is an Actuary, the other a Statistician,though we know not which. We have already determined that Y isan Actuary.

324 R. J. SQUIRES

In the above example it is easier to use logical deduction thanalgebra. In particular, the fact that Y is an Actuary becomesapparent immediately, without the need of multiplying out thederived equation. However, it may be seen that this was in factan example of a simple and useful theorem, namely

EXAMPLES FOR PRACTICE

1. Anthony, Brian, Charles and Douglas competed for ascholarship. 'What luck have you had?' someone asked. SaidAnthony: 'Charles was top. Brian was second.' Said Brian:'Charles was second and Douglas was third.' Said Charles:' Douglas was bottom. Anthony was second.' Each of the threeboys had made two assertions, one at least of which was true. Whowon the scholarship?

2. Four members of my club, Messrs Albert, Charles, Frederickand Dick, have recently been knighted, so that now their friendshave had to learn their Christian names. Now, the surname of eachis the Christian name of one of the others. Dick is not the Christianname of the member whose surname is Albert. The Christian nameof the member whose surname is Frederick is the surname of themember whose Christian name is the surname of the member whoseChristian name is Charles. Find the Christian names of each of thefour members.

[(A B).(A B)] [B=1].Further, there are many cases where the logical chain is so complexthat the algebraic method will be easier than the process of deduc-tion, and, most important of all, it will not let us deduce facts thatwe should not; that is to say, logical deduction will sometimesappear to point to a conclusion which we are anxious to make:algebra will not let us make it!

There are some further operators in use, which I would notconsider worth using, but list for the benefit of anyone who mightencounter them whilst reading on this subject:

Symbol EquivalenceNameRing Sum ('Exclusive or')Scheffer strokeDouble Stroke

BOOLEAN ALGEBRA 3253. Mrs P; Mrs Q; Miss W; Miss X; Miss Y and Miss Z are

sitting round a circular table. They are, not necessarily respectively,the Admiral's wife, an ATS girl, an NFS girl, a Red Cross worker,a WRAF girl, and a WRNS girl. The Admiral's wife sits oppositeMiss X. The ATS girl sits on the left of Mrs P. Miss Z sits on theleft of Miss Y. The WRAF girl sits opposite Miss Y. Miss Wsitsopposite the Red Cross worker. Mrs Q sits on the left of theWRNS girl. Mrs P is not the NFS girl.

Who represents each service and how do they sit round thetable?

Solutions to these examples are given in the Appendix.

SWITCHING ALGEBRA

This is a comparatively modern application of Boolean Algebra,and demonstrates well its possibilities. It also helps to bring homethe fact that there is no 'Act of God' about either normal or BooleanAlgebra, but that observing certain phenomena leads us to formulaterules for our algebra, which provided they do not lead to contra-dictions, will yield useful results. The main application is inanalysing circuits, chiefly in connexion with telephones, but also inconnexion with mechanics of computers.

F(a, b)=a+bFig. 4

F(a, b)=a.bFig-5

The basic idea is very simple. We denote a given switch by 'a'and write a = 1 meaning that the switch is closed; a = o meaningthat it is open. We denote by a any switch which is open when a is.closed and vice versa. It is then a natural step to designate switchesin parallel by addition, and switches in series by multiplication.Consider Figs. 4 and 5. In Fig. 4 the circuit between X and Y is.closed if either a or b is closed; in Fig. 5 the circuit is only closedwhen both are closed.

In the figures the function F(a, b) has been introduced. This is

336 R. J. SQUIRES

normally referred to as the 'Switching Function' of the circuit,meaning that it is a function that will take the value 1 when there isa path through the circuit and 0 when there is no path. It followsthat the functions may be combined in the same manner as theoriginal variables. As an exercise in writing down switching func-tions, consider the circuit in Fig. 6. This may be seen to be

abc+(a+b+c)x,by considering the various possible paths. (Switches are repre-sented by gaps.)

Fig. 6 Fig. 7

Fig. 7 is rather more difficult to analyse. The solution is toconsider separately the possible paths when x is closed and when xis open. If x is closed, there is a path if any one of a, b and c isclosed. If x is open there is only a path if a, b and c are all closed.Thus the switching function of the circuit in Fig. 7 is:

(a + b + c)x + abc.This is exactly the same function as that of the circuit shown inFig. 6, but, it will be noted, whereas the circuit in Fig. 6 contains7 switches, that in Fig. 7 contains only 5 (or 3 single and onedouble), thus achieving a saving in 'hardware'. This is hardlysignificant in one instance, but in the case of a relay in a telephonesystem, for example, the total saving could be very worth while.Now, the algebra itself does not point to the most economicalcircuit to achieve a particular purpose, but it will enable proof thata proposed simpler circuit will do the job it is supposed to. Tohelp convince yourself on this point, write down the switchingfunction of the circuit shown in Fig. 8, and then draw the circuithaving the same switching function but using only series andparallel connexions. (This type of circuit, and that in Fig. 7, areknown as 'bridge circuits'.)

As one last example, to show the use of the a notation consider

BOOLEAN ALGEBRA 327the common domestic arrangement where one light can be operatedfrom either of two switches. The circuit is shown in Fig. 9, and itsswitching function is ab + ab.

The switch a has two positions, one which connects the terminalsX and T; the other X and T. The switch b is similar. Each may beconsidered as a pair of complementary switches, one of which willbe closed when the other is open, and vice versa. We may therefore

Fig. 8

Fig. 9

designate them a and a, b and 5, where the algebraic effect of the baris that previously described. It will be seen that the circuit is closedif both a and b are closed, or if both a and 5 are closed, i.e. ifa = b = ab = 1 or a = 5 = a.b = 1.

APPLICATIONS TO INSURANCEBerkeley, in his paper, listed the following as fields in which thealgebra could provide useful techniques:

Contracts—applications in drafting clauses and provisions.Formulation of practice—application in stating rules and prac-

tices precisely and in reducing the number of omissions andconflicts.

Record keeping—application in classification and systematicfiling.

Punched card machines—the furnishing of a mathematicaldescription for the operations and results of sorting and classifyingpunched cards.

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328 R. J. SQUIRES

Underwriting—applications in developing and codifying rulesfor selection of risks.

Actuarial mathematics—applications in statistics, in probability,and in the analysis of joint life and multiple decrement statusesand contingencies, and in the patterns of contingencies and actionsin beneficiary settlements.

This may sound an ambitious list, and to the best of my know-ledge nothing has in fact been published showing that any head-way has been made in developing these possibilities. However,punched-card machinery and computers would seem to be a dis-tinctly possible field, and I would be interested to know if anycurrent methods of programming embody algebraic methodssimilar to those described. As to the last suggestion, consider thefollowing hypothetical question:

Under a settlement, a sum of money is to be paid to the family ofJohn Doe, on his death if: (i) his wife is then alive; (ii) his son isthen alive and under 21; (iii) his daughter is then alive and under2 1; (iv) his son is then dead, and either his wife or daughter is alive.What contingent assurance would be needed to complement thiscontingent expectation?

Solution: Denote the various statuses as follows:

a: wife is alive at death of J.D.;b: son is alive at death of J.D.;c: daughter is alive at death of J.D.;d: son is 21 at death of J.D.;e: daughter is 21 at death of J.D.

Then the status of defining payment due under the policy at thedeath of J.D. may be written:

BOOLEAN ALGEBRA 329Thus the policy must provide for payment of the sum assured on

the death of J.D. if his wife is dead, and, either:(i) The son and daughter are both dead,(ii) The son is alive and 21 and the daughter is dead.(iii) The son and daughter are both alive and both have attained

2 1 .

Now, it would have been possible to reach this conclusion with-out use of the algebra, and some may think it would have beensimpler, but this much may be claimed:

(i) The algebra must lead to the required answer.(ii) It is just possible that general reasoning might lead us to a

solution which missed an obscure possibility in a complicated case.Another application is in reviewing rules to see that there are no

cases where two rules contradict one another, or similarly when wewish to replace a complicated system of rules by a simpler set, thealgebra will enable us to see which cases would be treated dif-ferently, and decide whether these are sufficiently few to justify thechange. An example of this, taken from Berkeley's paper, follows:

In a certain company, on a request for change in mode ofpremium payment, either action C or action P will be taken.Basically action C consists of going back to a convenient date andallowing against premiums required from that date, premiums paidsince that date. Action P consists of starting the new mode ofpremium payment from the next convenient date and charging aproportionate premium in the meantime. The present rule is:

I. Where the date of the request is within two months of thedate of issue of the policy, take action C.

II. Where the date of the request is more than two months afterthe date of issue of the policy:

A. Where there are no premiums falling due on policy anni-versaries, under the existing mode of payment, take action P.

B. Where some premiums fall due on policy anniversaries underthe existing mode of payment:

(i) Where the assured requests paying annual premiums oneach policy anniversary, the existing paid-to date is not apolicy anniversary, and the date of the request is within twomonths following any policy anniversary, take action C.

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330 JR. J. SQUIRES

In other cases:(ii)Where the date of request is within the grace period ofthe last premium paid, take action C.

Where the date of request is after the end of the graceperiod, take action P.

(b)

(a)

This cumbersome rule has evolved over the years and it is nowsuggested that it be replaced by the rule:

1. Where the request for change is made within two months ofthe issue date or any anniversary (that is, two months after) takeaction C.

2. In other cases, take action P.We wish to know which cases will be dealt with differently on

the basis of the new rule. If we designate by C1 cases dealt with byaction C according to rule 1, with similar definitions for C2 P1 andP2, the cases we wish to find are represented by

C1C2 + C1. C2 + P1P2+p1. P2.

Now, since each rule yields one action only on each case, we have

P1 = C1, C1 — P1 P2 = C2, C2 = P2.

Hence we can immediately simplify the above expression toC1P2+P1C2, which is intuitively acceptable.

We wish to express this in terms of the criteria set out above, andthe next step is to translate these into symbols; thus:

a represents cases where some premiums under existing mode ofpayment fall on policy anniversaries.

b—cases where the assured requests that future premiums shall bepayable annually on the policy anniversary.

c—cases where the existing paid-to date is a policy anniversary.

d—cases where the date of request is within two months of theissue date.

e—cases where the date of request is within two months followingany policy anniversary.

f—cases where the date of request is within the grace period of thelast premium paid.

BOOLEAN ALGEBRA 331

Since with each rule all possible cases are covered, we may write

Thus the required function is

This may now be translated back into the four cases representedby the four terms in the expression when multiplied out. Theyhave been grouped as above for simplicity in translating into words.Thus we see that the condition 3 applies to all four cases; the mainsubdivision is then a or a, the next ef or ef, and the last b or c.

OTHER APPLICATIONS

Berkeley also mentions an interesting application in formingHighest Common Factors and Lowest Common Multiples withthe factors of a base number. If, for example, 30 is our base num-ber, each variable may take the values, 1, 2, 3, 5, 6, 10, 15 and 30.Here a+b represents the L.C.M. of a and b; a.b represents theirH.C.F. Also a = 30 a, and we have to replace our symbol 1 byanother. The definition of a leads us to replace 1 by 30, the uni-versal multiple; o we can keep as the 'Null class multiple'. Thus30.a = a; o.a = o. It will be found that all the other rules workalso. Perhaps the most interesting is (20), which says (x +y) = x.y.If x = 2 and y = 3, for example, this says that the H.C.F. of 30/2and 30/3 or 15 and 10, is equal to 30 divided by the L.C.M. of thesenumbers, or 30 ÷ 6 = 5, which is easily seen to be correct. I do notremember having encountered this theorem before, however, andbearing in mind that the base number 30 need not have been fixedin advance, we have a theorem of very wide scope. The only limiton the base number is that the factors must not recur.

332 R. J. SQUIRES

As one last example, consider the problem:A certain man, X, left in his will three objects to his friends A, B

and C, and three statements, necessary and sufficient for them todecide the order in which he intended them to choose. The state-ments were:

1. No one who has seen me in a green tie is to choose before A.2. If B was not in Oxford in 1920, then the first person to choose

never lent me an umbrella.3. If C chooses 2nd, then B chooses before the person who first

fell in love.A, B and C will remember the incidents referred to and will

ascertain the order in which they are to choose. What is this order?In this question, the fact that the statements are necessary and

sufficient for A, B and C is vital information to us. The problemwhich I have not been able to solve is how to put this informationinto algebraic form. I shall be very interested to hear from anyonewho manages this. The problem is quite interesting in itself: if yousolve it easily, go back and check on your working. The chancesare that you have made an invalid assumption.

REFERENCESBERKELEY, E. C. (1937). Boolean algebra (the technique for manipulating' and',

'or', 'not', and conditions) and applications to insurance. R.A.I.A. 26, 373.BIZLEY, M. T. L. (1957). Probability: an Intermediate Text Book. Cambridge

University Press.

APPENDIXSolution I

Let C1 represent the assertion Charles was first , etc.Then C 1 +B 2 = 1, C2+D3 = 1, D4 + A2 = 1 from the given

facts.Also c1. = o, etc., and C1B2 = o, etc.From the given relationships (C1+B2)(c2+D3) = 1.

Hence Ds = 1..'. Since D4+A2 = 1 , A 2 = . 1 ,

and since C1 +B2 = 1, C1 = 1; finally B4 = 1 by elimination.

.VI = (CtC2+ C1D)3+B2C2+B2D)S) = C1D8+.B2D3

= (C1+B2)D3

BOOLEAN ALGEBRA 333

Solution 2

Let Ad represent the assertion' The Christian name of Mr Albertis Dick', etc.

Solution 3

Number the seats clockwise, starting with Mrs AD at 1.Let Yr represent the proposition that Y sits at seat r

(r = 1 6).Then:

(Mrs AD)1 = 1, X4 = 1.(ATS)r = Pr_1, .-. P6 = o, since (ATS)1 = o.Zr = Yr-1 . · . Ys = oandZ5 = o, since Y4 = Z4 = o.(WRAF)r = Yr+3, . y4 = o, which we already know.Wr = (REDX)r+3, .-. W4 = o, which we already know.Qr = (WRNS)r-1, . Q2 = o, since (WRNS)1 = o.Pr.(NFS)r = o.

also

(i)(i)(iii)(iv)(v)(vi)(vii)

(viii) P1 + Q1 = 1, since Admiral's wife is a Mrs..". from (viii)

.'. from (iii)

.'. Z1+Z2+Z4+Z5 = o,

Z2 = o.

W1 = X1 - Y1 - Z1 = o.

(Ff = o). .'. (Fa.A1f+Fe.Cv+Fa.Dt,).Ye = 1.

Similarly, replacingy by a, d andf(Cc = o) we have

Fa.(Aa.Ac+Ad.De+A,.Fe) + Fe.(Ca.Ae+Cd.Dc+Cf.Fc)

Whence, multiplying out and remembering that such terms asAa, Fa.Fe and FC.AC are zero, we obtain

Fa.Ad.De+Cf.Fc+Fd.Da.Ac = 1.Now Cf.Fe D a . .A d but Ad = o (given).Hence Cf,Fe = o and Fd = Da = Ac = Cf = 1.

Then Ad = o; Fx = 1; X , = 1; yc = 1 from the given facts.Hence Fx.Xy.Yc = 1.Now if x = a, X = A, and x can take the values a, c and d

+Fa.(Da.Ae+Dd.De + Df.Fc) = 1.

and hence 2 3 + Z6 = 1.

334 R. J. SQUIRES

Likewise

Now

I.e.

(ATS)2. (WRAF)5. [(WRNS)4. (RED X)3 + (WRNS)5. W5]+ (WRNS)6. (WRAF)5. (ATS)6. W6 + (ATS)2. (WRAF)2. Q3W2

+ (WRNS)6. (WRAF)2. [(ATS)3. (RED X)6

+ (ATS)4.(REDX)5] = 1.

I.e.(ATS)2,. (RED X)3. (WRNS)4. (WRAF)5. (NFS)6

+ (WRAF)2. (NFS)3. (ATS)4. (RED X)5. (WRNS)6 = 1.Le.P1. (RED X)3. (WRNS)4. (WRAF)5. (NFS)6

+ (WRAF)2. (NFS)3. P3. (RED X)5. (WRNS)6 = 1.

.·. (Mrs AD)1. (ATS)2,. (RED X)3. (WRNS)4. (WRAF)5. (NFS)6

= 1.X4.P1.W6.Q5.Y2.Z3=1. I .