JSS 17 (4) (1964) 314-334
AN INTRODUCTION TO BOOLEAN ALGEBRA
by
R. J. S Q U I R E S
A T the Annual General Meeting of the Students' Society in 1961,
a motion was passed that articles of a less serious nature than those
normally 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 does
not pretend to advance actuarial science in any way, and its only
purpose is to introduce members of the profession to a technique
which is interesting in that it is different from our normal techniques and to show that it has possibilities in the actuarial field by
demonstrating its applications in other fields.
This is not the first time that the subject has been written about
in an actuarial journal. Edmund C. Berkeley (1937) wrote a paper
for the American Institute of Actuaries going into the subject
in some detail, and it is from this paper that I have taken my
examples of applications to life office practice.
At this point it would seem meet to record my thanks to him for
permission to use this material, and also to F. M. Goldner of
Alleyn's School who first introduced me to this subject, and to
whom 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 to
be the first person to demonstrate the possibility of paraphrasing
logical argument by algebraic manipulation. The following quotations sum up the power that is thereby added:
' The more complex forms of inference cannot be studied until
a specially devised symbolism is introduced.. .which enables us
to keep different meanings distinct, to concentrate upon what is
essential, to show the form of propositions and to save labour
and thought.' M. R. Cohen and E. R. Nagel.
'Arithmetic at first lacked any more appropriate medium than
that of ordinary language. Ancient Greek mathematicians had
BOOLEAN ALGEBRA
315
no symbol tor zero and used letters 0f the alphabet for other
numbers. As a result it was impossible to state any general rule
for division—to give only one example. Operations which any
fourth grade child can accomplish in the modern notation, taxed
the finest mathematical minds of the age of Pericles. Had it not
been for the adoption of new and more versatile ideographic
symbols, many branches of mathematics could never have been
developed, because no human mind could grasp the essence of
their operations in terms of the phonograms of ordinary
language.' 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 its
first field, that of logical deduction, there has been no advance, as
far as I know, for many years. There is, however, no reason to think
that further advances cannot be made, and what more can a field of
research offer?
Turning to the details of the subject, I present it here as an
abstract algebra where the rules are to be learnt as they stand, and
there is no question of understanding the relationships. However,
most people will find it easier to skim through these rules and pass
rapidly to the applications and examples. For those brave souls
intending to try to master the conditions of manipulation first, it
may be some help to mention that for most applications the variables have two values only, o and 1, and these are not to be thought
of as integers, but rather symbols of 'nothingness' and 'everythineness'.
RULES OF THE ALGEBRA
Some of the rules of our normal algebra we take so for granted that
we do not even realize their existence. The first group of laws,
given below, hold both for our normal algebra and for Boolean
algebra:
Addition
Multiplication
Commutative Law
x+y - y+x
(2)
xy = yx
(z)
Associative Law
(xy)z = x(yz) (4)
(x+y)+z = x+(y + z) (3)
x(y+z) = xy+xz (5)
Distributive Law
R. J. SQUIRES
316
The Algebra of Classes, which is a Boolean Algebra, has the
following additional laws:
Multiplication
Addition
Law of Absorption
x+x = x
X.X
(6)
= X
(7)
Distributive Law
x+yz = ( x + y ) ( x(8)+ 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 'universal class', that is the total of the class we are referring to, and o
to 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).
lix.y = i,
then x = 1 and y = 1 (19).
(xy) x+y (21).
The last two laws, (20) and (21), are sometimes referred to as De
Morgan's Laws.
It will be noticed that the laws go in pairs; for each law of addition there is a corresponding law of multiplication (Law (5) pairs
with Law (8)). Further inspection yields the Principle of Duality.
This says that having proved a certain relationship, if we substitute multiplication for each addition, addition for each multiplication, and replace each variable or constant by its complement,
the resulting relationship will be true also. For example, Law (5) is
x.{y+z) = xy+xz. Following the above procedure we arrive at
{x+y)
x.y
(20);
BOOLEAN ALGEBRA
317
the relationship x +y. z = (x +y)(x+z) which is Law (8) written in
terms of the variables x, y and z.
The expression may be paraphrased in terms of x, y and z if
required. The question may then be asked, what is the need to take
the complements of the variables? The answer lies in the relationship between laws involving o and 1; only by this device do they
translate correctly. (To make this position quite clear, perhaps it
should be pointed out that in the transformation, o becomes 1 and
1 becomes o.)
We are now ready for the important step that since this transformation holds for Laws (1)-(21) and since any other functions or
relationships we form will be based on these laws, these functions
and relationships may also be transformed according to the principle of duality. This is evidently a powerful technique.
THE ALGEBRA OF CLASSES—VENN DIAGRAM
So far, we have developed rules for an algebra on a completely
abstract basis except for the definitions of complement, o and 1.
Even here it was not necessary to introduce terms of the algebra of
classes, but it was convenient to do so to avoid difficulties in defining abstract values for these. We now proceed to define the variables and operators in terms of this algebra.
' x' represents the class of objects having the property px, out of a
given reference class.
'y' represents the class of objects having the property py, out of
the 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 in
the 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 wholly
contained by the class b\ This immediately leads us to the equation ab = a, so that the operator is not indispensable although useful in the translation of statements into algebraic form.
To see how the algebra operates on classes, let us take as an
example the Distributive Law of Addition—Law (8). This states:
x+yz = (x+y)(x+z), and to the mind trained to normal algebra
is probably the most strange-looking of all the laws. Translated
into terms of classes it states:' The class of objects having either the
propertypx or both the propertypy and the propertypz is equal to
the class of objects having (i) either the property px or the property
py, and (ii) either the property px or the property pz.' To see that
this is true consider first objects having the property px. These are
included by either statement of the rule. Next, consider those
objects which do not have this property. To be included according
to either statement of the rule, these must have both the property
py and the property pz. Hence the two statements are equivalent.
These conceptions can be represented diagrammatically, and for
many people the principles will be easier to grasp if presented in
this way. The diagrams below, known as Venn diagrams, illustrate
some of the points above. The rectangle in each case is supposed to
represent the universe or reference class with which we are dealing,
and the classes which we are considering are represented by irregular shapes. It will be realized that although, in the diagram below,
each class is represented by a single area contained by a continuous line, this is not necessary, and a class could consist of one
object represented by a point in the top right-hand corner and
another represented by a point in the bottom left-hand corner, or
indeed anywhere else within the rectangle. Perhaps at this point it
would be helpful to point out that the components of the classes
need not necessarily be objects. I have worked in terms of objects
because I feel the ideas are easier to grasp if put this way; it is not
necessary to confine the discussion to objects.
In Fig. 1 the shaded area represents x+y, that is the class of
objects having either the propertypx or the property py. In Fig. 2,
the shaded area represents x.y, that is the class of objects having
both 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 white
area 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 be
verified by reference to Fig. 3.
Fig. 1
Fig. 2
Case I.
Case II.
Fig. 3.
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 consider
the classic question posed by Lewis Carroll:' In a very hotly fought
battle, at least 70 % of the combatants lost an eye, at least 75 % an
ear, at least 80 % an arm and at least 85 % a leg. How many lost all
four members?'
For convenience, let us suppose the total number involved was
100: we may revert to percentage form at the end.
Let
Let
Let
Let
(/)]
(E) represent the class of combatant
(A) that lost an
(L),
eye; / m l n . = 70.
ear; E min = 75.
arm;A mln. =80.
leg; Lmln_ = 85.
R. J. SQUIRES
320
Then (/+E) represents the class of combatant that lost either an
eye or an ear, etc.
Now,(/+£)
••• (/£)«!«.
= (I) + (E)-(IE),
= (I)min. + (E)min.-(I+E)max.'
= 70+75-100 = 45.
Similarly, (^L)mm. = (A)min. + (L)mln.-(A+ L)max. = 65,
and (IE. AL)mln. = (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 should
convince 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 some
other text-book, but was unable to trace it.
THE ALGEBRA OF PROPOSITIONS
In this algebra our variables represent propositions. Thus x
might represent the proposition ' I am now writing with a pen' and
y, ' I am now writing with a pencil'. Addition represents propositions taken in disjunction; that is one or other or both are asserted.
Multiplication represents propositions taken in conjunction, that is
both 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 a
pencil, or both', and xy represents the proposition ' I am now
writing with a pen and with a pencil'. If pens and pencils were the
only writing implements we could write x +y = 1: for most normal
people we could write xy — o. In this algebra x represents the
opposite assertion; that is for our example, 'I am not now writing
with a pen'.
The algebra follows Laws (1)-(21) above: most solutions come
from Laws (18) and (19). Very often the most difficult part of the
BOOLEAN ALGEBRA
321
problem is to derive a satisfactory notation. Once this has been
done, and all the information made use of, the solution follows
without 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, but
not necessarily respectively. One only of the following statements
is 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 disjunction
correctly. Since only one is true any conjunctive assertion of two
statements will be false.
In algebra:
Now a..br = o, since if A is red, B cannot be red, and vice versa.
(Law 5).
But
(Law 9).
(Law 15).
(Since if c is blue, one of A or B must be red.)
Also 1
But since
(Law 16).
(Law 15).
(Laws 5 and 9).
(Law 15).
(Law 16).
(Law 14).
(Law 14).
But
(Law 14).
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. But
who? Harry said 'Charles and George'. James said, 'Donald and
Tom'. Donald said, 'Tom and Charles'. George said, 'Harry and
Charles'. Charles said, 'Donald and James'. Four of the boys
interrogated named just one miscreant correctly. The fifth had lied
outright. Who stole the apples?
Here the notation suggests itself immediately: Let C denote the
proposition that Charles stole the apples, etc.
Since in each case there was at least one lie, each of the products
CG, DT, CT, CH and DJ equals zero.
Since in exactly one case there was a double lie, all the sums equal
one, save one. Thus the product of all five sums is zero, and the
product of one set of four sums is one, so that the sum of all five
possible 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) = 0
by continued application of Law 8. Now since only two boys stole
apples, any product of three terms is automatically zero, so that the
above equation becomes:
C{D+TJ) = 0,
and ultimately
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) = 1
or
(remembering X+X = X, and products of three terms are zero).
Multiplied out this yields: CD+CT + CJ+CD = 1. But
CD = CT= 0. Hence CJ = 1.
Translation: Charles and James stole apples.
BOOLEAN ALGEBRA
323
There is a further conception and operator that is sometimes
useful, but like the operator 'inclusion' in the algebra of classes is
not 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.' This
may alternatively be phrased 'Either A is not true, or else B is
true' which leads us to our alternative form A+B = 1. The type
of case where this operator is most useful is that where we wish to
express 'The fact that A implies B implies that C implies D;
where we can write {A B) (C D), and proceed by easy
stages to the algebraic form. Let us consider another simple
example:
In a certain strange community, everyone is either an Actuary or
a Statistician. Actuaries always tell the truth. Statisticians always
lie (in this strange community!). Three men meet and X introduces
himself, mentioning his profession. Y then remarks to Z, 'He says
he is an Actuary'. Z replies, 'No, he is a Statistician'. How many
were there of each profession?
Solution: Let XA be the proposition that X is an Actuary, and
similarly for Y and Z.
YA, (since X will have made a true statement, and Y
Now XA
correctly reported it).
YA (since X will have made an untrue statement and
Also XA
Y correctly reported it), i.e.
Also
Therefore either ZA.XA = 1 or XA.ZA = 1. (Not both in this
case.)
Hence one of X and Z is an Actuary, the other a Statistician,
though we know not which. We have already determined that Y is
an Actuary.
324
R. J. SQUIRES
In the above example it is easier to use logical deduction than
algebra. In particular, the fact that Y is an Actuary becomes
apparent immediately, without the need of multiplying out the
derived equation. However, it may be seen that this was in fact
an example of a simple and useful theorem, namely
[(A B).(A
B)] [B=1].
Further, there are many cases where the logical chain is so complex
that the algebraic method will be easier than the process of deduction, and, most important of all, it will not let us deduce facts that
we should not; that is to say, logical deduction will sometimes
appear 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 not
consider worth using, but list for the benefit of anyone who might
encounter them whilst reading on this subject:
Name
Ring Sum ('Exclusive or')
Scheffer stroke
Double Stroke
Symbol
Equivalence
EXAMPLES FOR PRACTICE
1. Anthony, Brian, Charles and Douglas competed for a
scholarship. 'What luck have you had?' someone asked. Said
Anthony: '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 three
boys had made two assertions, one at least of which was true. Who
won the scholarship?
2. Four members of my club, Messrs Albert, Charles, Frederick
and Dick, have recently been knighted, so that now their friends
have had to learn their Christian names. Now, the surname of each
is the Christian name of one of the others. Dick is not the Christian
name of the member whose surname is Albert. The Christian name
of the member whose surname is Frederick is the surname of the
member whose Christian name is the surname of the member whose
Christian name is Charles. Find the Christian names of each of the
four members.
BOOLEAN ALGEBRA
325
3. 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 opposite
Miss X. The ATS girl sits on the left of Mrs P. Miss Z sits on the
left of Miss Y. The WRAF girl sits opposite Miss Y. Miss Wsits
opposite the Red Cross worker. Mrs Q sits on the left of the
WRNS girl. Mrs P is not the NFS girl.
Who represents each service and how do they sit round the
table?
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 home
the fact that there is no 'Act of God' about either normal or Boolean
Algebra, but that observing certain phenomena leads us to formulate
rules for our algebra, which provided they do not lead to contradictions, will yield useful results. The main application is in
analysing circuits, chiefly in connexion with telephones, but also in
connexion with mechanics of computers.
F(a, b)=a+b
F(a, b)=a.b
Fig. 4
Fig-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 meaning
that 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 switches
in 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 closed
when 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 is
a path through the circuit and 0 when there is no path. It follows
that the functions may be combined in the same manner as the
original variables. As an exercise in writing down switching functions, 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 represented by gaps.)
Fig. 6
Fig. 7
Fig. 7 is rather more difficult to analyse. The solution is to
consider separately the possible paths when x is closed and when x
is open. If x is closed, there is a path if any one of a, b and c is
closed. 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 in
Fig. 6, but, it will be noted, whereas the circuit in Fig. 6 contains
7 switches, that in Fig. 7 contains only 5 (or 3 single and one
double), thus achieving a saving in 'hardware'. This is hardly
significant in one instance, but in the case of a relay in a telephone
system, for example, the total saving could be very worth while.
Now, the algebra itself does not point to the most economical
circuit to achieve a particular purpose, but it will enable proof that
a proposed simpler circuit will do the job it is supposed to. To
help convince yourself on this point, write down the switching
function of the circuit shown in Fig. 8, and then draw the circuit
having the same switching function but using only series and
parallel connexions. (This type of circuit, and that in Fig. 7, are
known as 'bridge circuits'.)
As one last example, to show the use of the a notation consider
BOOLEAN ALGEBRA
327
the common domestic arrangement where one light can be operated
from either of two switches. The circuit is shown in Fig. 9, and its
switching function is ab + ab.
The switch a has two positions, one which connects the terminals
X and T; the other X and T. The switch b is similar. Each may be
considered as a pair of complementary switches, one of which will
be 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 bar
is that previously described. It will be seen that the circuit is closed
if both a and b are closed, or if both a and 5 are closed, i.e. if
a = b = ab = 1 or a = 5 = a.b = 1.
APPLICATIONS TO INSURANCE
Berkeley, in his paper, listed the following as fields in which the
algebra could provide useful techniques:
Contracts—applications in drafting clauses and provisions.
Formulation of practice—application in stating rules and practices precisely and in reducing the number of omissions and
conflicts.
Record keeping—application in classification and systematic
filing.
Punched card machines—the furnishing of a mathematical
description for the operations and results of sorting and classifying
punched cards.
21
ASS 17
328
R. J. SQUIRES
Underwriting—applications in developing and codifying rules
for selection of risks.
Actuarial mathematics—applications in statistics, in probability,
and in the analysis of joint life and multiple decrement statuses
and contingencies, and in the patterns of contingencies and actions
in beneficiary settlements.
This may sound an ambitious list, and to the best of my knowledge nothing has in fact been published showing that any headway has been made in developing these possibilities. However,
punched-card machinery and computers would seem to be a distinctly possible field, and I would be interested to know if any
current methods of programming embody algebraic methods
similar to those described. As to the last suggestion, consider the
following hypothetical question:
Under a settlement, a sum of money is to be paid to the family of
John Doe, on his death if: (i) his wife is then alive; (ii) his son is
then alive and under 21; (iii) his daughter is then alive and under
2 1; (iv) his son is then dead, and either his wife or daughter is alive.
What contingent assurance would be needed to complement this
contingent 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 the
death of J.D. may be written:
BOOLEAN ALGEBRA
329
Thus 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
21.
Now, it would have been possible to reach this conclusion without use of the algebra, and some may think it would have been
simpler, 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 we
wish to replace a complicated system of rules by a simpler set, the
algebra will enable us to see which cases would be treated differently, and decide whether these are sufficiently few to justify the
change. An example of this, taken from Berkeley's paper, follows:
In a certain company, on a request for change in mode of
premium payment, either action C or action P will be taken.
Basically action C consists of going back to a convenient date and
allowing against premiums required from that date, premiums paid
since that date. Action P consists of starting the new mode of
premium payment from the next convenient date and charging a
proportionate premium in the meantime. The present rule is:
I. Where the date of the request is within two months of the
date of issue of the policy, take action C.
II. Where the date of the request is more than two months after
the date of issue of the policy:
A. Where there are no premiums falling due on policy anniversaries, under the existing mode of payment, take action P.
B. Where some premiums fall due on policy anniversaries under
the existing mode of payment:
(i) Where the assured requests paying annual premiums on
each policy anniversary, the existing paid-to date is not a
policy anniversary, and the date of the request is within two
months following any policy anniversary, take action C.
21-2
330
JR. J. SQUIRES
(ii) In other cases:
(a) Where the date of request is within the grace period of
the last premium paid, take action C.
(b) Where the date of request is after the end of the grace
period, take action P.
This cumbersome rule has evolved over the years and it is now
suggested that it be replaced by the rule:
1. Where the request for change is made within two months of
the issue date or any anniversary (that is, two months after) take
action 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 by
action C according to rule 1, with similar definitions for C2 P1 and
P2, 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 = C 2 , C2 = P 2 .
Hence we can immediately simplify the above expression to
C 1 P 2 +P 1 C 2 , which is intuitively acceptable.
We wish to express this in terms of the criteria set out above, and
the next step is to translate these into symbols; thus:
a represents cases where some premiums under existing mode of
payment fall on policy anniversaries.
b—cases where the assured requests that future premiums shall be
payable 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 the
issue date.
e—cases where the date of request is within two months following
any policy anniversary.
f—cases where the date of request is within the grace period of the
last premium paid.
33 1
Since with each rule all possible cases are covered, we may write
BOOLEAN ALGEBRA
Thus the required function is
This may now be translated back into the four cases represented
by the four terms in the expression when multiplied out. They
have been grouped as above for simplicity in translating into words.
Thus we see that the condition 3 applies to all four cases; the main
subdivision 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 forming
Highest Common Factors and Lowest Common Multiples with
the factors of a base number. If, for example, 30 is our base number, 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 their
H.C.F. Also a = 30 a, and we have to replace our symbol 1 by
another. The definition of a leads us to replace 1 by 30, the universal multiple; o we can keep as the 'Null class multiple'. Thus
30.a = a; o.a = o. It will be found that all the other rules work
also. 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 30/2
and 30/3 or 15 and 10, is equal to 30 divided by the L.C.M. of these
numbers, or 30 ÷ 6 = 5, which is easily seen to be correct. I do not
remember having encountered this theorem before, however, and
bearing in mind that the base number 30 need not have been fixed
in advance, we have a theorem of very wide scope. The only limit
on 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 to
decide the order in which he intended them to choose. The statements 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 problem
which I have not been able to solve is how to put this information
into algebraic form. I shall be very interested to hear from anyone
who manages this. The problem is quite interesting in itself: if you
solve it easily, go back and check on your working. The chances
are that you have made an invalid assumption.
REFERENCES
BERKELEY, 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.
APPENDIX
Solution 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.
.VI = (CtC2+ C1D)3+B2C2+B2D)S) = C1D8+.B2D3
= (C 1+ B 2 )D 3
Hence
Ds = 1.
.'. Since D 4 +A 2 = 1 , A 2 = . 1 ,
and since C1 +B2 = 1, C1 = 1; finally B4 = 1 by elimination.
BOOLEAN ALGEBRA
333
Solution 2
Let Ad represent the assertion' The Christian name of Mr Albert
is Dick', etc.
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
(Ff = o).
.'. (Fa.A1f+Fe.Cv+Fa.Dt,).Ye
= 1.
Similarly, replacingy by a, d andf(C c = o) we have
Fa.(Aa.Ac+Ad.De+A,.Fe)
+ Fe.(Ca.Ae+Cd.Dc+Cf.Fc)
+Fa.(Da.Ae+Dd.De
+ Df.Fc) = 1.
Whence, multiplying out and remembering that such terms as
Aa, 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.
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:
(i) (Mrs AD)1 = 1, X4 = 1.
(i) (ATS)r = P r _ 1 , .-. P 6 = o, since (ATS)1 = o.
(iii) Zr = Yr-1 .·. Ys = oandZ 5 = o, since Y4 = Z4 = o.
(iv) (WRAF)r = Yr+3,
. y4 = o, which we already know.
(v) Wr = (REDX) r+3 , .-. W4 = o, which we already know.
. Q2 = o, since (WRNS)1 = o.
(vi) Qr = (WRNS) r-1 ,
(vii) P r .(NFS) r = o.
also
(viii) P1 + Q1 = 1, since Admiral's wife is a Mrs.
.". from (viii)
W1 = X1 - Y1 - Z1 = o.
.'. from (iii)
Z2 = o.
.'. Z 1 +Z 2 +Z 4 +Z 5 = o, and hence 2 3 + Z6 = 1.
334
Likewise
R. J. SQUIRES
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.
P 1 . (RED X) 3 . (WRNS)4. (WRAF)5. (NFS)6
+ (WRAF)2. (NFS) 3 . P 3 . (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.