Hugh MacColl.

# Symbolic logic and its applications online

. (page 3 of 11)
Online LibraryHugh MacCollSymbolic logic and its applications → online text (page 3 of 11)
Font size CD' + C'D + B'C' + B'D'.

Beginning with the first term we get

CD'(C'D + B'C + B'D')' = CD'(w + B'» 7 + B'e)'

= CD / (B / )' = BCD / .

Hence, the first term CD' must not be omitted. Taking
next the second term CD, we get

CTKCD' + B'C/ + B'D'/ = C'D( w + B'e + B',,)'

= C / D(BY=BC / D.

Hence, the second term CD must not be omitted. We
next take the third term B'C, getting

B^CD 7 -I- C/D + B'D'/ = B'C'(>iI)' + eD + eD'/

= B'C'(D + D , / = B / C'>/ = >/.

This shows that the third term B'C can be omitted as

24 SYMBOLIC LOGIC [§§ 29-31

redundant. Omitting the third term, we try the last
term B'D', thus

B'D'(CD' + CD)' = B'D'(Ce + C>,)' = B'D'C.
This shows that the fourth term B'D' cannot be omitted
as redundant if we omit the third term. But if we retain
the third term B'C, we may omit the fourth term B'D',
for we then get

B'D'(CD' + CD + B'C 7 )' = B'D'(Ce + C'n + eC')'

= B / D , (C + C / ) , =B / D / J? = i7.

Thus, we may omit either the third term B'C, or else
the fourth term B'D', as redundant, but not both.

30. A complex alternative may be said to be in its
simplest form* when it contains no redundant terms, and
none of its terms (or of the terms left) contains any re-
dundant factor. For example, a + ab + m + m'n is reduced
to its simplest form when we omit the redundant term ab,
and out of the last term strike out the unnecessary factor m' .
For a + ab = a, and m + m'n = m + n, so that the simplest
form of the expression is a + m + n. (See § 31.)

31. To reduce a complex alternative to its simplest
form, apply the formula (a + /3)' = a'/3' to the denial of
the alternative. Then apply the formula (a/3/ = a' + ft' to
the negative compound factors of the result, and omit
the redundant terms in this new result. Then develop
the denial of this product by the same formulae, and go
through the same process as before. The final result
will be the simplest equivalent of the original alternative.
Take, for example, the alternative given in § 30, and
denote it by (p. We get

cp = a + ab + m + m'n = a + m + m!n.

(p' = (a + m + m'n)' = a'm'(m'n)' = a'm'(m + nf) = a' m'n'.

<P = (cp')' = (a' m'n')' = a + m + n.

* What I here call its " simplest form " I called its " primitive form "
in my third paper in the Proceedings of the London Mathematical Society ;
but the word " primitive" is hardly appropriate.

§§31,32] METHODS OF SIMPLIFICATION 25

As another example take the alternative

AB'C + ABD + A'B'D' + ABD' + A'B'D,

and denote it by (p. Then, omitting, as we go along, all
terms which mere inspection will show to be redundant,
we get

r/> = AB'C + AB(D + D') + A'B'(D' + D)

= AB'C + ABe + A'B'e = AB'C' + AB + A'B'.
<£ / =(AB / C / ) , (AB)'(A / B , ) /

= (A' + B + C)(A' + B')(A + B)
= (A' + B'C)( A + B) = A'B + AB'C.
i p = «p')' = (A'B + AB'C)' = (A + B')(A' + B + C)
= AB + AC' + A'B' + B'C.

Applying the test of § 29 to discover redundant
terms, we find that the second or fourth term (AC or
B'C') may be omitted as redundant, but not both. We
thus get

(p = AB + A'B' + B'C = AB + AC + A'B',

either of which may be taken as the simplest form
of (p.

32. We will now apply the preceding principles to
an interesting problem given by Dr. Venn in his " S} T m-
bolic Logic" (see the edition of 1894, page 331).

Suppose we were asked to discuss the following
set of rules, in respect to their mutual consistency and
brevity.

a. The Financial Committee shall be chosen from
amongst the General Committee.

/3. No one shall be a member both of the General
and Library Committees unless he be also on the Finan-
cial Committee.

y. No member of the Library Committee shall be on
the Financial Committee.

26 SYMBOLIC LOGIC [§32

Solution.

Speaking of a member taken at random, let the
symbols F, G, L, respectively denote the statements
" He will be on the Financial Committee," " He will be
on the General Committee," " He will be on the Library
Committee." Putting >;, as usual, for any statement that

a = (F:G); /3 = (GLF' : >,) ; 7 = (LF:#,);
so that

a/3 7 = (F:G)(GLF / :i?)(FL:i7)
. = (FG':*i)(GLF':T,)(FL:r,)
= FG' + GLF / + FL:>/.
Putting (J> for the antecedent FG' + GLF' + FL, we get
^(F' + GXG' + L' + FXF' + I/)

(See § 25, Formulae (4) and (5))
= (F' + GL')(G' + L' +■ F) = F'G' + F'L' + GL' + FGL'
= F'G' + GL';

for the term FGL', being a multiple of the term GL', is
redundant by inspection, and F / L / is also redundant,
because, by § 29,

F / L / (F , G / + GL')' = F'L'(eG' + G<)' = F'L'(G' + G)' = n .

Hence, finally, we get (omitting the redundant term FL)

<£ = (<£')' = (F'G' + GL')' = FG' + GL,

and there I ore

a/3y = (f>:ri= (FG' + GL : rf) = (F : G)(G : L')-

That is to say, the three club rules, u, (3, 7 , may be
replaced by the two simple rules F : G and G : L', which
assert, firstly, that " If any member is on the Financial
Committee, he must be also on the General Committee,"
which is rule a in other words ; and, secondly, that " If
any member is on the General Committee, he is not to be
on the Library Committee."

§33] SOLUTIONS, ELIMINATIONS, LIMITS 27

CHAPTER V

33. From the formula

(a : b)(c : d) = ah' + cd' : »/

the product of any number of implications can always be
expressed in the form of a single implication,

« + fi + 7 + &c : i],

of which the antecedent is a logical sum (or alternative),
and the consequent an impossibility. Suppose the im-
plications forming the data of any problem that contains
the statement x among its constituents to be thus re-
duced to the form

Ax + B,v' + C:tj,

in which A is the coefficient or co-factor of x, B the co-
efficient of x', and C the term, or sum of the terms, which
contain neither x nor x'. It is easy to see that the above
data may also be expressed in the form

(B:asX«:A')(C!:9) J

which is equivalent to the form

(B^iA'XCm).

When the data have been reduced to this form, the ^iven
implication, or product of implications, is said to be solved
with respect to x ; and the statements B and A' (which are
generally more or less complex) are called the limits of x;
the antecedent B being the strong * or superior limit ; and
the consequent A', the weak or inferior limit. Since the

* When from our data we can infer a: /3, but have no data for inferring
|8 : u, we say that el is stronger than /3. For example, since we have
AB:A:A + B, we say that AB is stronger than A, and A stronger than
A+B.

28 SYMBOLIC LOGIC [§§ 33, 34

factor (B : x : A') implies (B : A'), and our data also imply
the factor (C : >j), it follows that our data imply

(B:A')(C:>,),

which is equivalent to AB + C : *). Thus we get the
formula of elimination

(Ac + B,/ + C:>,):(AB + C:>;),

which asserts that the strongest conclusion deducible
from our data, and making no mention of x, is the im-
plication AB + C : >/. As this conclusion is equivalent to
the two-factor statement C^ABy, it asserts that the state-
ment C and the combination of statements AB arc both
impossible.

34. From this we deduce the solution of the follow-
ing more general problem. Let the functional symbol
<J)(x, y, z, a, b), or simply the symbol (p, denote data
which refer to any number of constituent statements
x, y, z, a, b, and which may be expressed (as in the
problem of § 33) in the form of a single implication
a + 18 + y + &c. : rj, the terms a, /3, y, &c, being more or
less complex, and involving more or less the statements
x, y, z, a, b. It is required, firstly, to find successively in
any desired order the limits (i.e., the weakest antecedent
and strongest consequent) of x, y, z; secondly, to eli-
minate x, y, z in the same order ; and, thirdly, to find
the strongest implicational statement (involving a or b,
but neither x nor y nor z) that remains after this
elimination.

Let the assigned order of limits and elimination be
z, y, x. Let A denote the sum of the terms containing
the factor z ; let B denote the sum of the terms contain-
ing the factor z , and let C denote the sum of the terms
containing neither z nor z . Our data being (p, we get

(j, = kz + B/ + C : 17 = (B : z)(z : A , )(C : r,)
= (B : z : A')(C : >;) = (B : z : A')(B : A')(C : >/).

§ 34] SOLUTIONS, ELIMINATIONS, LIMITS 29

The expression represented by Az + Bz' + G is understood
to have been reduced to its simplest form (see §§ 30,
31), before we collected the coefficients of z and z'. The
limits of z are therefore B and A' ; and the result after
the elimination of z is

(B : A')(C : >/), which = AB + C : n .

To find the limits of y from the implication AB + C : >/,
we reduce AB + C to its simplest form (see §§ 30, 31),
which we will suppose to be By + Ey' + F. We thus get,
as in the previous expression in z,

AB + C : r\ = By + E/ + F : r, = (E : y : D')(E : D')(F : >/)•

The limits of y are therefore E and D', and the result
after the successive elimination of z and y is

(E : D')(F : >,), which = ED + F : >/.

To find the limits of x from the implication ED + F : >/,
we proceed exactly as before. We reduce ED + F to its
simplest form, which we will suppose to be Gx + Hx + K,
and get

ED + F : n = Gx + Ha/ + K : r, = (H : x : G')(H : G')(K : >;).

The limits of x are therefore H and G', and the result
after the successive elimination of z, y, x is

(H : G0(K : >/), which = HG + K : >,.

The statements z, y, x having thus been successively
eliminated, there remains the implication GH + K : }j,
which indicates the relation (if any) connecting the
remaining constituent statements a and b. Thus, we
finally get

(/) = (B : z : A')(E : // : D')(H : x : G')(GH + K : ,,).

in which A and B do not contain z (that is, they make no
mention of z) ; D and E contain neither z nor y ; G and
H contain neither z nor y nor x ; and the expression K

30 SYMBOLIC LOGIC [§§ 34, 35

in the last factor will also be destitute of (i.e., will make
no mention of) the constitutents x, y, z, though, like G
and H, it may contain the constituent statements a and b.

In the course of this process, since >) : a and a : e are
certainties whatever the statement a may be (see § 18),
we can supply >/ for any missing antecedent, and e for any
missing consequent.

35. To give a concrete example of the general prob-
lem and solution discussed in § 34, let (p denote the data

e : xyza + xyb +xy z +y z a .

We get, putting (p for these data,

<p = {xyza + xyb' + xy'z' + y'z'a')' : r\

— x'y + bijz + y'z + abz + ax : »/,

when the antecedent of this last implication has been
reduced to its simplest form by the process explained in
§ 31. Hence we get

(j> = (y'+ ab)z + (]jy)z + {x'y + ax') : n

putting A for y' + ab, B for by, and C for x'y + ax'. As
in § 34, we get

(B:s:A')(AB + C:>7),

so that the limits of z are B and A', and the result after
the elimination of z is AB + C : »/. Substituting their
values for A, B, C, this last implication becomes

{ab + ,c)y + ax' : ?/,

which we will denote by *Dy + Ey' + F : n, putting J) for
ab + x, E for n, and F for ax. Thus we get

(f> = (B : z : A')(Dy + E/ + F : >/)
= (B :z : A0(E : y : D')(ED + F : »;).

Having thus found the limits {ix., the weakest ante-

§§35,36] SOLUTIONS, ELIMINATIONS, LIMITS 31

cedents and strongest consequents) of z and y, we proceed
to find the limits of x from the implication ED + F : n,
which is the strongest implication that remains after the
elimination of z and y. Substituting for D, E, F the
values which they represent, we get

DE + F : n = {ah + J)n + «J : n = Gx + BJ + K : n,

in which G, H, K respectively denote >/, a, n- We thus
get

DE + F : >i = (H : x : G')(HG + K : tj) ;

so that our final result is

<\$> = (B : z : A')(E : // : D')(H : a; : G0(HG + K : i,)

= (by : z : a'y + b f y){n : y : rt ; ;« + b'x){a : £C : e)(>/ : »/)
= (/>// :z:a'y + b'y)(y : a'x + b'x)(a : x).

To obtain this result we first substituted for A, B, D, E,
G, H, K the values we had assigned to them ; then we
omitted the redundant antecedent >/ in the second factor,
the redundant consequent e in the third factor, and the
redundant certainty (»/ : »/), which constituted the fourth
factor. The fact that the fourth factor (HG + K:>/)
reduces to the form (n : rj), which is a formal certainty
(see § 18), indicates that, in this particular problem,
nothing can be implicationally affirmed in terms of a or
b (without mentioning either x or y or z) except formal
certainties such as (ab : a), (aa f : >;), ab(a + b') : >i, &c, which
are true always and independently of our data (p.

36. If in the preceding problem we had not reduced
the alternative represented by As + Bz' + C to its sim-
plest form (see §§ 30, 31), we should have found for the
inferior limit or consequent of z, not a'y + b'y, but
x(a'y + b'y). From this it might be supposed that the
strongest conclusion deducible from z (in conjunction
with, or within the limits of, our data) was not A' but
xk'. But though xh! is formally stronger than A', that

32 SYMBOLIC LOGIC [§§ 36-38

is to say, stronger than A' token we have no data but our
definitions, here we have other data, namely, <p ; and <p
implies (as we shall prove) that A' is in this case equiva-
lent to xA', so that materially (that is to say, within the
limits of our particular data <p) neither of the two state-
ments can be called stronger or weaker than the other.
This we prove as follows : —

<p : (z : A 7 : y : D' : x) : (A' : x) : (A' = x A') ;

a proof which becomes evident when for A' and D' we
substitute their respective values a!y + b'y and a'x + b'x ;
for it is clear that y is a factor of the former, and x a
factor of the latter.

37. In the problem solved in § 35, in which (p denoted
our data, namely, the implication

e : xyza' + xyb' + xy'z' + y'z'a',

we took z, y, x as the order of limits and of elimination.
Had we taken the order y, x, z, our final result would have
been

(j> = (z:y: b'x + xz)(z + a : x){z : a' + b').

38. The preceding method of finding what I call the
" limits " of logical statements is closely allied to, and
was suggested by, my method (published in 1877, in the
Proc. of the Lond. Math. Soc.) for successively finding the
limits of integration for the variables in a multiple
integral (see § 138). In the next chapter the method
will be applied to the solution (so far as solution is pos-
sible) of Professor Jevons's so-called " Inverse Problem,"
which has given rise to so much discussion, not only
among logicians but also among mathematicians.

§39] JEVONS'S "INVERSE PROBLEM" 33

CHAPTER VI

39. Briefly stated, the so-called "inverse problem"
of Professor Jevons is this. Let tp denote any alternative,
such as abc + a'bc + aV V '. It is required to find an im-
plication, or product of implications,* that implies this
alternative.

Now, any implication whatever (or any product of
implications) that is equivalent to <p% or is a multiple
of <p e , as, for example, e : cp, or <p' : »y, or (abc : ab)(e : <p),
or (a : b)((f> f : rj), &c, must necessarily imply the given
alternative cp, so that the number of possible solutions
is really unlimited. But though the problem as enun-
ciated by Professor Jevons is thus indeterminate, the
number of possible solutions may be restricted, and the
problem rendered far more interesting, as well as more
useful and instructive, by stating it in a more modified
form as follows : —

Let cp denote any alternative involving any number of
constituents, a, b, c, &c. It is required to resolve the
implication e : cp into factors, so that it will take the
form

(M : a : N)(P : b : Q)(R : c : S), &c,

in which the limits M and N (see § 33) may contain

b, c, &c, but not a; the limits P and Q may contain

c, d, &c, but neither a nor b ; the limits R and S may
contain d, e, &c, but neither a nor b nor c ; and so on
to the last constituent. When no nearer limits of a con-
stituent can be found we give it the limits >; and e ;
the former being its antecedent, and the latter its con-
sequent (see §§ 18, 34).

* Professor Jevons calls these implications "laws," because he arrives
at them by a long tentative inductive process, like that by which scien-
tific investigators have often discovered the so-called " laws of nature "
(see§ 112).

C

34 SYMBOLIC LOGIC [§39

As a simple example, suppose we have *

(p = abc + a'bc + ab'c',

the terms of which are mutually exclusive. Reducing <p
to its simplest form (see §§ 30, 31), we get <p = be + ab'c',
and therefore

e : <£ = (f/ : ,, = (be)' {ab' c f )' : n = (b' + c')(a' + b + c):r 1
= a!b' + J'c + aV + be' : >/.

This alternative equivalent of cp' may be simplified (see
§ 31) by omitting either the first or the third term, but
not both ; so that we get

e : (p = b'c + a'c' + be' : rj = a'b' + b'c + be : 17.

Taking the first equivalent of e : <fi, and (in order to find
the limits of a) arranging it in the form Aa + Ba' + C : tj,
we get (see §§ 33, 34)

e : (p = tja + c V + (6'c + W) : »/
= (c 7 : a : e)(c : b : c)(t] : c : e).

Thus, we have successively found the limits of a, b, c (see
§§ 34, 35). But since (a : e), (>; : c), and (c : e) are all
formal certainties, they may be omitted as factors, so
that we get

e : <p = (c' : «)(c : 6 : c) = (c' : a)(c = b).

The first of these two factors asserts that any term of the
given alternative (p which contains c' must also contain a.
The second asserts that any term which contains c must
also contain b, and, conversely, that any term which con-
tains b must also contain c. A glance at the given alter-
native (p will verify these assertions.

* Observe that here and in what follows the symbol <j> denotes an
alternative. In §§ 34, 35 the symbol <j> denotes a given implication, which
may take either such a form ase:a + /3 + 7 + &c. , or as a + /3 + 7 + &c. : 7/.

§§39,40] JEVONS'S "INVERSE PROBLEM" 35

We will now take the second equivalent of e : <jj } namely,

a'b' + b'e + be' : tj,

and resolve it into three factors by successively rinding
the limits of a, b, c. Proceeding as before, we get

t -:^ = (6 / :rt )( c = &).

At first sight it might be supposed that the two ways of
resolving e : <p into factors gave different results, since
the factor (c : a) in the former result is replaced by the
factor (b' : a) in the latter. But since the second factor
(c = b), common to both results, informs us that b and c
are equivalent, it follows that the two implications c : a
and b' : a are equivalent also.

If we had taken the alternative equivalent of <p',
namely, a'b' + b'c + a'c' + be , in its unsimplified form, we
should have found

e:(p = (p':>] = (b' + c': a)(c = b) = {1/ : a)(c' : a)(c = b),

in which either the factor (b' : a) or the factor {c : a) may
be omitted as redundant, but not both. For though
the factor (c = b) alone neither implies (b' : a) nor (</ : a),
yet {b' :a)(c = b) implies (c' : a), and (c' :a)(c = b) implies
(b r : a). This redundancy of factors in the result is a
necessary consequence of the redundancy of terms in the
alternative equivalent of <ft' at the starting. For the
omission of the term a'b' in the alternative leads to the
omission of the implicational factor (a'b' : >/), or its
equivalent (b' : a), in the result ; and the omission of the
term a'c' in the alternative leads, in like manner, to the
omission of the factor (a'c' : rf), or its equivalent (c' : a), in
the result.

40. I take the following alternative from Jevons's
"Studies in Deductive Logic" (edition of 1880, p. 254,
No. XII.), slightly changing the notation,

abed + abe'd + ab'cd' + a'bed' + a'b'c'd'.

Let (p denote this alternative, and let it be required to

36 SYMBOLIC LOGIC [§40

find successively the limits of a, b y c, d. In other words,
we are required to express e : <fi in the form

(M : a : N)(P : b : Q)(R : c : S)(T : d : U),

in which M and N are not to contain a ; P and Q are
neither to contain a nor b ; R and S are neither to con-
tain a nor b nor c ; and T and U must be respectively
n and e. By the process of §§ 34, 35, we get

M. = d + b</ + b'c, N = bd + b'c, V = d, Q = c + d, R = >,,

S = e, T = r,, U = 6.

Omitting the last two factors R : c : S and T : d : U
because they are formal certainties, we get

e : (p = (d + be' + b'c : a : bd + b'c){d :b:c + d).

A glance at the given alternative <£> will verify this result,
which asserts ( 1 ) that whenever we have either d or be' or
b'c, then we have a ; (2) that whenever we have a, then we
have either bd or b'c ; (3) that whenever we have d, then
we have b ; (4) that whenever we have b, then we have
either c or d; and (5) that from the implication e -. (p we
can infer no relation connecting c with c£ without making
mention of a or b ; or, in other words, that c cannot be
expressed in terms of d alone, since the factor >/ : c : e is a
formal certainty and therefore true from our definitions
alone apart from any special data. The final factor is
only added for form's sake, for it must always have >/ for
antecedent and e for consequent. In other words, when
we have n constituents, if x be the n th or last in the
order taken, the last factor must necessarily be >; : x : e,
and therefore a formal certainty which may be left
understood. Others of the factors may (as in the case
of n : c : e here) turn out to be formal certainties also, but
not necessarily.

We have found the limits of the constituents a, b, c, d,
taken successively in alphabetic order. If we take the
reverse order d, c, b, a, our result will be

e : (p = (ab + ac' + bd : d : ab)(ab' + a'b : c : a + b),

§§ 40, 41] ALTERNATIVES 37

omitting the third and fourth factors >) : b : e and >; : a : e
because they are formal certainties. There is one point
in this result which deserves notice. Since every double
implication a : x : (3 always implies a : /3, it follows that
(in the first bracket) ab + ac' + he implies ab. Now, the
latter is formally stronger than the former, since any
statement x is formally stronger than the alternative
x + y. But the formally stronger statement x, though it
can never be weaker, either formally or materially, than
x + y, may be materially equivalent to x + y; and it must
be so whenever y materially (i.e., by the special data of
the problem) implies x, but not otherwise. Let us see
whether our special data, in the present case, justifies the
inferred implication ab + ac + be : ab. Call this implica-
tion \J/-. By virtue of the formula a + (3 + y : x = (a : x)
(/3 : x)(y : x), we get (putting ab for a and for x, ac for (3,
and be for y)

\|z = (ab : al)){ac' : ab)(bc : ab) = e(ac : ab)(bc : ab)
= (ac : a)(ac : b)(bc : a)(bc : b)
= e(ac' : b)(bc' : a)e = (ac : b)(bc : a).

This asserts that (within the limits of our data in this
problem) whenever we have ac we have also b, and that
whenever we have be we have also a. A glance at the
given fully developed alternative <p will show that this is
a fact (see § 41). Hence, the inferred implication
ab + ac + be : ab is, in this problem, legitimate, in spite of
the fact that its antecedent is formally weaker than its
consequent.

41. An alternative is said to be fully developed when,
and only when, it satisfies the following conditions :
Firstly, every single-letter constituent, or its denial, must
be a factor of every term ; secondly, no term must be a
formal certainty nor a formal impossibility ; thirdly, all
the terms must be mutually incompatible, which means that
no two terms can be true at the same time. This last con-
dition implies that no term is redundant or repeated.

38 SYMBOLIC LOGIC [§§ 41, 42

For example, the fully developed form of a+ft is
aft + aft' + aft. To obtain this we multiply the two
factors a + a and ft + /3', and strike out the term aft',
because it is equivalent to (a + ft)', the denial of the
given alternative a + ft. As another example, let it be
required to find the fully developed form of a + ft'y.
Here we first find the product of the three factors a + a,
ft + ft', and 7 + 7'. We next find that {a -{-ft'y)' is
equivalent to a' (ft'y)', which is equivalent to a'(ft + y'),
and therefore, finally, to aft + ay'. Then, out of the
eight terms forming the product we strike out the three
terms a'fty, a'fty, a'/S^', because each of these contains

Online LibraryHugh MacCollSymbolic logic and its applications → online text (page 3 of 11)