Complete Works of Lewis Carroll (161 page)

(3)
Fallacy of two Entity-Premisses.

Here the given Pair may be represented by either (
xm
1
†
ym
1
) or (
xm
1
†
ym
′
1
).

These, set out on Triliteral Diagrams, are

xm
1
†
ym
1

xm
1
†
ym
′
1

§ 4.

Method of proceeding with a given Pair of Propositions.

Let us suppose that we have before us a Pair of Propositions of Relation, which contain between them a Pair of codivisional Classes, and that we wish to ascertain what Conclusion, if any, is consequent from them.
We translate them, if necessary, into subscript-form, and then proceed as follows:—

(1) We examine their Subscripts, in order to see whether they are

(a) a Pair of Nullities;

or (b) a Nullity and an Entity;

or (c) a Pair of Entities.

(2) If they are a Pair of Nullities, we examine their Eliminands, in order to see whether they are Unlike or Like.

If their Eliminands are
Unlike
, it is a case of Fig.
I.
We then examine their Retinends, to see whether one or both of them are asserted to
exist
.
If one Retinend is so asserted, it is a case of Fig.
I (α); if both, it is a case of Fig.
I (β).

If their Eliminands are Like, we examine them, in order to see whether either of them is asserted to exist.
If so, it is a case of Fig.
III.; if not, it is a case of “Fallacy of Like Eliminands not asserted to exist.”

(3) If they are a Nullity and an Entity, we examine their Eliminands, in order to see whether they are Like or Unlike.

If their Eliminands are Like, it is a case of Fig.
II.; if
Unlike
, it is a case of “Fallacy of Unlike Eliminands with an Entity-Premiss.”

(4) If they are a Pair of Entities, it is a case of “Fallacy of two Entity-Premisses.”

[Work Examples §
4
, 1–11 (p.
100); §
5
, 1–12 (p.
101); §
6
, 7–12 (p.
106); §
7
, 7–12 (p.
108).]

BOOK VII.

SORITESES.

CHAPTER I.

INTRODUCTORY.

When a Set of three or more Biliteral Propositions are such that all their Terms are Species of the same Genus, and are also so related that two of them, taken together, yield a Conclusion, which, taken with another of them, yields another Conclusion, and so on, until all have been taken, it is evident that, if the original Set were true, the last Conclusion would
also
be true.

Such a Set, with the last Conclusion tacked on, is called a ‘
Sorites
’; the original Set of Propositions is called its ‘
Premisses
’; each of the intermediate Conclusions is called a ‘
Partial Conclusion
’ of the Sorites; the last Conclusion is called its ‘
Complete Conclusion
,’ or, more briefly, its ‘
Conclusion
’; the Genus, of which all the Terms are Species, is called its ‘
Universe of Discourse
’, or, more briefly, its ‘
Univ.
’; the Terms, used as Eliminands in the Syllogisms, are called its ‘
Eliminands
’; and the two Terms, which are retained, and therefore appear in the Conclusion, are called its ‘
Retinends
’.

[Note that each
Partial
Conclusion contains one or two
Eliminands
; but that the
Complete
Conclusion contains
Retinends
only.]

The Conclusion is said to be ‘
consequent
’ from the Premisses; for which reason it is usual to prefix to it the word “Therefore” (or the symbol “
∴
”).

[Note that the question, whether the Conclusion is or is not
consequent
from the Premisses, is not affected by the
actual
truth or falsity of any one of the Propositions which make up the Sorites, by depends entirely on their
relationship to one another
.

As a specimen-Sorites, let us take the following Set of 5 Propositions:—

(1) ”No
a
are
b
′
;

(2)   All
b
are
c
;

(3)   All
c
are
d
;

(4)   No
e
′
are
a
′
;

(5)   All
h
are
e
′
”.

Here the first and second, taken together, yield “No
a
are
c
′
”.

This, taken along with the third, yields “No
a
are
d
′
”.

This, taken along with the fourth, yields “No
d
′
are
e
′
”.

And this, taken along with the fifth, yields “All
h
are
d
”.

Hence, if the original Set were true, this would
also
be true.

Hence the original Set, with this tacked on, is a
Sorites
; the original Set is its
Premisses
; the Proposition “All
h
are
d
” is its
Conclusion
; the Terms
a
,
b
,
c
,
e
are its
Eliminands
; and the Terms
d
and
h
are its
Retinends
.

Hence we may write the whole Sorites thus:—

”No
a
are
b
′
;

  All
b
are
c
;

  All
c
are
d
;

  No
e
′
are
a
′
;

  All
h
are
e
′
.

         
∴
All
h
are
d
”.

In the above Sorites, the 3 Partial Conclusions are the Positions “No
a
are
e
′
”, “No
a
are
d
′
”, “No
d
′
are
e
′
”; but, if the Premisses were arranged in other ways, other Partial Conclusions might be obtained.
Thus, the order 41523 yields the Partial Conclusions “No
c
′
are
b
′
”, “All
h
are
b
”, “All
h
are
c
”.
There are altogether
nine
Partial Conclusions to this Sorites, which the Reader will find it an interesting task to make out for himself.]

CHAPTER II.

PROBLEMS IN SORITESES.

§ 1.

Introductory.

The Problems we shall have to solve are of the following form:—

“Given three or more Propositions of Relation, which are proposed as Premisses: to ascertain what Conclusion, if any, is consequent from them.”

We will limit ourselves, at present, to Problems which can be worked by the Formulæ of Fig.
I.  Those, that require
other
Formulæ, are rather too hard for beginners.

Such Problems may be solved by either of two Methods, viz.

(1) The Method of Separate Syllogisms;

(2) The Method of Underscoring.

These shall be discussed separately.

§ 2.

Solution by Method of Separate Syllogisms.

The Rules, for doing this, are as follows:—

(1) Name the ‘Universe of Discourse’.

(2) Construct a Dictionary, making
a
,
b
,
c
, &c.
represent the Terms.

(3) Put the Proposed Premisses into subscript form.

(4) Select two which, containing between them a pair of codivisional Classes, can be used as the Premisses of a Syllogism.

(5) Find their Conclusion by Formula.

(6) Find a third Premiss which, along with this Conclusion, can be used as the Premisses of a second Syllogism.

(7) Find a second Conclusion by Formula.

(8) Proceed thus, until all the proposed Premisses have been used.

(9) Put the last Conclusion, which is the Complete Conclusion of the Sorites, into concrete form.

[As an example of this process, let us take, as the proposed Set of Premisses,

(1) “All the policemen on this beat sup with our cook;

(2)   No man with long hair can fail to be a poet;

(3)   Amos Judd has never been in prison;

(4)   Our cook’s ‘cousins’ all love cold mutton;

(5)   None but policemen on this beat are poets;

(6)   None but her ‘cousins’ ever sup with our cook;

(7)   Men with short hair have all been in prison.”

Univ.
“men”;
a
 = Amos Judd;
b
 = cousins of our cook;
c
 = having been in prison;
d
 = long-haired;
e
 = loving cold mutton;
h
 = poets;
k
 = policemen on this beat;
l
 = supping with our cook

We now have to put the proposed Premisses into
subscript
form.
Let us begin by putting them into
abstract
form.
The result is

  (1) ”All
k
are
l
;

  (2)   No
d
are
h
′
;

  (3)   All
a
are
c
′
;

  (4)   All
b
are
e
;

  (5)   No
k
′
are
h
;

  (6)   No
b
′
are
l
;

  (7)   All
d
′
are
c
.”

And it is now easy to put them into
subscript
form, as follows:—

  (1)
k
1
l
′
0

  (2)
dh
′
0

  (3)
a
1
c
0

  (4)
b
1
e
′
0

  (5)
k
′h
0

  (6)
b
′l
0

  (7)
d
′
1
c
′
0

We now have to find a pair of Premisses which will yield a Conclusion.
Let us begin with No.
(1), and look down the list, till we come to one which we can take along with it, so as to form Premisses belonging to Fig.
I.
We find that No.
(5) will do, since we can take
k
as our Eliminand.
So our first syllogism is

  (1)
k
1
l
′
0

  (5)
k
′h
0

∴
l
′h
0
… (8)

We must now begin again with
l
′h
0
and find a Premiss to go along with it.
We find that No.
(2) will do,
h
being our Eliminand.
So our next Syllogism is

  (8)
l
′h
0

  (2)
dh
′
0

∴
l
′d
0
… (9)

We have now used up Nos.
(1), (5), and (2), and must search among the others for a partner for
l
′d
0
.
We find that No.
(6) will do.
So we write

  (9)
l
′d
0

  (6)
b
′l
0

∴
db
′
0
… (10)

Now what can we take along with
db
′
0
?
No.
(4) will do.

(10)
db
′
0

  (4)
b
1
e
′
0

∴
de
′
0
… (11)

Along with this we may take No.
(7).

(11)
de
′
0

  (7)
d
′
1
c
′
0

∴
c
′e′
0
… (12)

And along with this we may take No.
(3).

(12)
c
′e′
0

  (3)
a
1
c
0

∴
a
1
e
′
0

This Complete Conclusion, translated into
abstract
form, is

“All
a
are
e
”;

and this, translated into
concrete
form, is

“Amos Judd loves cold mutton.”

In actually
working
this Problem, the above explanations would, of course, be omitted, and all, that would appear on paper, would be as follows:—

  (1)
k
1
l
′
0

  (2)
dh
′
0

  (3)
a
1
c
0

  (4)
b
1
e
′
0

  (5)
k
′h
0

  (6)
b
′l
0

  (7)
d
′
1
c
′
0

  (1)
k
1
l
′
0

  (5)
k
′h
0

∴
l
′h
0
… (8)

  (8)
l
′h
0

  (2)
dh
′
0

∴
l
′d
0
… (9)

  (9)
l
′d
0

  (6)
b
′l
0

∴
db
′
0
… (10)

(10)
db
′
0

  (4)
b
1
e
′
0

∴
de
′
0
… (11)

(11)
de
′
0

  (7)
d
′
1
c
′
0

∴
c
′e′
0
… (12)

(12)
c
′e′
0

  (3)
a
1
c
0

∴
a
1
e
′
0

Note that, in working a Sorites by this Process, we may begin with
any
Premiss we choose.]

§ 3.

Solution by Method of Underscoring.

Consider the Pair of Premisses

xm
0
†
ym
′
0

which yield the Conclusion
xy
0

We see that, in order to get this Conclusion, we must eliminate
m
and
m
′
, and write
x
and
y
together in one expression.