486 104 18MB
English Pages 105 [110] Year 1966
ORIGINAL TITLE: “EXACTE LOGICA” De Erven F. Bohn N.V., Haarlem, 1961
T
L ;
En

THE LANGUAGE of LOGIC by
HANS FREUDENTHAL Professor of Pure and Applied Mathematics,
Utrecht University, Utrecht, The Netherlands
ELSEVIER PUBLISHING COMPANY AMSTERDAM/LONDON/NEW YORK 1966
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——— CONTENTS
1.
Setsand mappings
...... see eee ee
2.
Propositions .
3,
Subject and predicate... 2... ee
En
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2 1. we
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ee
;
42 i
4.
Formallogic. ..... eee eee eee eee
5.
Language and metalanguage. .... 1... 2. ee ee eee
6»
Solutions. …. … «4 «a «se ren eee ee ee » »« 102
90
 
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BE
1SETS AND MAPPINGS
1.1  Examples of Sets The set of all people (now alive). — Theset of all hydrogen atoms. — Theset
ofall positive integers (i.e., 1, 2, 3, ...). — The set of all integers (i.e.,..., — 3, —2, —1,0, 1, 2, 3,...). — The set of all rational numbers (i.e, in addition to v2
=i
6
i 2
8
the integers, also the fractions of integers, e.g., 4, —2°, etc.). — The setof all
real numbers(i.e, also such numbers as /2, log 7, zr, infinite decimalfractions);
this setmay also be conceived as the socalled number scale, i.e, a horizontal
straight line on which the positive numbersare set out as lengthsto theright,
and the negative numberstotheleft, from a point 0. — Theset of all complex
numbers (these are numbers of the form a+ bi with real a and b and a symbol i
representing the nonreal square root of —1). — Theset of all points X which
have the same distance from twofixed points A and B (in plane geometry we speak of “locus” instead of “set’””).— The set of all real x which satisfy the condition —3 pi) A (pi > p2)] > (ps > pe), P1Ps, Pe pr, Pe — (ps — Po).
is
Kk
Ch. 3
“ALL” AND “THERE IS”
43
The propositions pa Pe,
Ps A Pe
are never true. (The example in the secondlineisstill a true proposition,if we replace pi, pa, ps by any other proposition. In the otherlines, we have toinsert for pi, pe, ps, the propositions indicated above.)
As soon as we have to deal with propositions in which a variable occurs, we need symbols that express whether a proposition is always, sometimes or never true.
3.2  A proposition like F(x), in which there is a variable x, represents a whole „system of propositions in the sense of chapter 2; we obtain this system, by substituting for the variable x the individuals a, b, c,..., over which x is running:
F(a), F(b), F(c), ... « For example, we may substitute for x in pi(x) after each other all people; we obtain then propositionslike in 2 such as: The worldchampionjiujitsu lives in London, Brigitte Bardot lives in London,
The Lord Mayor of Londonlives in London,
some of which are true and someare not. With a constant proposition p, like we know them from 2, we had two possibilities: it was true or false. We could even satisfy ourselves with only one truthvalue (e.g. “‘true’’); the falseness of p being expressed by the truth of —p. Do we need now a differentiated concept of truth for these variabledependent propositions, or can weinsert this “always”, “sometimes”, “never” with the word “true” into the proposition, just like we replaced “‘p false” by ““—p true’’?
If a proposition F(x) like
x lives in London — x lives in England is to be always true, then this means that,
F(a), FB), F(c), … are true, or in other words that the conjunction F(a) A F(b) A Fo a...
is true. This is a cumbersome formula, especially if the variables x in F(x) are
44
SUBJECT AND PREDICATE
Ch. 3
C
allowed to run over infinitely many individuals. Instead of this we therefore write
(
AzF(x), and weread this as for all x, F(x). F(x) alwaystrue, i.e. true for all x, is now expressed by Aa F(x) is true. Herewith we have inserted the “always”’differentiation with “true” into the
proposition, of which the truth is asserted. Similarly we do this with ““sometimes”’. If a proposition F(x) like x lives in London is to be true sometimes, then this meansthatat least one of the propositions
F(a), F(b), F(c), ... has to be true, or in other words, the disjunction F(a) v F(b) v F(c) v . is true. Here too we simplify this notation to
VaF(x), which weread as there is an x, so that F(x). F(x) true for some x, is now expressed by
VaF(x) is true. Herewith we have inserted the “sometimes’’differentiation with “true” into the proposition, the truth of which is being asserted. (Note: “there is an x, ...” does not mean,that there is only one such x; there may be more. This may even be all. “All” is a special case of “some”’.) F(x) never true, is clearly the same as, there is no x for which F(x) is true, i.e. VeF(x) is not true, or formulated in still another way —_V2 F(x)
is true.
ER
ka Ch. 3
CALL” AND “THERE IS”
45
3.3 A and V are called quantifiers, because they indicate for how many (quantum) valuesof the variable the propositionis true. In
AaF(x) and Ve) x no longer occurs as a variable but is bound. There are also other ways to bind variables. We know already the method of doing this by substitution of an individual for the variable, e.g., substitution of “Mars” or “Sirius” for x in x is a planet
(in the formercaseit is true,in thelatter false). Note that
(AzF(x)) > Fa), F(a) > Va F(x) are true.
3.4  In a proposition different variables may be bound in different ways. Take x is motherofy,
abbreviated xMy. Then for
Vy(xMy) we mayalso read: x is a mother. Onevariable has been eliminated by binding; the otheris still free. We can also bindthis x: VaVy(xMy)
means: there is a mother. AgVy (xMy)
means: everyoneis a mother. Of course,this is false. But the followingis true: —AaVy(xMy).
Wecan, however,also start with the x: Va(xMy) means: y is a child.*) *) Always supplementthis by “‘of a mother”.
46
SUBJECT AND PREDICATE
Ch. 3
Vy Va(xMy)
means: there is a child. AyVa(xMy)
(*)
means: everyoneis a child. Thisis true. We canalso start with the universal quantifier:
Ay(xMy) means: x is everyone’s mother. AaAy(xMy) means: everyone is everyone’s mother.
Va/\y(xMy)
(**)
means: there is a motherofall.
Or, again starting with x:
Aa(xMy) means: y is everyone’s child.
Ay/\a(xMy) means: everyoneis everyone’s child.
VyAa(xMy) means: there is some one whois everyone’s child. Note more particular that (*) and (**), which differ only in the order in which the quantifiers occur, have different truth values.
3.5  On the other hand,in the case of similar quantifiers the order in which in which they occur does not matter: VaVy(xMy) VyVa(xMy) Va,y(xMy)
(there is a mother), (there is a child), (there is a pair “motherchild””)
are equivalent. Generalizing, it is true that VaVyF(, y),
are equivalent.
VyVak(x, y),
Vary F(x, y)
Ch. 3
“ALL” AND “THERE IS”
47
(The notation fx, y! had been agreed uponfor the directed pair of x and y, but
this has here been simplified for typographical reasons.)
Howcan we verify the equivalence of the above propositions? For the variables we substitute all possible individuals a, b, c,... and write down F(a, 4),
F(a, b),
F(a, €),
aes
Vy F(a, y)
F(b, a),
F(b, b),
F(b, €),
eee
Vy Fb, y)
F(c, a), Valk, a),
F(c, b),
VaF(x, b),
F(c, ©),
ees
VyF(c, y)
Valk(x, €),
After the ith row is written the disjunction of what is stated in theith row; and under the jth column is written the disjunction of what is stated in the jth column. The disjunction of the last column is Ve VyF(x, y).
In order to verify the truth of this, we must find something that is true in the last column, and in order to verify the truth of something in the last column, we must find something that is true in the row thereof. It therefore all boils down to finding a pair Fx, y! for which F(x, y) is true, i.e, Var F(x, y).
The same applies to
VyVe Fx, y).
. . . In this case we have to run over the schemein the opposite way, hence first the columns and then the rows.
3.6  We can similarly show that
AaAvF,y), AvAaF,y), Aa vF(x, y) are equivalent. An analogous proof can be given for more than two variables.
3.7  Ifp is independent ofx, then of course and
Vap [Va(p A FOX), [p Vv AzF(x)] > [Asp v F(X))). To prove this, write down: [p A (F(a) v F(b) v Fc) v le (p A F(a)) v (p A Fb) v v (p A Fc) v …) etc.
3.9  There exists a similar analogy with (27) and (28) in 2.10; true are
[Ac(F(x) > p)] @ (V2 F(x) > pl, [Az > F(x))] @ [p > Ac FOX). 3.10  Whatis the negation of “all men are mortal”? Not “all men are immortal’,
for the first statement is refuted even if only one immortal human being can be found. The true negation of the above statement is therefore
“there is a man who is not mortal”. In general we seek an equivalent of
—Ar F(x). If for this we write again
—(F(a) A F(b) A F(c) A …) and apply (26) (see 2.10), we find —F(a) v —F(b) v —F(c) v …, hence
Va—F(x). The negation of “F(x) holds for all x” is therefore “there is an x for which notF(x) holds”; hence
Az FG) > (VanF(X)).
Ch. 3
PROBLEMS
49
If we take more particularly for F(x) (xis aman) — (xis mortal), we then obtain the true proposition {AA 2[(xis aman) > (x is mortal)]}
(x is mortal)]},
or applying (29) (see 2.10) = {V2[(xisaman) A —(xis mortal)]}.
3.11  Similarly is true
(AV2 F(x)
(AzF(X).
The statement ontheleft is: there is no x for which F(x) holds. Ontheright: for all x, notF(x) holds.
The laws in 3.1011 can be summarized in the following words: Wemayinterchange — and a quantifierif the latter is reversed at the sametime. This recalls the reciprocity laws stated in 2.10 (2526) and similar laws applicable to operationsonsets.
3.12
Problems
The following notation will be employed: M(x): xis male,
V(x):
xis female,
xJy: xKy:
xis younger than y, xisachild ofy,
xGy: U(x): A(x):
xis married to y, xlives at Utrecht, xlives at Amsterdam.
Sa eee
where all the variablesrelate to people. Now write down the symbolic expressions for: Everyone hasa father and a mother. Whoever has a father, also has a mother. Everyone is younger than his or her parents. Everyone is younger than his or her grandparents.
x is married. There is a man with a daughterinlaw who is older than heis.
SS 50
SUBJECT AND PREDICATE
Ch. 3
7. x and y are brothers(i.e., full brothers). 8. If there is woman at Utrecht with a brother at Amsterdam then there is a man at Amsterdam with a sister at Utrecht.
9.
A married man need not live at Utrecht.
10. Not every woman at Utrecht has no son at Amsterdam. 11. Allthe children of x are married. 12. There is some one, all of whose children are married. 13. Every child of x is married to a child ofy. 14. There is a child of y who is not married to a child of x. 15. There are two people, each child of one of them is married to a child of the other.
16.
There are two people, no child of one of them is married to a child of
the other.
17.
Ifyisachild of x, then every child of y is a grandchild ofx.
Andfurther:
18.
Read Ay {[Aa(xKy > U(Q))] © [(V2xKy) > UQ)]}}

Ch.
31,
32. 33.
34,
35.
36.
thi
37. 38.
see 39.
suc
Ne
al
Re
and find out whatthis has to do with 3.9.
40.
Read the following formulae and find out on the basis of which laws they are true (independently of the meaning of K, G, U,etc.).
42
19,
=aAz{lVy(xGy)] => U(x)} > ValVy(xGy) A —U(X)).
20.
[AAyVaeKy)] > [VyAernlxKy)].
U.
VaeArlyKx > Velzky)) > Aa VolyKx A Azl—alzKy))].
The following notation will now be introduced: Z(x, t): Isee the thing x at the instant ¢. P(x, t): I seize the thing x at the instant¢. t’ (2 =3).
46.
Vay[(x>y>0) A (*+y=0)].
48. 49. 50.
Az{(x2>x)< [(x> 1) v (x (b2 — 4ac = 0)}. Aa,b,c{[Az(ax? + bx + c>0)] > [(b2— 4ac 0)]}.
52. 53. 54.
Ag{[x>2) aA &D v &a) A (x (a 0)] > (b > 0}.
57.
AvVaAalx? +axtb>0).
58.
As{[AaVa2(x? + ax+b=0)] (b £0}.
59.
VoAaVa(x2 +ax+b=0).
43. VaVy(x+y = 3).
47. N\af{[Valax = 6)] > (a #0}
51. Agf{I(x>2) A A> 3)] Oo 2 [(x=1) v (x=21}
> [(a= —3) A (b=2)}).
60. VaAoVa(x? + ax+b=0).
52
Ch. 3
SUBJECT AND PREDICATE
3.13  One of the above problems should by rights have been accompanied by a warning. “If from x2?+ax+b=0 follows x=1 or x=2, then a= —3 and b=2 (a, x, b are assumed to run over all real numbers)” Is this true? No! Because, for example, “‘x? + 1 =0” also implies “x= 1 or x = 2” (for any x) —
Ch
T
a
h c
since x2 + 1 =Ois false (for any x) and we know that p — q is true when p is false. Hence
3
Nal? + ax+b=0) > [(x=1) v &=2)]}}
(*)
is, for example, also true if a= 0, b= 1. We, therefore, may not infer from (*) that (a= — 3) A (b= 2) holds, as was asserted in 3.12.55. Instead of x2 + 1=0 we could have chosen, as another counter example, any other quadratic equation that has no (real) solution. Even for example x2—2x+1=0 or x2—4x+4=0,for
Aal(x? — 2x + 1=0) > (x= 1}, hence certainly
i
i
L T
H
Aci? — 2x +1) =0) ~> (x=1 v x=2)], and analogously with the other example. We maytruly assert:
I
Aa,o(An{(x? + ax +b =0) > [(x=1) v (x =2)]} > {[@=3) a 0=2)] v AVaylx4y) A (2 +ax+b=y?+ayt+
+ b=0)]}), or
A
L
Aaw(Vaylx#y) A (2 +ax+b=y2+ay+b=0)] > (Ac{(x? + ax +b=0) > [x= 1) > (x = 2)]})
> [a= —3) A
A (6=2))).
I I
t
For those whostill refuse to believe that forall x (x2+1=0) > [(x=l) v (x=2)] is true: Let x? + 1 =0. Then hence
P
t
0 G(x))] > [tz FQ) < (te G@))]. Proof: Suppose
Az(F(x) > G(x)
(1)
to be true. Nowlet
ae (tc F(x).
(2)
54
SUBJECT AND PREDICATE
Ch. 3
We must show that also a € (Îx G(x)). From (1) followsin particular that
F(a) > G(a); from (2) follows, having regard to the definition of Î: F(a). Hence (modus ponens) G(a).
Hence(definition of f)
ae (TzG(x)). Herewith one half of the statement has been proved. The reader is invited to
prove the other half for himself.
ta(F(x) A G(x)) = (Te FQ) 0 (Te G(x). Tto(F(x) V G(x) = (Te F(x) U (fx GC). Heis likewise invited to prove these formulae!
Ch
10
ll
12
13
Le
14
Le
15
16
17
Le de
In 2.10 we had our attention called to the similarity in laws (19)(24) for, U, n, on one hand, and, on the other hand, v, A. This is not an accidental simi
larity, as appears from the last two formulae. The similarity also extends to — and * (complementforming) respectively (2.10.2526). Where did we just encounter the relationship between — and *?
3.15  Problems
Seer
Using the notation indicated in 3.12, write down: Theset of the inhabitants of Amsterdam.
Theset of the married inhabitants of Amsterdam. The set ofthe children ofy. Theset of all married couples. Theset of all instants at which I see things without seizing them. Theset of all things that I seize withoutfirst having seen them.
Check whether fra,y1xKy, tx Ty xKy, ty texKy are the same. Check whether(for real numbers) the following propositionsare true: 8.
[fraverVelax? + bx + e=0)}= {If ra,v,e1(b? — 4ac 2 0)]
A Tra,v,c1[(a = b =0) > (c= 0} 9.
Ta (x2 + 1 = 0) = fz(x2 ++ 2=0).
18
19
20
21
In
d
2 2
I b m
2
2
2
m
Ch. 3
PROBLEMS
55
10. teAy({ae [fz(xy =O} > (py =0)) = Îa Volab = 1). 11. Tal — DD— 2)— 3x — 4) > 0] = {[te(x [W=1) v y= x)1})? In 1.22 we divided a set 2 into equivalence classes. Let an equivalence relation be given in the set 2; the elements of @ are designated by small letters; ““x ~ y” means “‘x equivalent to py”. 24.
Whatis the meaning of fz(x ~ y)?
25. Is Ay,2([z € tee ~ y] > [tee ~ y) = Ta~ 2))) true? 26.
Verify that
Tta{VolA = To~ y)]} means: thesetof all equivalenceclasses(in 2).
56
27.
SUBJECT AND PREDICATE
Ch. 3
Ch.
we
Verify (see 1.25):
i inte
[A, B] = teVay[(x eA) A (ye B) A (z="x, y)]. 3.16  In the first chapter we considered unions and intersections of finite numbers of sets only. We did, however, encounter several instances of“sets of sets”,
Thi the
Simi
Let N be the set of natural numbers and R be the set of real numbers. Let M,, be the set of the numbers < 5 hence

1 Mn=fel(xe BR) A (x S pl
 
Com

The
The intersection of all M‚ is the set
Tel(xe B) A (x S 0)]. This intersection is also denoted by
1.
A nen Mn.
2.
Let Pa be the set of the real numbers Anl(ne N) > (ce Ma)]
;

3.1
Note the relationship between f) and A. Wecansimilarly define the union of the sets M, (n € N) by
the is o
© € Une Mn Val(n€ N) > (ce M‚)].
F(x
This is the relationship between Wand V.
If there can be no misunderstanding as to the set to which the subscript belongs, the notation may be simplified to U, M‚ and (\n Mn respectively. A relation with 3.14 is: Let F(x, y) be a proposition dependent upon x and y; here again x can run over atotal T, and yrunsover a set N whichis not related to T. Now for each n e N
'
(No this
The
Ch. 3
THE ARTICLE
57
we consider the set M, of all x for which F(x, n) is true, and we take their intersection:
Nn fe Fx, 1) = TeAnF(x,n). Think of the case where N consists only of two elements and compare this with the penultimate formula in 3.14. Similarly Unto F(x, n) = fzVn F(x,n).
Compare this with the last formula in 3.14.
3.17  Problems The notation used in 3.15 will be employed.
1. Whatis Nytx(X¥ ec Ye A)?
2. Whatis Urfr(Xe Ye A?
If small letters denote natural numbers, then:
3. Whatis ls 1yx Div y?
4. Whatis U‚frx Div y? 5. Whatis UnfeVeAol(z= x")) A {y Div x > [(y=1) v (y= 2)J})? 6.
Whatis Pls tes (x =ay+1)?
The Article
3.18  We speak ofthe solution of 3x = 2, butnot of the solution of 0x = 3 or of the solution of x? — 5x +6=0. Wespeak of the x with the property E if there is one and only one x that has the property E. If E is the property, to turn F(x) into a true proposition then the condition for speaking of the x is
VaF(x) A Azyl(F) A FO)) > x=). (Note that we must have a relation “=” at our disposal to be able to write
this down.)
The x with the property F(x) is designated by Vz F(x).
58
SUBJECT AND PREDICATE
Ch. 3
Let us, just for the present, confine ourselves to real numbers. How do we
write: the largest x for which x2 — x < 5? Answer:
Le{@? — x35) A AsO? yS 5) > WS >)]}. How do we write: the smallest set of which the numbers 288 and 120 are members and of which, with two numbers a and b, a — b is also a member? Answer: Wedefine a proposition F(X) about sets X by:
F(X) {(120 eX) a (288 € X) A Aawl([(ae X) A (bE X)] > > (a—be X))}. Are there sets X that satisfy F(X)? Of course, the set of all numberssatisfiesit. If X is such a set, then 288e X, and 120 e X, hence 168 = 288 — 120 e X, hence 168 —120=48eX, hence 120 —48= 72e X, hence 72—48= 4e X.
Furthermore 0 =24 — 24e X, —24=0 — 24e X. All multiplies 24 m (where m is an integer) of 24 are members of X (why?). Let A denote the set of all these multiples. Then
F(A) holds (why?). We now just proved:
Ch
H
n
is
B 1 T
e
i
h al se (
t
Ax[F(X) > (X¥ > A)]. Hence: F(A) is true, and every X for which F(X) is true will include A. Therefore A is the smallest set Y with the property F(X). We mayask ourselves the following question: in general let M be a set of sets X
determined by a proposition F(X) about sets X, i.e.,
3
Azl(Xe M) = F(X)]. Is there a smallest set A in M, i.e., an A such that
(AEM) a Ax[(Xe M) = (X > A)]
I
is true? In general, something of this kind need not exist. (Counterexample:
e
the set M of all nonempty subsets of the set (a, b); M will then consist of (a), (B) and (a, b).) An A as required must be contained in every Xe M; hence also
E t
Ac nzx, But since A must also occur among the Xe M, Nxem XeA
must be true and therefore A= Mlxeu X.
o
t R
s
i f
c t
Ch. 3
PREDICATES, RELATIONS
59
Hence it immediately follows that there is at most one such a set. (Maze X) e M
need not always be true (see the counterexample). But if it is true, then Mlzeu X is the smallest set of M, because
(NxemX)< Y forevery
YewM.
Byvirtue of this criterion we can again show thatthere is a smallest set of which 120 and 288 are members and of which, with a and b, a—b is also a member. This is apparent from the following. Let M betheset of all these sets. For every Xe Mis 120 X, hence 120 €f) x. X. Similarly for 288. Furthermore:
if ael\xeuX and be (\ xem X, then ae X for all Xe M, andbe XforallXe M; hence a— be for all Xe M, and therefore a — bef) xew X. Thus () xem Xmeets
all the requirements on thesets belonging to M. Hence () xem X is the smallest set that is a member of M.
(In this last section the word “all” has repeatedly been used. Rewrite it, using the logic symbols as muchas possible.)
Predicates, Relations
3.19  In this chapter, we often used schemeslike .. is mortal,
... lives at Amsterdam, .. isachild of..., ... is equivalent to... In each case the dots represent something that must be filled in (humans, elements ofa certain set, etc.). We thus obtain propositions which maybe false
or true.
Expressions of this kind with one free place are called properties or predicates;
those with two or more free places are known asrelations. Whatwefill in at the free places are called subjects. Relations may, however, also be conceived as predicates, namely, predicates of subjectpairs, subjecttrios, etc. ‘‘... lives at Amsterdam” is a property of
individual people; “... is a child of ...” is a property of pairs of persons, in fact, directed pairs. “x is a child of y” is a proposition which maybe true for certain pairs x, y and false for others(if it is true for the pair 'x, y', it is certainly false for the pair ly,x1).
60
SUBJECT AND PREDICATE
Ch. 3
The predicate ““.… lives at Amsterdam” is fully known to me, if I know the set of all those x which, when filled in, turn the predicate into a true proposition, and vice versa. Similarly, the relation ““.… is a child of …”is fully known to me
Ch.
Th
if I knowthesetall those pairs Fx, y! which, on being consecutively filled in, turn the relation into a true proposition. Let F(...,...) be a relation with two free places. Suppose that elementsofa set
M may be filled in at the first place, and elements of a set N at the secondplace. (For example Z(..., ...) from 3.12; at the first place we mayfill in things, at the second, instants.) Considerthe set
Th
P=IM,N\, consisting of all "x, y! with x e M, y € N (see 1.25). We can now conceivetherelation F(..., ...) as a predicate G(...) where elements of P maybefilled in at the free place and for which holds:
3.
st
3.20  The definition of a mapping can similarly be reformulated (see 1.12). A mapping f of A into B is fully known to meif I know all those members Ta, bl of [A, B] which have the property that b is the fimage of a. I may even
(1 ( ( ( ( ( ( ( ( (1 (1 (1 (1 (1 (1 (1 (1
regard f as the set of these pairs "a, b’ hence f Fx, 9). 1.24 contains a definition: “A set is said to be ordered by meansofa relation “ (fb, a! & DI} A Aave{l(ae Z) A (be Z) A (ce Z) A (fa, ble Y) A (fb, cle Y)] — (fa, cle Y)}.
are then mappings of A into B? Weshall designate the set of the mappings of A into Bby:
A B.
@ ( ( ( (
NN Ch. 3
PREDICATES, RELATIONS
61
Then (fe A= B) [f < IA, Bl
n Aallae 4) > (Vo (a, Def)]
A Aa,be{l(fa, bef) A (Fa, lef)] > (b=0)}]. Thefimage of a is then
fa=\o(la, Bef). 3.21  With the aid of the now available symbols we shall give the successive steps in the proof of the theorem stated in 1.19:
(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15) (16) (17)
Q=tx(Xc A). Ao [== A.
fe (Ao A 2).
Ax{(Xe Q) > [Vallxe Ao) A (f) = X)]}U=frl(xe Ao) A (x £f(9))1.
Az{(x € U) @ [(x € Ao) A (x Ef) ]}Val(x € Ao) A (f(x) = U)). (ue Ao) A (fu) = U). ue An.
fu) =U.
we U) [we Ao) A UES). (we U) > [we Ao) A UE U)].
we Ao) > {[we Ao) A WEU) ut U}. [we Ao) A u& U)] > (u & U).
we U) > (u U). (ug U) > (ue U). we U) > (ue U).
Explanation:
(1) (2) (3) (4) (5)
Definition of 2 as theset of the subsets. Assumption of the Ao and of the
f, which has to map Ao into 2 and even onto Q. Definition of U.
62
SUBJECT AND PREDICATE
Ch
Ch. 3
(6) The same with other words. (7) Uhas an inverseimage with respect to f, which is denoted by (8) u. (9)(10) Follow from (8). (11)
By substitution of u for x into (6).
(13)
q > [(q A p)
(12)
4) 5)
b
In fo Th
Replacingf(u) by U according to (10). pl] is always true.
Th
(14) Modus ponens from (9) and (12). (15) From (12) and (14), as [(p q) A (q > r)] > (p «> r) is always true. (16)
Th
Definition of¢.
(17) Acontradiction, because p «> —p is alwaysfalse.
de
The reader is invited to treat the theory of equivalence (see 1.22) himself by means of symbols.
T
d x
Functions
T
3.22 In 1.11 we considered the terminology for mappings and functions. Notations such as
1) 2)
thefunction f(x), thefunction y=f(x),
3) 4) 5)
thefunction thefunction thefunction
d 3 I f a
x?—3x+2, y=x?—3x+2, ax?+bx+c
are objectionable and will eventually drop out of use. The objections to them will be illustrated with the aid of some examples. 1) In the sentence “The function f(x) assumes the value f(a) for x =a” f(x) is evidently meant to be a function and f(a) a number. Why? The letter aisno better and no worse than theletter x.  2) If y=f(x) is a function, then y — f(x) =0 must, according to the rules of 
algebra, be the same function. Now consider for example f(x) =2x. Then be the “independent” and y the “dependent” variable?
speak of the function x and of the function 7. This is in itself already not too
nice. But, in addition, the terminology 3) leads directly to the terminology 5), which is unsatisfactory for other reasons.
f m
s
f
T
o
p T
y — 2x =0 must therefore be a function. But which function? Must x perforce 3) This notation is the least disputable. Only, it will not get us very far. Instead of the seconddegree polynomial, take for example x or 7. We must then
A

d
NE Ch. 3
FUNCTIONS
63
4) This is open to the same objection as 3).
5) Whatis meantis that this is a function of x. Why? Why nota function ofa,
b or c or all of them? In physics a function which is derived from “‘the function f(x)” by substituting for x “the function g(t)” is very frequently referred to as “the function f(t)”.
This is very confusing.
The correct terminology is illustrated by the following examples: The equation F(x) =3x26x—9
(for all x)
defines a function (mapping) f. The equation F(x) =3x26x9
(for all x with 0< x (rc P), we V) > ve P)
holds and the requirements 16 are met. 4.3  As a matter of fact, we already encountered axiomatic systems ofthis kind in earlier sections of this book. For instance, from the axioms for the equivalence concept, namely,
tho We dir re fo Ou oc
The fo th ab
Ho th
Am
Aa(a ~ a),
Aa,ve{[(b ~ a) A (c~a)] > (c~d)} we can deriveall assertions concerning the equivalence concept which weregard as being generally true.
Similarly, from
pl “
me “v
\a,b{(a # b) [(a0),
p
(AB {[A(B>O](4C)},
A t f C
3. ((—A)) >A
are axioms. Monus ponensis an operation whereby from two formulae
t
A and
( (
AB
the formula
B
is obtained. If I’is a set of formulae and D is a formula, then we understand by
(
Proof of D from I’ a chain of formulae with D as the last formula, in such a manner that each link of that chain is
(a) either an axiom (b) or taken from I’,
(
(c) or obtained by modus ponens from an appropriate pair of previous links in that chain. Provablefrom I’is said of a formula D if there is a proof of D from I’. To denote this we use the symbol
IED.
(
(
o
A
(
Itheorem is any formula which is provable from I’. Itheory is the set of all formulae provable from I’. If T'is empty, then we simply speak ofprovable, theorem, theory.
B
kD means: D is provable.
*
d
m
ze.
>. Ch. 4
DEDUCTION THEOREM
73
4.6  Deduction Theorem
If I, AtD. then also
IFAD. Proof*): The supposition states that there is a proof of D from I’, A. This proof is a sequence of formulae, X. We replace each link C of that chain 2’ by A —C. Wethus obtain a new chain ’. The last member of that chain will then be A —> D,i.e., that which, according to the assertion, should be provable
from I’. However x” is not yet a proof. We must insert some additional links.
Corresponding to the definition of “proof from I” we have to distinguish
the following casesfor the links of 2 (see 4.5, “proof of D from [””). (a)
Cisanaxiom,
(b)
Cistaken from I’, A, (b’) CistakenfromJ,
(ce)
(b”) Cis A, Chas been obtained from previous links of2 by means of modus ponens.
(a)
Before the link A > Cin” weinsert the axioms C,C (A= C)
(See axioms1.). A > C is obtained from this by modus ponens.
(b’)
Before A > Cin” weinsert the links C,C > (A > C),
of which thefirst is taken from J" and the second is an axiom. A — Cis again obtained by modus ponens. (b’’)
By substituting A for Cin axioms2, we obtain the axiom
(A > B) > {[4 > (B> 4)] > (A > AD}. By substituting A — A for B, we obtain from this axiom the axiom
(A > (4 > A) > {[A > ((4 > A) > 4] > (A > 4}.
(1)
*) Note that this word “proof” has a different meaning from the one used before. There it denoted a formal operation within the propositional language, but here it has the usual meaning.
aa gp 74
FORMAL LOGIC
Ch. 4
Before A > A we now insert in” this axiom, furthermore the axiom
A>(4— 4)
or w
(2)
obtained from axioms1 with C instead of A and
[A > ((A > A) > A)] > (4 > 4),
Hen
(3)
Hen
which is obtained from (1), (2) by modus ponens, furthermore the axiom
A((A > 4) > A)
(4)
also obtained from axioms 1 with A instead of C and with A — A instead of A. Then finally A — A is obtained from (3) and (4) by modus ponens. (c)
Let C have been obtained by modus ponens from
B BoC,
(5) (6)
Fr
The
A — B,
(59
A > (B > C),
(6)
which therefore occur before the A — C under consideration. Before this A — C insert: axiom 2
He cha
Fr (7)
and furthermore [A > (B> CY] > (4 > CO),
bel
is a
which occur before Cin A. In 4’ the corresponding formulae are
(A > B) > {[A > (B > CO] > (A > CO},
Ch.
He (8)
which follows from (5’) and (7) by modus ponens. Then
He
AC is justified at its place by modus ponensfrom (6’) and (8).
an
With all these insertions X” has become a proof of A — D from I’, as was our
intention.
4.7  A few more theorems. (6)
F(—B) > (BC).
Proof: Axioms 1 include
® > (A > ®)
wh
(7) Pr
He
DEDUCTION THEOREM
Ch. 4
75
or with a different notation ® > (HA). Hence also (with —C instead of A):
® >0). Hence ®
—(—C).
Ga) > C belongs to axioms 3.
(0),0) => ©, C is a proof of C from ®. Hence
® FC From the false ® follows everything.
Therefore (deduction theorem) F@C.
Hencethere is a proof for ® — C. Henceforth we may include @ — C in proof chains: if desired, the proof for @ — C can beinserted beforehand.
From B > @, Ba prooffor C is @,@C,C. Hence
Bo®, BEC.
Hence (deduction theorem) Bo@tFBOC,
and therefore (again applying the deduction theorem)
FB >®) > (B> CO), which was to be proved.
(7)
FB [C)>((B > C)1.
Proof: B, nC, B> CFC, C>®, ®.
Hence (deduction theorem)
—
76
FORMAL LOGIC
(again): (again):
Ch. 4
Ch.
BF(—C) > (a(B > O1, HB [(—C) > (—(B > CO].
4,9 the the
4.8  In the proof of 6 from 4.7 we encountered ® FC
Val
and
“va
F@ OC.
pai
Hence: if ® is in I’, then the J’theory consists of all formulae. Stated more
generally: if © is provable from J’, then every formula is provable from I.
A theory of this kind is not interesting. We prefer theories in which not every formula is provable. Wecall I consistent if ® is not provable from I’ or — what amountsto the same thing — if not every formula is provable from I’. If /’is consistent, then not both A and —A can be provable from I’, because: with A and A > @ also ® would beprovable, so that I’ would be inconsistent. If I’ is consistent and J”+ A, then I’, A is also consistent, because otherwise I’, A+ @, hence (deduction theorem) "+ A — ®, and therefore both A and
—A would be provable from I.
If is consistent, then either [', A or I’, 4A is consistent. Otherwise I’, A + @
In Pr
Ax
va
(2)
as well as I’, =A + @, hence J’ SA as well as [’+ —(—A), and therefore ITtA. A consistent I’ can be extended to a maximally consistent I’, i.e., to a set of formulae which, with any further extension, becomesinconsistent.
To this end, we shall conceive all the formulae as being arranged in a row in accordance with some appropriate ordening principle*). We seek the first formula which is not yet in J’ and which can be added to [’ without disturbing the consistency. We add this formula to I’. Continuing in this way, we obtain a set A of formulae which comprises I’ and to which, on penalty of inconsistency, no further formula can be added. From the foregoing follows:
IfA is maximally consistent, then precisely one memberof the pair A, 4A will belong to A, and everything that is provablefrom A will belong to A. *) This is directly possible if the language in question has only a countable numberof primi
tive formulae, and therefore also only a countable number of formulae. The reader should
bear this case in mind. In general, the desired object can be achieved with what is known as a
wellordering of the formulae. This cannot be dealt with here.
Bu
(3
va ha
In
Fo s th
in
a
7
Ch. 4
VALUATION OF PROPOSITIONAL LANGUAGE
77
Valuation of the Propositional Language 4.9  In the foregoing it was already announced that we were going to interpret the propositional language. This is what we shall do now. Weshall “‘valuate”’ the formulae. We speak of a Valuation of the propositional language if to each formula precisely one of the
“values” 0 (false) or 1 (true) is so assigned that @ has the value 0 and for each pair of formulae B, C the value of B > C is determined by the table of values
=
—
Of
Of,
_

0 ©
0
—
B Cc B>C
In a valuation all axioms (see 4.5) automatically have the value 1. Proof: Axioms(1) can have the value 0 only if 4 — C has the value 0 and C hasthe value 1; but these two are contradictory.
(2) can havethe value 0 only if the successive values are A—B
1
A7>~(BC) AC
1 0
(A>(B>C))(4>C) 0 A
C B BoC A
1
0 1 (because of Ist and 5th line) 0 (because of 6th and 7th line) 0 (because of 3rd and 8thline).
But then there is a contradiction between the Sth and 9th line. (3) can have the value 0 only if A has the value 0 and (A > ®) — ® has the value 1; but then A — ® on the one handhasthe value 0, and on the other hand hasthe value 1.
In a valuation all theorems have the value 1.
For let D be a theorem and C be a link of a proof for D. Let us furthermore
suppose thatall the links preceding C have the value 1. We need only show
that this is also true for C. If C is an axiom, then C hasthe value 1 (as proved
in the foregoing); if C has been obtained by means of modus ponensfrom earlier
mg
78
FORMAL LOGIC
Ch. 4
Ch.
B, B > C having the value 1, then the table of values shows that C must also have the value 1. In accordance with 2.4 we call a set of formulae I"
“di ing
satisfiable if there is a valuation for which all formulae from I" obtain the value 1; a valuation ofthis kindis called a satisfaction of I’. If there isno sucha valuation, then I’ is said to be unsatisfiable or false.
A formula A is said to be I’true if, for every valuation, which is a satisfaction of I", A gets the value 1.
A is called
true (pure and simple) if A gets the value 1 for every valuation. If, for some valuations, A gets the value 0, then A is therefore not true and is accordingly
said to be
disputable.
“True” and “indisputable” are therefore the samething. For each valuation 4 and —A havedifferent values, because, according to the
tru
4.
ea Th
pos
a) b) c)
Fo
th th mo mu
He
table of values for 4 — B, the value 1 would follow for ® from equalvaluesfor
Aand A — ®, and this is contradictory to the definition of valuation. Hence it follows that: If A is true, then —A is false. If A is false, then —A is true.
Fo
4.10  Every language has a syntactic aspect and a semantic aspect. If we translate the French “il pleut” by “he is raining” or by “it are raining”, we are committing a syntactic error. If we translate it by “‘it is snowing”, we are com
ha T
with the formal aspects of a language, whereas semantics is concerned with meanings. “Verb”, “endings’’, “sentence”, “punctuation mark” are syntactic concepts. “Synonym’’, “metaphor”’, “‘a carefully worded statement”, “true” are semantic concepts. In 4.58 we were concerned with syntactic matters. In 4.9 we were in thefield of semantics. By evaluating formulae we give them a meaning; the values 0,1 have nothing to do with the vocabulary of the propositional language. “‘Formula”, “proof”, “provable”, “consistent” did not, it is true, belong to the propositional language either but had been defined onthebasis of that language; they correspond to concepts such as “verb”, “endings”, > “sentence”, “‘punctuation mark” with regard to ordinary language. “Satisfiable”, “false”, “true”,
P c i
mitting a semantic error (thoughit is syntactically correct). Syntax is concerned
T
v ®
N
I
w
v
(
Ch. 4
EVERY SATISFIABLE J' IS CONSISTENT
79
““disputable” correspond to “synonym”’, etc. To be on the safe side we accordingly say semantically true or false if these concepts are meant; the syntactic truth aspect is provability.
4.11  Let I’ bea satisfiable set of formulae. Let D be provable from J’. For each satisfaction of I’, D gets the value 1.
This can be shown as follows. For a link C of the proof of D there are three
possibilities, namely, C is
a) b) )
an axiom, taken from 7’, obtained by modus ponens from two preceding links of the proof: B,BoC.
For case a) we know from 4.9 that C hasthe value 1, and in case b) C will have
the value 1 by definition of valuation andsatisfaction of I’. If we already know that all the links of the proof preceding C have the value 1, then we know this
more particularly for Band B — C. The table of values for B — C shows that C must then also have the value 1. Each link is therefore a formula with the value 1. Hence D, as the last link, must also have the value 1.
4.12  Every satisfiable I’ is consistent. For, according to 4.11, everything provable from I" has the value 1. Now ® has the value 0 andis therefore not provable from I’. Hence J’ is consistent. Thereverse of this theorem is also true: Every consistent I is satisfiable.
Proof: According to 4.8 we determine for J’ a maximally consistent A which comprises J. Furthermore, according to 4.8, of each pair A, —A precisely one
isin A.
To all members of A weassign the value 1, to all nonmembers we assign the value 0. We shall verify that this is a valuation. ® hasthe value 0, for because of the consistency of 4, ® does not belong to 4.
Now check the table values (4.9): In the case of the first two lines, —Bis in A; since theorem 6 (from 4.7)is in A,
we have A  (B > C); hence B > C is in A, so B + C does indeed have the value 1. In the case of the third line, B is in A and —C is in A; since theorem 7
(from 4.7) is in J, we have A + —(B > C); hence —(B > C)in A, andsoB>C
=
=
Ch. 4
Ch.
is indeed not in 4; hence B — C has the value 0. In the case of the second and the fourth line, C is in 4; since C > (B >C) occurs amongthe axioms 1, we have 4 hk B —C, and therefore B —C is in 4; hence B > C hasreally the value 1.
Pri
80
FORMAL LOGIC
Thus J" is satisfied, for 4 contains I’ and every formula in A gets the value 1. 4.13  By contraposition we obtain from theprevious theorem: Every unsatisfiable (false) I" is inconsistent. Wenowarrive at the mainresult, the
THEOREM OF COMPLETENESS: In propositional language every true formula is provable.
Proof: Let A be true. Then —A is false (see end of 4.9) and therefore unsatisfiable and — byvirtue of the previous theorem — inconsistent. Hence
AA ® and therefore k (HA) > ®. Wenow successively write down:
a proof for —(—A), axiom 3, A. This is a proof for A. Hence A is provable. This theorem shows that our set of axioms is adequate for proving any true formula.
This explains why it is called the “completeness theorem”’.
Axioms of the SubjectPredicate Language 4.14  The propositional language so far considered in this chapter is a rather
poor language. In the third chapter we were introduced to much more powerful linguistic resources. These will now be subjected to axiomatization. A Subjectpredicate language
is composed of
Subjects among which an infinite number of variable and (perhaps also, but not necessarily) constant subjects,
pre for
Co Th
Fo
a) th
b)
€) d)
a)
In
Fr D
th a)
b) B) c
b
a o
F c
A A W
A
Ch. 4
SUBJECTPREDICATE LANGUAGE
81
Primitive*) predicates, of degree 0, 1, 2, ..., which must include the Othdegree predicate © (the Othdegree predicates are what we formerly called primitive formulae — propositions), Connectives: >,(,), /. The subjects and predicates together are called the vocabulary of the language. Formulae: a) If fis a primitive predicate of degree n>0 and if x1, ..., xn are subjects, thenf(x1, ..., Xn) is a formula. Every primitive predicate of degree 0 is a formula. b) If Aand Bare formulae, then (A — B)is also a formula.
c) If Aisa formula and x is a variable subject, then AA is a formula. d) All formulae are obtainable by repeated application of the principles a), b), c). Instead of “variable subject’ we may refer more briefly to “variable”. Free and bound variables. Depending on the way — a), b), c) — in which a formula originates, we define the “free” or “bound” occurrence of a variable. a) Inf(x1, ..., xn) the x; (if it is a variable subject) occursfreely.
b)
Any free or bound occurrence of the variable x in A (and similarly in
B) is also called free or bound occurrence, respectively, in A — B. c) Any free or bound occurrence of a variable y in A is also called free or bound occurrence, respectively, in AA if the variable y differs from the vari
able x; any occurrence of the variable x in (A is called bound. (The presence
of the x as subscript to A is not regarded as “‘occurrence”’. For any occurrence ofa variable in an arbitrary formula it is, by meansof these
conventions, determined whetherit must be regarded as free or as bound. A formula with n different free variables is called a formula of degree n. A formula of degree 0 is called a proposition. Wewrite —A instead of A > ©. Axioms: \f A, B, C are formulae, then the axiomsare: 1, 2, 3 (see 4.5),
4.
A(A > B) > (A > AzB), if the variable x does not occur freely
in A; 5. AzA— A’, if A’ is obtained from A by substitution of a subject y for x wherever x occurs, while it is understood that x does not occur
freely in any subformula of A of the form A yC. *) See footnote p. 71.
82
FORMAL LOGIC
Ch. 4
Besides the
M
Modus ponens we shall here accept another meansto establish a proof, namely.
Binding: an operation whereby a formula B is replaced by a formula (zB. Proof ofDfrom I is defined as in 4.5 with the addition (d) or obtained by binding with Ax from a previous formula in that chain, while it is understood that such an x does not freely occur in any formula of I. Provable from I’ is then defined as in 4.5. The same applies to the concepts: F=theorem, I’theory, provable, theorem, theory. The proofof the Deduction Theorem now requires a supplement, corresponding to the addition (d) to the definition of proof:
If
B, AaB
are links in a proof from I’, A, then we haveto insert extra links in this proof between
AB, AaB. As such we shall take
Az(4 > B), which is obtained from A — B by binding, and the axiom
Az(4 > B) > (A > cB). (Note that x was not allowed to occurfreely in J’, A.) From these two formulae
we then obtain A — /xB by modus ponens. We shall use the abbreviation VazA
Ch
for
—_Az(—4).
The proofs of 6 and 7, as considered in 4.7 remain unchanged. One easily proves
(8)
FAz(—4) >(—Vz4).
(9)
t+—Acd > Va(— A).
Consistency is defined as in 4.8 . 4.15  Formulae without free variables are called propositions, since with these
formulae we can do everything that was done with propositions in 4,513.
ma
si th O as V t st o
v
g
a
m
f v
(
W
i
T 4 t T a
b
A
Ch. 4
SUBJECTPREDICATE LANGUAGE
83
More particularly, we can extend every consistent set of propositions into a maximally consistent set of propositions, to which belongs exactly one proposition out of any pair of propositions A, 4A. We are going to do more than this, however. Our vocabulary need not contain a single constant subject. The predicates are, as it were, floating in the air; they do not refer to anything. It may occur that VA is provable and there is nevertheless no “example” of “there is an x so that A” in the vocabulary, i.e., there is no fixed subject which, on being substituted for x, turns A into a provable formula. Now we shall enlarge on one hand the vocabulary by the addition of constant subjects and on the other hand a given consistent set Io of propositions by adding fresh propositions, with the aim that the set of propositions 2 eventually obtained shall be maximally consistent and that for every proposition \/zA from 2 a constant subject a can be found in the vocabulary, so that the formula A’, which is obtained from A by substitution (according to axioms5) ofa forx, likewise belongs to Q. Weshall construct such a 2 stepwise. We shall keep a supply of things in readiness which we shall gradually add to the vocabularly as constant subjects. To start with, we shall extend the given 1’) to a maximally consistent Ao (see
4,8). We can conceive the propositions from Áo arranged in a row and we take
the first proposition of type VxA (where x is a variable and A is a formula). Then we take a thing a from the available supply, add it to the vocabulary as a constant subject, and add to Ap the proposition A’ that is obtained from A by substitution of a for x (in all places where x is free). We can show that Ao will then remain consistent: Suppose that Ao, A’ is inconsistent, i.e.,
Ao, A’ t ®, hence (deduction theorem) Ao F A’.
In the proof X' for A’ from Ao we shall replace a, wherever it occurs, by a variable y which does not occur in the proof (here the infinity of the number of variables is used). We assert that the new chain X* is again a proof. To see whythis is so, consider each link of the proof C that is changed into a formula C* as a result of the substitution of y for a. Having regard to the list of means
to execute a proof there are four possibilities: a)
Cis an axiom,
ee.
84
FORMAL LOGIC
b)
Chas been taken from Ao,
c)
Chas been obtained from previous formulae of X' by modus ponens,
d)
Ch. 4
Chas the form (Band has been obtained from previous B by binding.
C case b) cannot occur because a has played nopartat all in the constitution
of Ap. In the case a)it is evident that C* is again an axiom,andalso in the cases c) and d) wesee at once that C* is obtained by the same procedure from previous formulae of X* as that whereby C was obtained from corresponding formulae of2’. Hence also Ao F =A",
where A” is derived from A’ by substitution ofy fora, i.e., from A by substitution of y for x in all places where x occursfreely. By binding of y we then also obtain
Mo F Ay(A") and by virtue of axioms(5)
Ap t A; whence weobtain by binding of x Ao
k
Az(—4),
or A 0 F (Vz A) => ® 7
which, in combination with the given
Áo + Vad would lead to the inconsistency of Ao. This contradiction is a consequence of the assumption that Ap, A’ was incon
sistent. Hence it follows that Ao remains consistent after the addition of A’.
A’ is now inserted behind \/zA, in the sequence in which the propositions of
Ao were placed. We shall now seek the second proposition of the sametype,
say VyB, take a thing b from the available supply, add it to the vocabulary as a constant subject, and add to Ao, A’ the proposition B’ which is obtained from
B by the substitution of b for y. In this operation the consistency is again pre
served. Proceeding in this manner, this process yields a vocabulary and a
consistent set 1 with the property: for each formula of the type V‚K in [ia
constant subject k can be found in the vocabulary, so that K’ (obtained by
substitution of k for every free occurrence of z in K) belongs to 15. I; is again extended into a maximally consistent set 41 of propositions; Ai is
nn
Ch. 4
SUBJECTPREDICATE LANGUAGE
85
treated as Ap; this leads to a I’, with properties analogous to those of J%,etc. The ultimate result is an extension of the vocabulary and an extension of
I’ to Q with the desired properties: Q is a maximally consistent set of propositions; for every proposition VA from @ there is an “example” a of “there is an x so
that A”. This property of 2 can also be formulated asfollows: Q is a maximally consistent set of propositions; if all the propositions obtained from the formula A by the substitution of any
particular constant subject for x belong to 2, then A\zA belongsto 2.
To prove this, suppose that this were not so. Then, because of the maximal
consistency of 2, \zA would belong to Q, and so would therefore \V 2(—A).
There would then be a constant subject a so that —A’ belongs to 2 (where A’ is
derived from A by substitution of a for x). Because of the consistency of 2
however, A’ would then certainly not belong to 2, and this would be contrary to the supposition. — The converse can be provedin similar fashion. 4.16  The problem of the interpretation of the subjectpredicate language introduced axiomatically in 4.14 is solved in principle with the construction
of Q.
What do we mean by such an interpretation? We want the “subjects”, “predi
cates’’, “—”’, “A” of the given subjectpredicate language to correspond to something that is in agreement with our ideas thereof, as envisaged in Chapter 3.
The subjects present no problems. But what did we mean in Chapter 3 by a predicate? A predicate such as “... is a man” was completely given if, with regard to anything filled in at the place of the dots, it was fixed whether it would makethe proposition, thus obtained, true or false. The predicate “... smaller than ...”’ was determined if it was settled for which pairs it would becometrue and for which pairs it would becomefalse. An nthdegree predicate F was determinedif, for every ntuple of things a1,..., dn, it was certain whether
F would thereby be madetrue orfalse. Annthdegree predicate was therefore a mapping ofthe set of the mtuples of things into the set consisting of the judge
ments “true” and “false” (also called 1 and 0 respectively). The said set of things will be noted by J; the notation J” will be used for the set [J,..., I of the n
tuples from J. Hence we shall speak of an interpretation of the given subjectpredicate language if
with every constant subject a is associated an elementofJ, called Ra, with every nthdegree predicate F is associated a mapping,called RF, of
I” into theset (0,1).
Cn
86
FORMAL LOGIC
Ch. 4
Ch
In this way, toa formula F(x1,..., xn) corresponds an expression (RF)(x1,..., Xn) which, on substitution of elements of J for x1,..., Xn, assumes the value 0 or 1, A substitution of elements of J for the subjects of the subjectpredicate language will be called a cover w. Hence w is a mapping of the set of subjects into the set I. The interpretation R should be chosen fixed, but we shall, of course, consider all covers that agree with the interpretation (i.e. the covers which map every constant subject according to the interpretation R). We must, however, require something moreof an interpretation. As before, we
b) to c)
shall require that (the Othdegree predicate) © always gets the value 0; furthermore, that for “—” the table of values in 4.9 shall apply and shall do so for
every cover, that is not in conflict with R. But now there is an additional requirement with regard to “A”. In Chapter 3 we meant by AxA that A, for every substitution for x, is true. To conform to this we must now demand: If the interpreation of A has the value 1 for every cover, then the interpretation of AxA will also have the value 1. If these requirements are fulfilled, we shall call the interpretation “valuable”. This analysis can now be summarized in a definition.
co
In no
On Th fr Bu Le pr
sa
ov
un
A
T
4.17  Consider a given subjectpredicate language. Let J be a set, J” the set [J, ..., J] of the ntuples from J, and 7° a set consisting of one element io. A cover is a mapping w which associates with every subject an element ofJ. An interpretation over I is a mapping R with the following property: to every constant subject corresponds an elementof J,
to every primitive predicate of the nth degree corresponds a mapping of J” into the set (0, 1). Let R be an interpretation. The cover w agrees with the interpretation R if w and R are the same for
every constant subject. The only covers w that will be allowed are those which agree with R. R is called valuable if with every cover w, R causes a valuation; i.e. to each
formula C is assigned a value C = 0 or 1 (dependent on w) so that the follow
ing requirementsare fulfilled: a) If Cis of the formf(x1, ..., xn) (fbeing an nthdegree primitive predicate, and X1, ..., Xn being subjects), then:
[Cla = (Rf) o(x1), ...,@(xn)") for
= (Rf) (io)
for
n>0,
n=O,
T
tr tr
“
4
T
T W t o t
p
4
4
E
r
E
Ch. 4
SUBJECTPREDICATE LANGUAGE
87
b) If Cis of the form (A > B), then Cq will depend on [Alo and Blw according
to the table of values in 4.9. ce) If C is of the form A74, then [Clo = if, and only if, A= 1 for every cover w’ which only differs from w in z.
In view of the origin of the formulae (see 4.14) such a valuation is possible in not more than one way(if the interpretation R is given). On verifying the cases a), b), c) we see: The value of C in w depends only upon whatw assignsto the variables that are
free in C. More particularly, the value of a proposition does not depend on w.
But it does of course depend on the interpretation R. Let I" be a set of propositions. A valuable interpretation overJ, in which every
proposition from gets the value1, is called a
satisfaction of I over I. If has such satisfaction, then I’is said to be satisfiable over I; otherwiseit is said to be unsatisfiable or false over I. A proposition A is said to be I’true over Tif, for each satisfaction of I over J,it gets the value 1, Itrue if it is Itrue over J for every J, true over Tif, for every valuable interpretation overJ, it gets the value1, true if, for every J, it is true overJ. “‘Disputable’’, etc. can be defined similarly. 4.18  Let I’ be a set of propositions satisfiable over J. Let D be provable from
I’. Foreach satisfaction of I’ over I, D gets the value1.
This is analogous to the theorem in 4.11 and it is proved in the same fashion. We must only take into account the possibility that the link C has been obtained from a precedinglink by binding, i.e. it has the form Ax B. If it is known of the links preceding C that they have the value 1 for the valuable interpre
tation R and every cover w that agrees with R, then this will be known more particularly for B. According to the definition of a valuable interpretation (see 4.17e), Ax B will then also have the value 1 for every cover.
4.19  Henceit follows (see also 4.12): Every set of propositions that is satisfiable over a suitable I is consistent. The reverse of this is also valid: Every consistent set ofpropositionsis satisfiable overa suitable I.
Ch. 4
Ch.
Proof: We extend the set of the constant subjects and the given consistent
so sub of
88
FORMAL LOGIC
system of propositions I’ according to 4.15. We obtain a maximally consistent system of propositions 2 which comprises J’ and in which, for every proposition VA, a constant subject a of the enriched language is to be found,so that the
formula A’, derived from A by substitution of a for x, also belongs to Q.
For J we shall take the set of the constant subjects of the enriched language. The
interpretation R is defined:
for every constant subject a we have Ra=a, for every primitive predicate f of degree n and every ‘a1, ... ay! from I” we have
(Rf) (fa, ...5 An') = or 0, depending on whether from/(%1,..., xn), after substitution of a;forx;(i = 1,...,7) >
a proposition arises which does or does not belong to 2. Ris valuable and C» is defined for every formula C, and for every cover w agreeing with R, by the definition Clo=1
or 0,
depending on whether, on substitution of w(x) for every variable x thatis free
in C, from C a proposition is obtained which does or does not belong to 2. To
see this, we must only verify that the requirementsa), b), c) in 4.17 are met. a) Let C be of the form f(x1, ..., xn) (see 4.17), n >0. According to definition,
(Rf)(@(xa), ..., @(Xn)") is calculated by substituting w(x;) for x; in f(x1, ..., Xn) and checking whether or not the result belongs to 2. But accordingto the definition this is exactly the same as the calculation of f(x1,..., Xn)lo. For n = 0 this is also evident. ® = 0 follows from the consistency of Q (i.e., ® not in Q@). b) See the proof in 4.11. c) Let C be of the form AA (see 4.17). Let [Clo = 1. Let w’ be a cover which,if at all, only differs from w in z. Let C’ be whatis obtained from C on substitution of w(x) for every free x; let A’ be whatis obtained from A on substitution of w’(x) for every free x different from z; and let A’’ be what is obtained from A on substitution of w'(x) for every free x. A” is obtained from A’ on substitution of w'(z) for z. Now C’ is of the form AzA', and this belongs to @ as C» = 1. From axioms 5 (see 4.14) we infer that, on substitution of w'(z) for z, something is obtained that belongs to 2.
Therefore A’’ belongs to Q, hence  Alo’ = 1. Let Cw@ = 0; let C’ and A’ be defined as in the foregoing. Then C’, and therefore
also \zA’, does not belong to @. Hence — A24’ belongs to 2, and therefore
co A
4.
Co is
Th
L.
4. Q
Su ab
Th
th on al
If fi
In nu Th co
nn nn
Ch.4
SUBJECTPREDICATE LANGUAGE
89
so does \/z(—4’). So there is (see first paragraph of this proof) a constant subject a so that for the formula A’’, which is obtained from A’ on substitution
of a for z, the following holds: 4A’ in Q, hence A” not in Q. Let w’ be the cover with w’ (z) = a and equal to w everywhere else. Then 4’ is obtained from A by substitution of w'(x) for every free x. Hence  Al = 0.
4.20  As in 4.13, there follows from the last theorem the Completeness Theorem: in the subjectpredicate language every proposition that is true is also provable. This theorem is due to K. G6DEL (1930). The proof given here is essentially L. HENKIN’s (1949).
4.21  The theorem arrived at in 4.19 can be strengtheneda little. Consider how Q was obtained (see 4.15). Suppose that the vocabulary is countably infinite. Hence there is only a countable numberof formulae; so Ao has also only a countable number of formulae. Thusto arrive at I’, we only have to add a countable numberof formulae, and the vocabulary, too, will at most be enriched by a countable number. And so
on. Finally, the enriched vocabularyis still found to be countable. Hence J is also countable. So:
If the vocabulary is countable, then every consistent set of propositions is satis
fiable over a countable I.
In general it holds that it is enough for J to be a set with the same cardinal number as the given vocabulary. This is, for reasons that will be explained in the next chapter, a paradoxical
conclusion.
Ch 5LANGUAGE AND METALANGUAGE
But
poi
set
5.1  Althoughitis richer than the propositional language, the subjectpredicate languageis still insufficient. Even ordinary speech not only has predicates of subjects but also predicates of predicates. Take the sentences: This roof is red. Redis a colour. Colour is an optical phenomenon. If we accept that “this roof” is a subject, then “red” is to be regarded as the predicate. But in the next line this predicate appears as a subject of a proposition with “colour” (or “.. is a colour”) as the predicate (of the second stage). And then in the third line this predicate (of the second stage) is the subject of a proposition formed with a new predicate (of the third stage).
We must be cautious with those predicates of various stages. In ordinary
speech, predicates of the second stage will not directly apply to subjects — “‘this roof is a colour” is not actually felt to be false, but meaningless. At the present time, there is a preference to confine oneself to predicates of the first stage. To do this, “red” in the second line must be interpreted differently
from “red” in the first line. The first “red” (or “.… is red’’) will then really be a
predicate, whereas the second has to be a genuine subject. But in that case a connection will have to be established between the two. Now in logistic language we already know such a connection between predicates and subjects. Associated with the predicate F (of first degree) is the set of all x with true F(x), Tr F(x),
which werepeatedly interpreted as a subject. Hence it stands to reason also to regard sets as subjects. The second “red” is then to be interpreted asthe set of
all red things. Similarly, “green” is to be the set of all green things. “Colour”
in the third line is therefore the set of colours, i.e., a set of sets. If we want to consider sets as subjects, then we need notonly a predicate that should be interpreted as “‘...is a set”, but also a relation between subjects that should be interpreted as ...isan elementof ...,
us
th set We
« ki
be
5.
ma us
ab
We na
or
Le
is A
1) 2) 3)
4) 5) 6)
or
In
SI
Ch. 5
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91
But this is still not enough. Just as it is necessary to lay down the properties of points, lines, etc. as the basis of geometry, we here need axiomsrelating to sets. These can be formulated in various ways. But we must ensure that the usual processes of set theory whereby new sets are created, are reflected in the present theory — e.g., that there is an empty set, that the union ofa set of sets is again a set, that for every set there exists the set of subsets, etc.
Wethusget a system I’ of propositions in which the predicates “... is a set” and “*... is element of ...” occur; furthermore, an equality relation, ...=.... This kind ofthing is called a system of axiomsfor set theory. The set theory will then be the I’theory, i.e., the whole of the propositions that can be proved from I’. 5.2  But this, too, is still not enough for the purpose of mathematics. Mathematics is based on the natural numbers, which can be introduced by their usual representation. We shal! attach to them the predicate ... is a natural number,
abbreviated to Z. Hence the following holds:
Z(1), Z(2), Z(3), ... We now have to axiomatize the known connection between the individual natural numbers. This can be done by meansofthe relation ... is successor of ..., or in short
Vlas zee) Le.,
va, 1), VG, 2), V4, 3), …
is true. A system of axioms for the natural numbers was given by PEANO. 1)
Z(I.
2) Ax{Z(x) > [Vy(ZO) A Voy, x))]}3) —Ve VL, x).
4) Az, v{[Z) A 20) A ZO > [W, 7) A VAI) > =D 5) Az, v{[Z@) A ZX!) A ZO) > VW, 2) A VO, x’) = x= 2) 6) (FW) A Ar{Z(y) > Ael(Z(») A F(x) A VD, x)) > FO)I)) — AzlZ(z) > F(z)].
In particular, the last statementis important; it is the
92
LANGUAGE AND METALANGUAGE
Chis
Ch
Principle ofcomplete induction: Let F be a property of natural numbers. Suppose
Fo
that 1 possesses the property F and that whenever a natural number possesses
the property F, its successor also possesses that property. Then all the natural numbers will possess the property F.
With theaid ofthis principle addition and multiplication of natural numbers can be defined. x + 1 is the successorto x,
x+(y+ 1) is the successor to x + y,
ly=y,
(x+]l)y=xp+y. 5.3  As axioms for sets and for natural numbers we chose statements which are regarded as being true in conventional mathematics. The system I" which they constitute is satisfied by the sets and the natural numbers in the normalsense.
One might therefore assume that I’ is consistent. In the constitution of J” we
may confine ourselves to a countable number of subjects. According to 4.21,
therefore, J’ would be alreadysatisfiable over a countable J.
This is very strange — for the I’theory comprises things which we normally interpret as noncountable sets. For example, the set of the subsets of the set of the natural numbersis certainly not countable (see (1.19). This conclusion is known as the paradox of LOWENHEIM and SKOLEM. Howare weto account for this apparent contradiction?
Such a noncountable set M must be derivable from I" with the aid of the re
sources of the subjectpredicate language. But, at the same time, no oneone mapping f of M into the set of natural numbers must be derivable from J” with these resources. J’ is satisfiable over a countable J — hence, in a valuable interpretation, M will be interpreted by a countableset. What emerges from this? The oneone mapping of M, whose existence we can establish by speaking about the subjectpredicate language,is not derivable from I’ by meansof the subjectpredicate languageitself.
5.4  This is the most suitable momentto realize that, in talking about the subjectpredicate language, we often have gone far beyond the scope ofthat language.
The subjectpredicate language, as defined here, can be manipulated by a
machine. In accordance with the extremely simple rules that we have laid down, a machine can successively write down the formulae. Also, a machine can be made to apply the rules of inference systematically and thus to prove one theorem after another. Will every Itrue proposition eventually emerge from the machine?
th
ve a n W
is
n in
p
t s r
f
o H
a
( i p c
s I
o p w t p
5 u
H s e
n
e Ch. 5
LANGUAGE AND METALANGUAGE
93
For every series of signs that we feed into it, the machine could check whether that series meets the requirements for a formula. But can the machine also verify the truth or falseness of a given proposition otherwise than, in constantly applying the rules of inference, waiting to see whether that proposition (or the negation thereof) will perhaps emerge? We have shownthat every I’true proposition of the subjectpredicate language is provable from I (at least if [’is consistent). But here we made use of resources not available to the machine. We accounted for the use of these resources, not in the subjectpredicate language, but in English. In order to demonstrate the provability of a formula we reasoned as follows on several occasions: suppose that there exists no proof for A; then this or that contradiction will arise. We showed the provability by indirect means (reductio ad absurdum). But by
refuting “there is no proof for A”, however, we do not yet produce a proof
for A. But to the machine the real production of a proof for A would be the only method to demonstrate the provability of A. Hence it would not be inconceivable that propositions which, in talking about the subjectpredicate language, we recognize to be true, cannot be proved (and not refuted either) by the machine manipulating in this language. According to a famous theorem of K. G6DEL (1931) there are indeed such indecidable propositions in every language in which the sequence of the natural numbers can be formed. This theorem is known as the incompleteness theorem. Later we shall outline the reasoning that leads to this theorem. It should be noted that the difficulties discussed here do not occur if we confine ourselves to the propositional language. The completeness theorem for the propositional language can be proved with very elementary means, or in other words: the machineitself can prove the completeness theorem for the propositional language with its own resources. In general the machine cannot accomplish the same task with regard to a subjectpredicate language. 5.5  During the proof of the completeness of a subjectpredicate language, we
used English as the medium of communication for talking about that language. How can we make the machine, which only has the conventionally agreed subjectpredicate language at its disposal, talk about this language, so thatit for example can establish the provability of something? We shall consider the answerto this question later on. Meanwhile I want to point out a pitfall that threatens us if we discuss the phenomena of a language in that same language. There is the famous paradox of the Cretan who asserts that all Cretans are liars. We are asked to decide whether, in saying this, he is lying ortelling the truth. Formulated in this way the paradox is not watertight. It is better to ask: If someone says “I am
94
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Ch. 5
lying”, is he telling a lie or speaking the truth? If he is lying, thenheis lying that he is lying and therefore speaking the truth. And if he is speaking the truth,
thenit is true that he is lying, and so heis lying.
The cause of the difficulties in which the liar involves us and himself can be traced to the fact that we are speaking about linguistic phenomena in the same language as that in which those phenomena occur. If a person whose native language is English uses nothing but French for speaking about phenomena of the English language, he cannot be trapped in paradoxes of this kind. Such a
person should not say ““dog’ is a noun”, but “‘‘dog’ est un substantif”. Not
“the assertion ‘Abel killed Cain’ is false’ but “la proposition ‘Abel killed Cain’ est fausse’’. He can never say “I am lying’’, for this relates to the falseness of his direct assertions made in English; instead, the assertion “I am lying” as an assertion about assertions is made by him in French, namely, in the form “‘tout ce que je dis en anglais est faux”’, and this assertion is now notaffected by the valuation made in it, since it is stated in French (and not in English). Of course, the speaker may also wish to talk about phenomena occurring in the French language. For doing this he could, for example, use German, i.e: “tout ce que je dis en anglais est faux” — “alles, was ich auf französisch sage, ist falsch”. These two assertions are formally not contradictory to each other (whether they are true is a different matter). A preciser formulation of the paradox of the liar is provided by the following sentence:
What is written in line 23 on page 94 of this book is false.
The readeris invited to analyse this for himself,
The following definition is known as J. RICHARD’s paradox: The smallest number, that cannot be defined in the English language with less than one hundred characters. If you count the used characters, you will find that this number has been defined
byalittle less than one hundred characters.
Analyse this too. Whathas been explained in the foregoing is known underthe slogan “language and metalanguage”. We talk about a given language in a metalanguage.
5.6  How can we nevertheless make the machine talk about the subjectpredicate language in the subjectpredicate language?
From now onwards wewill deal with a (countable) subjectpredicate language
which contains the concepts of set and equality, and the natural numbers.
The system J" consists therefore of the axioms for these concepts. From now on we shall omit the notation I’ in “‘J™true’’, “proof from J”’, etc. We devise a code for speaking about the subjectpredicate language which
C
m
j
s p
T
w
W t t t
t F i s
Ch. 5
LANGUAGE AND METALANGUAGE
95
makes use of the subjectpredicate languageitself. To the products of the subjectpredicate language we assign code numbers. To do this we utilize certain simple properties of the natural numbers.
p is called a prime numberif p #1 and p has no divisors other than 1 andp. The prime numbers 2,3, 55 Js 1ls 135 17, will be consecutively numbered. Pi, P2, P3, Pa, P5, Pe, Pv, +++
We now assign code numbers as follows: to the numerals 1, 2, 3,..., the code numbers 21, 22, 23, ... to the other constant subjects the code numbers 31, 32, 33, … to the variables x1, x2, x3, ... the code numbers 51, 52, 53, ...
to the primitive nthdegree predicates the code numbers 7”n, 7°», 7?n, ... Furthermore, if fis a primitive nthdegree predicate with code number [f] and z1, ..., zy are subjects with respective code numbers [zi], … , [zn], then we assign to
Ff (Z1y «++ Zn) the code number 11011371
(2,1,
?2
[291
tz)
FP *,
if A and B are formulae (with code numbers [A] and [B)]), then we assign to (4 — B) the code number 17/41  1918);
if A is a formula (code number[A]) and x is a variable (code number[x]), then we assign to \zA the code number23/4] 29I#], Thus, to every formula (at least) one code numberhas been attached. In order to ascertain whether a given number occurs as a code numberand, if so, to which formula it relates, we need merely resolve the number into prime factors and systematically “translate back’. A machine is perfectly capable of doing this
sort of thing. If, for example, there appears a 23 amongthe primefactors, then
the number must have the form 234 +295, if it is to be the code numberof a formula, and the formula must be of the form Ax4, where A and x have respectively codenumbers a and b. These a and b are processed in a similar way, and
so on, down to the subjects and the primitive predicates. “x occurs as a free
variable in 4” is, on coding, a numerical relation between [x] and [A] which the
machine is able to unravel. The property of a formula of being an axiom is
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LANGUAGE AND METALANGUAGE
Ch. 5
essentially a numerical property of its code number, which can be verified by the machine. The fact that A1, Ag, eee Am
forms an inference chain is expressed, on coding, by a numericalrelation between the [Az], [Ae], ..., [4m], which can likewise be verified by the machine. To such a series of formulae X' we assign the code number
jz pg lfal … oo, Lm) That X' is a proof for C (from J’) is expressed as a numerical relation between [X'] and [C], which weshall denote by
C
c f W p
W
s
t
W o
[2] Pr [C]. Wecan start the machine andlet it systematically produce inference chains. Wecan wait and see what emerges, but we cannotgive the machinea particular
formula and commandto seek a proof for that formula (or the negation thereof). But with the available means, we can formulate in abstract terms that a formula is provable. Vax Pra
means that the formula with code number a is provable. The abbreviation of this is Epra. The means with which we have talked here aboutthe subjectpredicate language belong to the subjectpredicate language. We can therefore let the machine talk
about its own language according to this code. Thus, for a given formula the
machine can computeits code number, and for a code numberit can compute the associated formula. The machine can ascertain whether n is the code numberof a proposition, or whether 7 is the code numberof a formula with x; as the only free variable, etc. Whenthe machine has produced a proof, it can determine the code numbers of the proof and of the proved formula (say b and a respectively) and ascertain b Pra and also Epr a. 5.7  With the available meansit is not, however, possible to formulate in a
w p t
a 97
LANGUAGE AND METALANGUAGE
Ch. 5
consistent language what truth is. After A. TARSKI (1933), this can be shown as follows: Weexpress the fact that the formula with code number 7 is a true (more
precisely a Itrue) proposition by the validity of the formula T(n). Hence:
T(n) if, and onlyif, the formula with code numbern istrue.
(1)
We now showthat the rredicate T does not occur amongthe predicates of our subjectpredicate language (cannot be produced by the machine), at any rate if the language is to be J’consistent. Weshall consider the formulae in which x1 is the only free variable. By means of the validity of F(n)
we indicate that n is the code numberof such a formula. The predicate F can be
produced by the machine.Letn be the code number of a formula in which x1 is
the only free variable. If we substitute in this formula the number a for the free
variable x1, then a formula is obtained whose code numberwill be denoted by S(a, n).
(This is defined only for those n for which F(n) holds.) To calculate S(a, n), one only has to break down the code number 7, and replace at certain instants 5 (code number of x1) by 2% (code numberof a). The machineis certainly able to do this. The machineis also able to form S(n, n). Regard this now as a numerical function which we call R. Hence R(n) = code number of the formula, whicharises, if in the formula with code number n weinsert the numeral n at every occurrence of the only free
variable x1.
(2)
The function R is defined for all such code numbers n (ie. for all n, for which
F(n) is valid). The machine can form R and can calculate R(n) for every n, for whichit exists. Assume now,that the machine can also produce the predicate T, i.e. the formula T(x1). Then this is also the case for every numeral n with the formula
—_T(R(n)). With respect to this formula the machine can construct a predicate P (i.e. P(x)), so that for every n the formula
(AT(R(n)) — P(n)) is provable
(3)
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Ch. 5
We now put
p = code numberofP(x1).
(4)
C
W c
But then F(p)is valid. Replace now n in (2) and (3) byp. Hence R(p) = code number of the formula, which arises, if in the formula with
code number p (with the only free variable x1) the numeral p is substituted for the variable. (5) And:
(AT(R(p))) > P(p)) is provable.
(6)
If now in (5) the formula with code number p is explicitly substituted according to (4), then the result is R(p) = code number ofP(p).
(7)
Now(1) is used, with R(p) instead of n: T(R(p)) is true if and only if, the formula with code number R(p)is true, andthis is accordingto (7) T(R(p)) true,if and only ifP(p)is true.
(8)
But now the result is a contradiction between (6) and (8). At least if we demand that nothing that is false is provable. Hence the assumption of a truthpredicate that can be produced by the machine has been proved false. If we demandofTinstead of (1) the stronger T(n) — Nis provable,
(1)
again, if N is the proposition with code number 7,then theresultis also T(R(p)) & P(p) is provable.
(8)
Hence in that case the machineis even able to infer the inconsistency. 5.8  We know that the predicate Epr can be formed by the machine. Now substitute in the text of 5.7 the T by Epr. Then a priori there is no reason for conditionslike (1) and (1’). But we do know that the machine can transform a proof B for A into one for b Pr a
and vice versa,
if a and b are the code numbers of A and B.
(1)
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Write now q instead of R{p). q is then the code number of P(p), whichis also called Q. With Epr in stead ofT theresult is
—Epr(q) > Q provable,
(6’)
g = code numberof Q.
(7)
In the course of turning out proofs, the machine can never produce a formula b Pr g. Forif this should occur, then according to (1’’) it can immediately transform
this into a proof for Q: hence according to (6’) into a proof for —Epr(q), whichis contradictory to b Pr q.
Onthe other hand, the machine can verify for every chain of formulae, if it is a proofof Q, hence for every b whether b Pr g or —(b Pr q).
Wesaw just now that the former will never emerge from the machine. Hence the machinewill find all formulae —(b Pr q):
We should therefore expect it also to find the generalization hereof,i.e., Az (x Pr q),*)
or
= Va(x Pr q). But this would be —Epr q,
and this would mean that accordingto (6’) a proof for Q must emerge from the machine.
The proof B of Q could then be transformed by the machine accordingto (1’’) into a proof of b Pr q, where b is the code number of B, and this is impossible, as has been proved above. Assuming consistency, we have found a (by the machine producible) predicate *) Inaccurately written; one should add the requirement Z(x).
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® ofnatural numbers with the property that the machine can prove
e
D(a) for every a but not
As DG). (G(x) is—(* Pr q).)
This is the principal content of K. Göper’s incompleteness theorem,already referred to in 5.4. The proposition /\z®(x) is not provable by the machine,
although for every a there exists a machinemadeproofof D(a).
The fact, that this unprovable proposition in a way asserts its own unprov
ability, should cause us to regard it as being true.
It seems obvious, to add A #®(x) to ['as a new axiom.
Will the system then stay consistent? One should think: yes. For Ax®(x) — ® means also Vz—@(x). If we could obtain from this (for a certain numeral a) the formula —®(a), then this is contrary to: ®(a) for all a. We should not however draw this conclusion so quickly, for from 4.15 we know,that it is not necessary that there exists an “example” A(a) for the provable VA. How
ever ® can be changed — weshall not do that here — so that the proposition A«®(x) can be added to I’ without detriment to the consistency (not so the proposition Epr Ax®(x), however). In the enriched language, however, it is again possible to find propositions like ®.
5.9  Anyway, — Az®(x) can just as well be added to the language without detriment to the consistency. (For a proof of ® from — Ax®(x) would be a
proof of the unprovable Ax®(x)). In the language thus enriched with axioms we now really have the situation that V2A without an “example” A(a)
C
(*)
(namely with —@(x) for A). According to the completeness theorem, we could add so many new constant subjects to such a language, that (*) does not occur any more. In our case these should be of course new natural numbers. But how? All natural numbers existed already in our language. Clearly PEANO’s axioms make the (wrong)
impression, that they completely fix the notion “natural number”. We have
been led here to a concept of “number’’, that does satisfy the axioms, andstill behaves differently from the intuitive concept.
5.10  Then why do we not prohibit those languages in which (*) occurs — for
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example, by adding as a rule of inference to every subjectpredicate language: if F(a) for every constant subject a, then AxF(»),
or
if Ve G(x), then thereis a constant subject a with G(a).
Well, such a rule of inference would be of no use to the machine, for if the number of constant subjects is infinite, the machine can never complete the
task of verifying F(a).
5.11  The moralofall of this is: If we formalize “all” by “/\”’, wetry to seize the infiniteness in a finite grasp. But we are only moderately successful.
C
6  SOLUTIONS
1
1.10 la. For xe A there are precisely two mutually exclusive possibilities: either xe B,ie.,xe AB, orx¢ B,ie., xe ANB.
b. A= (A\B) v (420 B)=(ANB) VU {40 [BXC) Vv (Bn Ol} = (ANB) v [40 (BXC)] VU (AN BNC) < (ANB) U (BAC) U (AN BOO), where la has been used twice and a distributivity has been used once. Analogous formulae for B and C show: lefthand memberof 1b c righthand memberof 1b. Thereverse is evident. c. Analogous to Prob. 1b. a’. See Prob.la. b’. Analogous. ce’. A=C, B=D=O. 2.a. Follows from Prob.1a, a’. b. Ditto. c. See 1.9.
d. Apply distributivity.
e. ANONBNO) =(A N CNB N C#) = (4.20 C4) 0 (Ba C8) =
(A 9 C*) 0 (B¥ UC) =A 2 C* ON B*=(4 29 B*) ON C*# = (AVBNC.
f. Follows from Prob.2e. 3.
See Prob. 1bc.
3.12
1. AalVy(MO) A xKy) A V2(V(2) A xKz))]. 2. NalVy(M(Qy) A xKy) > V2(V(@) A xKz)). 3. Aay(xKy — xJy).
4. Aa,y,zl(xKy A yKz) > xJz]. 5. Vy(xGy).
6. Va,y,z(M(x) A yKx A yGz A V(z) A xJz). 7. Vuw(M(x) A M(y) A xKu A yKu n xKo A yKo A uGw). 8» Vayz2 V(x) A U(x) A My) A Aly) A xKz n yKz) > Vaow(M(u) A Alu) A Viv) A Uw) A uKw A vKw).
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SOLUTIONS
9. sAzl(Ve(xGy) A M(x)) > U()].
10. —Az[(V(x) A U(x)) = AVy(yKx A M(y) A AQ)))11. Ay(yKx > V 2(yGz)).
12. VaAy(vKx > V 2(yGz)). 13. Az[zKx > Vu(uKy A uGz)]. 14. V2z[zKy A AV a(uKx A zGu)]. 15. Va,yAu,vl(uKx — VolpKy A uGp)) 0 (vKy > V q(qKx A vGq))]. 16. Va,y—Vuv(uKx A vKy A uGv).
17. yKx > {A2[zKy > Vu(uKx A zKu)]}. 18. True, independently of the meaning of K and U. 1921. True. 22. AtVaZ(%, ft).
23. VinV2Z(x, t). 24. V2VeZ(x,t).
25. AzVtZ(x,t). 26. A t,0(Z(x, t) > P(x,t)). 27. AtalZ(x, t) > Ve((t VeP(«, 1).
34. Amt{P(x,t) v Vell!