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Decidability and Essential UndecidabilityAuthor(s): Hilary PutnamReviewed work(s):Source: The Journal of Symbolic Logic, Vol. 22, No. 1 (Mar., 1957), pp. 39-54Published by: Association for Symbolic LogicStable URL: http://www.jstor.org/stable/2964057 .

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THE JOURNALOF SYMBOLICLOGIC

Volume 22, Number 1, March 1957

DECIDABILITY AND ESSENTIAL UNDECIDABILITY

HILARY PUTNAM

1. There are a number of open problems involving the concepts of

decidability and essential undecidability.' This paper will present solutions

to some of these problems. Specifically:

(1) Can a decidable theory have an essentially undecidable, axiomatiz-

,able extension (with the same constants) ?2

(2) Are all the complete extensions of an undecidable theory ever decid-

able ?

We shall show that the answer to both questions is in the affirmative.

In answering question (1), the decidable theory for which an essentially

undecidable axiomatizable extension will be constructed is the theory of

the successor function and a single one-place predicate. It will also be shown

that the decidability of this theory is a "best possible" result in the following

direction : the theory of either of the common diadic arithmetic functions

and a one-place predicate; i.e., of addition and a one-place predicate, or of

multiplication and a one-place predicate, is undecidable.

2. Before establishing the main result, it is convenient to give h simple

proof that a decidable theory can have an axiomatizable (simply) un-

decidable extension. This is, of course, an immediate consequence of the

main result; but the proof is simple and illustrates the methods that we

are going to use in this paper.

For this purpose, we select the theory of the successor function (call

Received October 4, 1956.1 The terminology of this paper is that of Undecidable theories, Tarski, Mostowski,

and Robinson, North Holland Publishing Co., Amsterdam (1953). This work will be

cited as "U.T.".

A solution to problem (1) has erroneously been announced by Myhill (Solution of a

problem of Tarski's, this JOURNAL, vol. 21 (1956), pp. 49-51). Actually Myhill hassolved a related problem; but an answer to (1) does not follow from what he proves,

as he asserts. (Cf. section 7., below).

(Except for section 7) attention in the present paper is confined to theories with a

finite number of constants. Kreisel has previously solved problem (1) for theories which

use an infinite number of constants. A statement of Kreisel's proof appears in hisreview of the article by Myhill just cited; see Mathematical reviews, vol. 17, no. 8

(Sept. 1956), p. 815.

2 This is easily seen to be a reformulation of a problem in U.T., p. 18. (Cf. the

discussion below). From now on, when the term 'extension' is used, 'with the same

constants' will be understood.

3 This problem was suggested by G. Kreisel whose interest and infectious enthus-

iasm have led me to work on these questions.

39



40 HILARY PUTNAM

it 'R'). This theory is decidable.4If to R we adjoin two new individual

constants, a and b,5 and no new axioms, we obtain another theory R'.A sentence of R' is valid only if its universal generalizationwith respect

to a and b is valid in R; so R' is likewise decidable. In the terminology

of U.T., R' is an inessential extensionof R.

Let us define 0 as a, 1 as S(a) ("the successor of a"), 2 as S(S(a)), etc.And finally, let us construct a theory L by adding to R' the followingaxioms:

(3) -_(b mJ (i = 1,2, . ..)

where mi, M2, M3, ... run through the membership of some set of positive

integers which is recursively enumerable but not recursive (for the sake of

definiteness, let us take the set of Gbdel numbersof theorems of quantifi-cation theory).

We now claim:

THEOREM 1. L is recursivelyaxiomatizable,and (simply) undecidable.

Proof:

i) L is axiomatizable: we have given a recursively enumerable set ofaxioms for L. But, according to Craig's theorem6every theory that can

be axiomatized with a recursively enumerable set of axioms can be axiom-

atized with a recursive set of axioms. So L is axiomatizablein the sense ofU.T. (recursively axiomatizable).

ii) L is undecidable: it suffices to show that b#n is not a theorem of Lunless it is an axiom; in other words, if n does not belongto the set {ml, M2,

in3, ... }, then b=n is consistent with all the axioms bomi. For if b-nwere inconsistent with the axioms of L, b=n would have to be inconsistent

by itself (sinceit entails all of those axioms), so (b n) would be a theorem

of R', and (x)'(x n) would be a theorem of R', which is absurd.

iii) L is not essentially undecidable: for if n is any integer not in {ml, m2,

M3, ... }, the theory obtained by adding b=n as sole new axiom to R'

4U.T., p. 64. Also, a decision method for the elementary theory of the successor

function and an arbitrary one-place predicate is given in section 3. below. This can,

of course, be used to decide sentences of R.I Logical signs are used as names of themselves throughout the present paper,

and not in their object-language use.

6 Craig's method is as follows (On axiomatizability within a system, this JOURNAL,

vol. 18 (1953) pp. 30-32): let {A,, A2, . ...} be a recursively enumerable but not re-

cursive set of axioms for a theory T. Replace each axiom Ai by the conjunctions

whose members are Ai repeated n times, for each n such that n is the number of a

proof that Ai belongs to the set {Al, A2, ... } (or rather that the Godel number of Aibelongs to the corresponding recursively enumerable set of integers). The recursiveness

of the new axiom set follows from the fact that the set of pairs (n, A) such that A

belongs to {Al, A2, .. . } and n is the number of a proof that this is the case (in a suit-

able formalism) is recursive.



DECIDABILITY AND ESSENTIAL UNDECIDABILITY 41

is a consistent extension of L, by the remark just made; and this theory

is decidable because it is a finite extension of R' and R' is decidable. q.e.d.

Thus we have that a decidable theory may have an undecidable and

axiomatizable extension: for R' and L are two theories that stand in just

this relation to one another.

3. In this section we shall establish the decidability of a certain simple

theory G: this will serve as a basis for showing, in the next section, that a

decidable theory (an inessential extension of G) may have an axiomatizable

and essentially undecidable extension.

G is the theory of the successor function and a single one-place predicate;

i.e., G has a single monadic operator S, and a single monadic predicate P.

The variables of G have arbitrary integers as values; S(x) means "the

successor of x"; and P designates an arbitrary class of integers. A sentence

of G is valid if it is true under this unterpretation (for all values of P).For convenience, let us also introduce into G the symbol Pr for "the

predecessor of x." Evidently Pr is definable in terms of S as follows:

y=Pr(x) =df x=S(y).

Instead of S(x) we shall henceforth write x+ 1; and likewise x+2 for

S(S(x)), x I for Pr(x), etc. Since Pr(S(x)) =x=S(Pr(x)), we can write x,

and not (x+l)-1; x+5, and not (x+7)-2; etc.

Also, we can "transpose," i.e., rewrite x+3-y-2 as x+5=y or x=y-5.,

etc.

The decision method for G will be given in four lemmas. We choose the

decision problem for satistiability as being a convenient form for our pur-

poses.

To introduce a terminology and notation we shall employ from now on:

P(x)P(x+ 1)P(x+2) will be written simply PPP, and similarly in similar

cases. A formula like PPP (with n terms) will be called a 3-series (n-series).

Our procedure in Lemma 1 is in part a variant of a decision procedure

for monadic quantification theory invented by Behmann: following himwe introduce the special symbols (3x)n(...) for "there are at least n x's

such that (. . )," and (3]x)n'( .. ) for "thereareexactlyn x's such that ( . )."LEMMA1: Every closed /ormula of G can be effectively reduced to-T or 7

7

or to a disjunction o/ formulas o/ the torms:

a) (3x)C (where C is an n-series)

7 T and 1 ("tee" and "eet") are Quine's symbols for truth and falsity (Methods

of logic, Harvard (1950)). They are always eliminated by elementary reduction

techniques; or else the whole formula reduces to T or 1. The logical notation in the

present paper follows Quine (e.g., in using juxtaposition for conjunction; in using both

- and - for negation), except that more dots than are strictly necessary for punctuation

are sometimes used to facilitate reading. Following U.T., identity is included in each

elementary theory as a logical constant.



42 HILARY PUTNAM

b) (3x) mC (where C is an n-series)

c) (3x) mC Sy ,, ,) Sy Spd) conjunctions of formulas of kinds a)-c) (with C's of the same length)

in which no two C's are the same.

Example:

(3x) (P(x) P(x + l)P(x +2)) . (3x) 1(P(x).P(x + 1).P(x+2)) . (3X) 2 (P(X) P(X+ 1)

P(x+2)) is a formula of the form d). (With n=3.)

Proof: Let a closed formula F of G be given. By employing Behmann's

procedure8a for monadic quantification theory, we confine each quantifier

to a conjunction of prime formulas9 containing only the variable of quanti-

fication and no identity formulas. In detail:

The formula F is first written in prenex normal form. The last quantifier

in the initial string of quantifiers will be either (3xn) or (3xn). In the

latter case, the second '-' is considered as part of the matrix; and (in eithercase) the matrix is put in disjunctive normal form. The existential quantifier

(3xn) is then distributed through the disjunction, say as (3xJ)B1 v (3xn)B2 V

.... Also, (3xn)Bi is replaced by B2 whenever x, does not occur in B1.

And quantifiers are confined, i.e., (3xn)Bi is rewritten as (3xJ)C,.Di where

C, is the conjunction of all the prime formulas in Bi containing xn. Thus

the quantifier (3xn) now stands only in front of conjunctions of prime

formulas, each one of which contains the variable x-.

Certain of these formulas(3xn)Ci

may contain variables y other than

xn (in identity formulas). If (3xn)Ci can be brought (e.g., by "transposing")

into the form (3xn)(xn-y+m . C'(xn)), where y is any variable other than xn,

the whole quantification is eliminated, and one writes instead C'(y+m).

Otherwise, y must occur in a negated identity formula, and (3xn)Ci has

the form (3xn)(xn#y+M.C'(xn)). In this case we write instead:

(4) C'(y-+-m) 3x ) (C'(x )) v C'(y+m) (3x )2(C'(x )).

If there is a variable other than xn in the scope of one of the resulting

quantifications, the procedure is repeated. In repeating, (3xn) m(x, 0y.

C"(xn)) would be replaced by

(5) C (y)(3Xn)m(C (xn)) vq(y)(3x)im+s(C (xns).

In this way,1? the quantifier (3Xn) is confined to conjunctions of prime

8 In the sequel, a single formula will also be called a "conjunction" (with one

member).

8a Beitrdge zuy Algebra der Logik, insbesondere zum Entscheidungsproblem, Mathe-

matische Annalen, vol. 9, pp. 1-36.9 "Prime formulas" are formulas of the forms P(x+m), P(x+m), x+m=y+n,

x+m #y+n, where m, n, may be positive, negative, or zero.

10 If the procedure eventuates in a formula of the form (3x)m(x My), this is reduced

to T. Formulas of the forms (3x)(x=y+m) and x=x+m are also eliminated.




formulas containing only the variable xa and no identity formulas. The

same procedure is then applied to the remaining quantifiers in turn. Finally,

the formula is placed in disjunctive normal form, and we insure that no

formulas of the form (ix,) eCoccur (with m> 1) by replacing such formulas

by

(6) (3x,)C v (3x,)1Cv (3X,)2 v ... v (3x1)mI1nC.

In this way we obtain a disjunction

(7) Bl v B2 V ... v Bk

where each Bi is a conjunction of quantifications of the forms (3x1)C,

(3x,)mC, (3x,)m'C; and where C is a conjunction of formulas of the forms

P(x,+n) and P(x,+n). In short, the B, have the form

(8) (x) Cl *(X) C2.*.*. 3X) "IVI l *(3X) "2C2s. .. (3X) m1lCniL'. (3x) mn2 ' ..

To get the B, into the form given in the statement of the lemma, we have

to put the C's into n-series form. This is done as follows:

A term occuring in one of the B, may be of the form x-m. But

(3x) C'(x+ml)C" (x+m2) ... C(k)(x+mk)) =(3x) (C'(x)C" (x+m2- ml) ...

C(k)(x+mk ml)); where C") is P or P and mlm2 ... ? mk; - e.g.,

(3x)(P(x-3)P(x+7)) _(3x)(P(x)P(x+ 10)). A constant is added or sub-

tracted in this way in each of the Bi to insurei) that no "negative" terms appear (i.e., we may have terms of the form

x+m, where m is a positive integer, but not x-m).

ii) that the variable of quantification (which will henceforth be written

as x) appears as a term once in every B, (without any "plus" or "minus").

Next, it may happen that some "in between" terms are missing from a

Bi. E.g., (3x)(P(x-3)P(x+7)) has been rewritten as (3x)(P(x)P(x+10)),

but the intervening terms x+ 1, x+2, . . ., x+9, are now missing. This is

remedied by replacing (3x)(P(x)P(x+ 10)) by (3x)(P(x)P(x+ 1)P(x+ 10)) v

t3x) (P(x)P(x+ 1I)P(x+ 1IO))

In this way the missing terms are brought in one by one: if we have

instead (3x)n(P(x)P(x+ 10)) we write first (3x)n(P(x)P(x+ 1)P(x+ 10) v

P(x)P(x+ 1)P(x+ 10)) and then we use:

(9) (3x)n(P v Q)a (3x)P(3x)nQ v (3x)l'P(3x)R-1QP v ...

v (3x)n-1P(3x)QP v (3x)'nP or

(I 0) (3X) n(P v Q) a_ (jx) P(3x) nQ v (3X)1P(3x) n-1Qpv..

v (3x)n'P(3x)QP.

Finally, we may assume that no two C's are the same: otherwise (8)

can be trivially simplified. E.g., (3x)m'C implies (3x)"C if m -a n, and



44 HILARY PUTNAM

the weaker formula may be dropped; and (3x)"mCis inconsistent with

(3x)"'C if m/n, and (8) may be reduced to L.

This completes the proof of Lemma 1.

Since a disjunction is satisfiable if and only if one of the disjuncts is

satisfiable, we have a general decision method provided we can find a

decision method for formulas of the form (8), where the C's are as describedin Lemma 1.

Let us make a list of all the n-series (for the given n) not excluded by

the negative existential quantifications in (8); i.e., of all the n-series except

C1, C2, etc. (If there are no negative existential quantifications in (9), this

list will comprise all the 2n n-series for the given n). The members of this

list will be called the permissible n-series.

For the sake of an example, suppose n=5, and that the following is the

list of permissible 5-series:

(11) (a) PPPPP

(b) PPPPP

(c) PPPPP

(d) PPPPP

(e) PPPPP

(f) PPPPP

(g) PPPPP

(h) PPPPP.

We now proceed to reduce this list in the following way: Suppose integers

n, n+ 1, . . ., n+4 realize pppPP.10a Then n+ 1, n+2, . .., n+5 must

realize either PPPPP or PPPPP. We will express thus by saying that (a)

in the list above can be succeeded only by (b) or (g). Similarly, we will say

that (d) can be preceded only by (c) (since PPPPP is not permissible);

and that (h) cannot be preceded by anything at all, although it can be

succeeded by (a). Formally, C1 can precede C2 (C2 can succeed C1) only

if C1 and C2 are both permissible, and C1 is obtained from C2 by deleting

the final P or P and prefixing an initial P or P.

We express these relations in the following diagram:

(12) (h) (a) (f) (b) (c) (d) (g) (e) (a) (f)

I \/ I I \/ I \(a) (b) (c) (d) (e) (f) (g) (h)

/\ I I ~~~~~~~~~~~I /\ I I(b) (g) (c) (d) (e) (f) (b) (g) (e) (a)

Branches going up from an element in the diagram indicate which

10i "Integers n, n-+ 1, n+2, ... n+4 realize PPPPP" if and only if the integer n

satisfies the formula PPPPP; i.e., if and only if P(n)P(n+ 1)P(n+2)P(n+3)P(n+4)

is true. - And similarly in similar cases.




permissible 5-series can precede it; branches going down indicate which

can succeed it.Now we reduce the list and the diagram by casting out any 5-series

(n-series)that cannot be preceded (no "branchesgoing up") or cannot be

succeeded (no "branches going down"). Thus we obtain as our reduced

diagram:

(13) (a) (f) (b) (c) (d) (g) (e) (a) (f)

\/ I I \/I \(a) (b) (c) (d) (e) (f) (g)

/'\ ~ ~~~I I I I /(b) (g) (c) (d) (e) (f) (b) (g) (e)

But in the reduced diagram, (a) cannot be preceded. So we apply the

reduction procedureagain (as often as is still

possible),obtaining in our

example:

(14) (f) (b) (c) (d) (g) (e) (f)

(b) (c) (d) (e) (f) (g)

(c) (d) (e) (f) (b) (g) (e)

Since no further reduction is possible, (b), (c), (d), (e), (f), and (g) com-

prise our final reduced list. (Henceforth this will be called simply the 're-duced list.')

The meaning of the reduction procedureshould be evident. If the state-

ment that only the n-series in the originallist are realized is true (and this

is the statement made by the conjunction of the negative existential

quantifications in (8)), then the stronger statement that only the n-series

in the reduced list are realized must likewise be true. For if (h), say, were

realized it would have to be precededby something or other; and since it

cannot be preceded by any permissible 5-series, this would falsify thestatement in question.

Certain corollaries are immediate: for instance, if the reduced list is

empty, or if one of the C7'2,C"i s not in the reduced list, then (8) is not

satisfiable. Moreover:

LEMMA 2: II there are no formulas of forms b) or c) in a conjunction of

formulas as described in Lemma 1, and the reduced list is not empty, the

conjunction is satisfiable.

Proof: Choosesome n-series from the reducedlist (say (b), in our exam-

ple) and interpret P with respect to the integers 1, 2, . . ., n so that this

n-series is realized (so that 1, 2, 4 belong to P, 3, 5, to P). Since this n-

series can be succeededby some element in the list (in fact, by (c)), we can

extend the interpretation to n+ 1 so that 2, 3, . .., n+ 1 realize this n-series



46 HILARY PUTNAM

(in the example, by letting 6 belong to P). Thus we can always go one step

further "to the right." Likewise, if we have interpreted P so that some

integers m, m+ 1, ..., m+n- 1 exemplify a permissible n-series C, and

we have not yet extended the interpretation to m-1, we simply pick an

n-series D that can precede C, and make mr I either P or P, - whichever

will cause mr-1, m, . . ., m+n-2 to realize D. Thus we can always go onestep further "to the left." Hence P can be interpreted in such a way that

every integer is assigned either to P or to P, and only n-series in the reduced

list are realized. q.e.d.

(In fact by the time we have assigned membership in P or in P to the

integer 2n+n we must have realized some permissible n-series twice; at the

earliest place where this happens, the sequence is made periodic.)

LEMMA3: If there are formulas of the form b) in a conjunction of/ or-

mulas as described in Lemma 1, but none of the form c), then the formula isgenerally satisfiable if and only if it is satisfiable over the integers 1, 2, . . ., m,

m L (2n+I)k-2n+n+1, in such a way that only C's in the reduced list

are realized. (Here k is the sum of the ni in sub-formulas of the form

(3x) "'C".)

Proof: Suppose the formula (8), minus quantifications of the form

(3x)rn'C'Jr, to be satisfied. Since each C"n s realized ni or more times in

the whole series of integers, there must be some finite subseries (say from

m to m+k) such that each C"tis realized ni or more times in the subseries,

and only C's in the reduced list are realized in the subseries. Conversely,

if there is a finite series (an interpretation of P with respect to some finite

sequence of integers 1, 2, 3, . . ., m) which satisfies these conditions, then

we can satisfy these conditions in the whole series of arbitrary integers.

To do this it is only necessary to extend the interpretation of P to m+ 1,

m+2, etc.; and to -1, -2, etc. But this can always be doue, as remarked

above.

Thus we can confine our attention to finite sequences of integers (and

interpretations of P with respect to these). Let k= in1. We wish to show

that it is sufficient to consider finite series with not more than (2n+ 1)k

2n+n+ 1 elements.

To prove this, suppose there is a finite series 1, 2, ..., m which satisfies

the conditions, under a certain interpretation of P. From this sequence,

it must be possible to pick out (for each i) ni occurrences of Ct"(i.e., ni sets

of n successive integers which realize Cn").Call these designated occurrences

of C"n. (There may also be other, non-designated occurrences of some

C~i; but what is important is that we designate exactly ni occurrences of Cii.)

We may drop all the integers to the left of the first designated occurrence

of any n-series in 1, 2, . . ., m; and also all the integers to the right of the

last designated occurrence, and the resulting series will still satisfy the




conditions. (It will no longer begin with 1, in general, but this can easily

be remedied by "renaming" the first integer '1', etc.) Thus we can always

obtain a new series which begins and ends with a designated occurrence.

Suppose that between two successive designated occurrences there is a

"repetition," i.e., some C1occurs twice.

Then we may drop all the C's between the two occurrences of C1, andalso drop the second of the two occurrences, and all conditions are still

satisfied. E.g., if we have:

(15) Cl*C4C3CCC6CC2*,

(the C's stand for successive n-series, not for integers, and the asterisks

indicate designated occurrences), we may put instead:

(16) Cl*C4C3CC2*,

and each C is still a possible successor for the preceding C.

Also, if the designated occurrences are of C1 and C2 respectively (as in

the example), we may suppose that no occurrence of either C1or C2 appears

between the two designated occurences, for similar reasons.

Thus, between the two designated C's we will have a series of at most

2n-2 C's. So given any finite sequence which satisfies the conditions, we

obtain a series which contains at most 2n 2C's between each pair of

designated occurrences; hence, a series with at most (k-1)(2n -2) +k,

or (2 -1)k-(2n-2) C's. Hence the series contains (2n- 1)k -(2"- 2)+

(n-1) integers. q.e.d.

Now we shall consider the last case; there are affirmative quantifications

of the form (3x) 'miC"". n this case, cast all the C"" out of the reduced list.

Call the result the 'special list.' We now reduce the special list in two

different ways:

1) Cast out any element in the special list which cannot be succeeded by

any element in the special list. (This is to be done as many times as possible,but an element is not cast out simply because it cannot be preceded by any

element in the list.) Let the elements thus obtained be E1,,EV. . ., Ek*

2) Cast out any element in the special list which cannot be preceded by

any element in the special list. (This is to be done as many times as possible,

but an element is not cast out simply because it cannot be succeeded by

any element in the list.) Let the elements thus obtained be D1, D2, . . ., DL.

If the first list (E1, . . ., Ej) is empty, no infinite endless series (of C's in

the reduced list) is possible that does not contain some C i". If the second

list (D1, . . ., DL) is empty, no infinite beginningless series (of C's in the

reduced list) is possible that does not contain some C""i. In either case,

there must be infinitely many occurences of some of the C> if only C's

in the reduced list are realized, and (8) is unsatisfiable.



48 HILARY PUTNAM

If neither list is empty, then we state:

LEMMA4: If there are formulas of the form c) in a conjunction of formulas

as described in Lemma 1, then the formula is generally satisfiable if and only

if it is satisfiable over the integers 1, 2, .. ., m, m?(2- -)r+21+n-1, in

such a way that only C's in the reduced list are realized, and the integers 1, 2,

. . ., n realize one of the Di, and the integers m-n+ 1, m-n+2, . . ., m realizeone of the E?. (Here r is the sum of the ni and mi in sub-formulas of the

forms (3x) iC'i and (3x) mi'Cmi.)

Proof: Suppose the formula (8) to be satisfied. Since each C" is realized

at least ni times, and each Cmi' is realized exactly m? times in the whole

series of integers, there must be some finite subseries (say from m to m+k)

such that each C'i' is realized ni or more times in the subseries, and each

Cmi' s realized exactly mi times in the subseries, and only C's in the reduced

list are realized at all. Moreover, no Cmioccurs to the "left" or "right" of

this subseries. But we have seen that it is possible to have an endless series

(which uses only C's from the reduced list) in which the Cmi'do not occur,

only if none but (some or all of) E1, . .., Ek are realized in the series. And

it is possible to have a similar series without beginning, only if none but

DI, . . *, DL are realized in it. Thus the integers m 1, m, ..., m+n- 2

must realize one of the Di; and similarly m+k-(n-2), .. ., m+k+l

must realize one of the E?.

Therefore, the whole series m- 1,m, ..., m+k+ is a finite series of

the kind whose existence is asserted in the Lemma. (It does not begin

with 1, in general, but as before this can be remedied by "renaming"

mi1 '1'.)

Let r= 'ni+2mi. WX'e ish to show that we can confine our attentioni i

to series with (2n 1)r+2n+n- 1 elements.

The argument exactly parallels the argument in the preceding case;this time we will have r+2 designated occurrences (the Cni, the Cmi, the

initial Di and the final EJ); and, since we can "condense" asin (5) -(16),

we can once again assume that there are at most 2n-2 C's between any two

designated C's. Thus our decision method is complete. If P can be inter-

preted with respect to 1, 2, . . ., m, m ? (2n- )r+2n+n- 1, so that the

conditions stated in the Lemma are satisfied, we can "continue to the left"

without ever using any Cm-',since this finite series begins with a Di; and

we can "continue to the right" from the final Ei in similar fashion. And

if none of the possible interpretations of P with respect to 1, 2, .

(2?' 1)r+21+n- 1 satisfies the conditions; (8) is not satisfiable. q.e.d.

THEOREM 2. G is decidable.Proo/: Immediate from Lemmas 1-4.

Since G is decidable, it is of course axiomatizable (the set of valid sen-




tences could be taken as the set of axioms, since these are recursive). For

our present purposes, we could use instead of G the theory of one permuta-

tion and one one-place predicate.

This theory has a single monadic operator, f, a single monadic predicate,

P, and the following axioms:

(17) (i) (x)(3y)(x f(y))

(ii) (X)(y)(/(X)=-(y) D x=y).

Since any model for this theory is isomorphic to the system of arbitrary

integers (when / is identified with S), or to the integers modulo m, or is a

sum of such models, the decision method just given for G is easily extended

to this theory. This yields an example of a finitely axiomatized and decid-

able theory from which (by the method given in the next section) an

essentially undecidable theory can be obtained by simply adding a recursive

set of additional axioms.

The decision method given also applies if there are any number of one-

place predicates. The major modification required is in the definition of

'n-series': if the formula being tested for satisfiability has m one-place

predicates, then an n-series is a possible character of n successive integers

with respect to these predicates. E.g., if there are two predicates, P1, P2,

then P1p2(x)plp2(x+ I)PP2(x+2) is one of the 2mn=64 3-series.

4. Let us construct an inessential extension of G (to be called G') byadding the single individual constant b and no new axioms. To obtain an

essentially undecidable extension of G' we make use of the fact that there

are disjoint recursively enumerable sets that cannot be separated by any

recursive set; i.e., there are a, b such that anb= A, but such that for no

recursive set c is it the case that a c c while bnc = A (Trachtenbrot11 has

even shown that the following two sets constitute such a pair: the Godel

numbers of theorems of quantification theory, and the Godel numbers of

formulas that fail in at least one finite domain.) Let {ml, M2, ... } and

{X1, n2, .}. I be two such sets, and define 0 as b, 1 as S(b), etc.; then we

obtain a theory H by adding to G' the axioms:

(18) (i) P(mi) (i 1, 2, . ..)

(ii) P(ni) (i =1, 2, .. .)

THEOREM 3. H is axiomatizable and essentially undecidable.

Proof:

(i) The axiomatizability of H follows again from Craig's theorem.

(ii) H is consistent (bearing in mind that the sets {Ml, M2,...} and

{nl, n2, . . . } are disjoint).

11 0 rekursivnoj oldelimosti, Doklady Akad4mii Nauk USSR, vol. 88 (1953),

pp. 953-956.



50 HILARY PUTNAM

(iii) H is essentially undecidable. For, assume some consistent extension

D of H is decidable. Then the set of integers n such that P(n) is a theorem

of D is recursive (call it k); k contains all the mi, because P(mi) is an axiom

of H (for i - 1, 2, .. .) and D is an extension of H; and k does not contain

any of the ni, because P(ni) is an axiom of H (for i = 1, 2, . . .) and D is a

consistent extension of H. Thus k is a recursive set separating {ml, M2, . . .}

and {nl, n2, ... }; which is impossible, since these were chosen as inseparable

sets. q.e.d.

5. Let us now turn to the question: whether the decidability of G is a

"best possible" result.12 Since the theory of the successor function and any

number of one-place predicates is decidable, we consider next the theory

of addition and a single one-place predicate. Let us call this theory K;

K has a single diadic function symbol + and a single monadic predicate P

as its extra-logical constants. The variables are interpreted as ranging overpositive integers; P stands for an arbitrary set of integers. A sentence of

K is valid if it is true under this interpretation (for all values of P).

THEOREM 4. K is undecidable.

Proof: It is known that one can formulate an essentially undecidable

and finitely axiomatizable theory in terms of + and the diadic relation

Sq(x, y) (read: "x is the square of yJ) 13 We shall show that it is possible

to define Sq(x, y) in terms of + and the particular monadic predicate

P(x) ("x is a square"). Thus there is an essentially undecidable and finitelyaxiomatizable theory T with the primitives + and P. Since P stands for

an arbitrary set of integers in K, all valid sentences of K must remain true

when P is interpreted as this predicate. Therefore T is compatible with K,

and K is undecidable.'4

The following is a possible definition of Sq(x, y) in terms of P: we first

define x < y and u=1, thus

(19) X<Y df (3Z)(X+Z Y)

(20) i= 1 =df -(3Y)(Y<u),

and then define

(21) Sq(x, y) df P(X) . (3Z)(P(z) .x<z. (W)(Wz<z.x<zW. D P()) .

z=x+y+y+ 1).

What (21) says is that x is a square and that the difference between x

and the next larger square is 2y+ 1. This is satisfied only if x=y2. q.e.d.

12 At the referee's suggestion, I stress the informal character of this question. The

terms 'immediately stronger,' 'just weak enough,' etc., at the beginning of section 6.

are also used informally.13 U.T., esp. pp. 77-80.

14 By Theorem 6 of U.T. (given below in section 7.).




K is undecidable even if the variables are interpreted as ranging over

arbitrary integers instead of just positive integers. For a possible definition

of Pos(x) ("x is a positive integer")in termsof + and P applied to arbitraryintegers is

(22) Pos(X) =df (3y) (3z) (3w) (3u) (P(y)P(z)P(w)P(U) .x=y+z+w+JU.

x+xOx).

(In words: "x is the sum of four squares, and x does not satisfy 2x=x."

This is true only if x is a positive integer.) Thus the theory of + and P

relativized to positive integers can be interpreted in the theory of + and P

for arbitrary integers; therefore so can the essentially undecidable and

finitely axiomatizable theory T referred to above.

Instead of K we might have considered M; the theory of multiplication

and a single one-place predicate Q. The argument for the undecidabilityof M is parallel to that for K. Here we use as our finitely axiomatizable

and essentially undecidable theory a subtheory of the theory of multipli-

cation and the particular one-place predicate "is an integer of the form

2a2." With this particular interpretation for Qwe define B ("is a power of

2"), thus:

(23) B(x)-df (y)((3z)(y.z=x) D (y.x=x) v (3w)(Q(w).

(3u) (w .u y))).

In words: "for every y, if y divides x, then y= 1 or y is divisible by some

w such that Q(w)." (If x is a power of 2, then every divisor of x is either 1

or is itself divisible by 2; and Q(2).)We establish a correspondence between the integers and the powers of 2,

thus: n corresponds to 2n, addition to multiplication, and P (i.e., "is a

square") to Q. This correspondence is clearly an isomorphism: n2n,

m42m, n+m-2n .2m-2n+m, P(n) = Q(2n). Thus the truth-value of any

sentence of the essentially undecidable and finitely axiomatizable theory T

mentioned above is unaltered when addition is replaced throughout by

multiplication, P by Q, and the quantifiers are relativized to B. In this way

there results an essentially undecidable and finitely axiomatizable theory

in the vocabulary Q.

6. To recapitulate: we have now shown that G is decidable, and that G',

an inessential extension of G, possesses the essentially undecidable, axio-

matizable extension H. However, we could not have used in place of G

either of the theories immediately stronger (i.e., M or K); for these are

undecidable, though not essentially so. Thus the peculiarity of G: that,

although decidable, it possesses an essentially undecidable and axiomatiz-

able extension, - depends on G being just weak enough. If G lacked some

notation for expressing the integers (the successor function) or some one-



52 HILARYPUTNAM

place predicates, we could not form H; on the other hand, any obvious

strengthening of G (beyond bringing in more one-place predicates) seems

to yield an undecidable theory.

We turn now to a kind of "converse" of this problem: instead of seeking

decidable theories with undecidable extensions, we will seek an undecidable

theory all of whose complete extensions are decidable.Let N be the following theory: there are no extra-logical constants;

and the axioms are

(24) (3x)m'i(x= x) (i = 1, 2, .).

As before (3x)m ' (x-x) means "there are exactly mi x's such that x=x";

and we assume that {m., M2, ... } is some recursively enumerable but not

recursive set.

THEOREM 5. N is undecidable; and all its complete extensions are decid-

able.Proof:

(i) N is axiomatizable. By Craig's theorem.

(ii) N is consistent: let N1 be the theory with the following single axiom

(where n is any integer not in {ml, M2, . . .):

(25) (3x) n(x=x) X

N1 is clearly consistent: and N1 is an extension of N. since (25) implies

each of the axioms (24). Thus N is consistent.

(iii) N is undecidable. For (3x)" (x-x) is not a theorem unless it is an

axiom; i.e., unless n belongs to the set {m,, M2, .. .}. And this is not a

recursive set.

(iv) Every complete extension of N is decidable. By Behmann's method

we can decide any sentence of any complete extension C of N provided we

know the truth-values of all sentences of the following kinds:

(a) (3x)n(x=x) (n1, 2, ...)

(b) (3X) nf(X ~X) (n~ l, 2, .)

Thus each complete extension is completely characterized by giving one

number: the number of individuals (oo, or some finite number). This infor-

mation enables us to assign a truth value to sentences of kinds (a)- (b);

and hence (since the method referred to can be used to express any sen-

tence of C as a truth-function of these'5) to any sentence of C.

7. In U.T. the following theorem is stated (p. 18): 'Theorem 6. Let T1

and T2 be two compatible theories such that every constant of T2 is also a

constant of T1. If T2 is essentially undecidable and finitely axiomatizable,

15 The method is in fact contained in the proof of Lemma 1 of section 3.; the only

modification is as follows: formulas of the form (3x)m(x #y) are not reduced to T

(as in fn. 10 above), but are instead replaced by (3x)m+1(x=x).




then T1 is undecidable, and so is every subtheory of T1 which has the same

constants as T1.'

Tarski goes on to remark (p. 19): 'From the point of view of applications,

it would be important to know whether Theorem 6 can be improved by

assuming that T2 is an arbitrary axiomatizable theory (which may not be

finitely axiomatizable). We could answer this question affirmatively if weknew that every essentially undecidable theory which is axiomatizable has

an essentially undecidable subtheory which is finitely axiomatizable.'

However, the question has in fact a negative answer (cf. Theorems 2

and 3 above). Myhill has provided an example of an essentially undecid-

able, axiomatizable theory without a finitely axiomatizable, essentially

undecidable, subtheory. Myhill erroneously claims that this answers Tarski's

question. It does not: for the finitely axiomatizable subtheories of the theory

constructed by Myhill are undecidable, though not essentially so. To show

that the answer to Tarski's question is in the negative, one must construct

an axiomatizable and essentially undecidable theory with a decidable

subtheory which uses all of the constants.

As remarked above (fn. 1), Kreisel has previously solved this problem

for theories with an infinite number of constants. Indeed, the use of an

infinite number of constants materially shortens the argument. The following

proof is my own, but the idea is related to that used by Kreisel:

Take as the decidable theory D, monadic quantification theory with an

infinite set of individual constants a,, a2, a3, . . ., and an infinite set ofpredicate constants P1, P2, P3, .. Let T(e, m, x) be the recursive pred-

icate: "m is the number of a proof (in a suitable formalism) that x belongs

to the recursively enumerable set whose Godel number is e." As the essential-

ly undecidable extension E take the theory with the following axioms:

I. P,(a") (for each i, n, such that the recursively enumerable set with

the Godel number i contains the integer n).

II. (X)(Pei,(x) D iiPes(X)) (for all i, j; where e,, is the Godel number

of the following recursively enumerable set: the set of all n such that(3m)(T(i, m, n) . (m')(T(j, i', n) v m' > m)).)

E is axiomatizable, by Craig's theorem. And every recursive set is de-

finable in E. For let R be a recursive set of positive integers, and let i

and j be G6del numbers of R and R respectively: then e21 is also a Gddel

number of R, and eji is a Godel number of R. Whenever n belongs to R,

Peg(an) is an axiom of E; and whenever n does not belong to R, Pe,.(an)

is also an axiom of E, and therefore -,~Pe,,(an) s a theorem of E, by axiom II.

But a theory in which every recursive set of positive integers is defined isessentially undecidable.16 To show this, let Sb(x) be the Gddel substitution

16 In U.T. only the weaker statement is given (p. 49) that a theory is essentially

undecidable if every recursive function is definable in it.



54 EHILARYPUTNAM

function: that is, the recursive function whose value for any n is the number

of the formula that results when an is put for the variable 'x' throughout

the formula whose number is n. Now suppose that a consistent extension

F of E were decidable. Let Prov(x) be the recursive predicate: x is the

number of a theorem of F. Then --iProv(Sb(x)) is also a recursive predicate,

and is therefore defined in E and hence in F, say as Pb.The formula Pb(x) has a number, say h. And the formulaPb(ah) is prov-

able in E and hence in F if the integer h has the property --iProv(So(h)),

and disprovable if it does not: i.e., the formula is provable if and only if

it is not provable, which yields a contradiction.

PRINCETON UNIVERSITY

putnam. decidability and essential undecidability

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