Download on partially conservative sentences and interpretability

PROCEEDINGS of the
AMERICAN MATHEMATICAL SOCIETY
Volume 91. Number 3. July 1984
ON PARTIALLYCONSERVATIVESENTENCES
AND INTERPRETABILITY
PER LINDSTRÖM
Abstract. A sentence <pis T-conservative over T if T + <pH \p implies T H $ for
every \¡i e F. In §1 this concept for T - 2°+, and Tl°+, is investigated. In §2 results
from §1 are applied to interpretability in theories containing arithmetic.
0. Introduction. Let T be a set of sentences. A sentence <pis T-conservative over a
theory T, <pe Cons(r, T), if T + q>\- \p implies rhif
for every ifef
This
concept was introduced by Guaspari [2] and the basic existence theorems were
established by him and Solovay (cf. [2]). The first example of a result of this type is,
however, due to Kreisel [7] who observed that if ConP is a "natural" normalization
of "P(eano arithmetic) is consistent", then -,ConP £ Consíil?, P). Related results
have also been obtained by Hájek [4], Jensen and Ehrenfeucht [6], and Kreisel and
Levy [8]. Subsequently the results of Guaspari and Solovay were somewhat improved and their proofs were simplified by Hájek [4], Smoryñski [14, 15], and the
author. The general method of proof applied in the papers mentioned as well as in
the present paper is, however, the same, namely a combination of partial truth
definitions and self-reference.
In §1 we present two groups of results. Those belonging to the first group
(Theorems 1-5 and Corollary 1) concern the existence of partially conservative
sentences and partially conservative extensions of theories satisfying various additional conditions. In the results of the second group (Theorems 6 and 7 and
Corollaries 2 and 3) it is shown that a number of sets occurring naturally in the
study of partial conservativity are complete at certain levels of the arithmetical
hierarchy. In §2 results from §1 are applied to (relative) interpretability using the
fact that if S and T are reflexive r.e. extensions of P, then S is interpretable in T iff
every IT° sentence provable in S is provable in T. In particular we answer two
questions raised by Orey [12] and Svejdar [17]. For further applications of partially
conservative sentences to interpretability see [10, 11].
1. Partially conservative sentences. In the following T is either 2°+, or n°+,. f is
the dual of T. A and B are consistent primitive recursive extensions of P. (By Craig's
theorem (cf. [1]), every r.e. set of (T) sentences is deductively equivalent to a
primitive recursive set of (T) sentences.) Th(T) = {<p: T \- <p).We write T \- X or
Received by the editors October 14, 1982 and, in revised form, May 16, 1983.
1980 MathematicsSubject Classification.Primary 03F25, 03F30.
©1984 American Mathematical Society
0002-9939/84 $1.00 + $.25 per page
436
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PARTIALLY CONSERVATIVE SENTENCES
Hito
437
mean that X ç Th(T). S is an X-subtheoryof T, S H x T, if Th(5) n X ç
Th(r). <p'is <pif / = 0 and -,(¡pif i = 1. The diagonal lemma (self-reference) will be
used repeatedly without further comment. For notation and terminology not ex-
plained here see [1].
As is well known, to each T there is a f formula r-true(^) s.t. for every <pe T,
Fr-œ«
r-true(œ).
Let a(x) be a PR binumeration of A and let {T)(x, y) be the formula
Vw <y(wisr
A Prfa(z)vz_x(t/,u)
-> r-true(i/))
(cf. [2, 4, 6, 8, 14]). The following lemma is then easily verified.
Lemma1. {LX*>y) is a T formula s.t.
(i)p \- (rxx, y) a / < y -»<rx*,y'),
(i\)A + q>\- {TX^, m), ally andm,
(iii) //^ is T and A + <ph \p, then there is a q s.t. P + {TX^, q) r- \p.
The next two lemmas serve to unify a number of proofs in what follows.
Lemma 2. Suppose x(x, y) is T. Then there is a T formula £(*) s.t.
(i)A + t(k~)^x(k,fñ),
(ii) ifB -\A, then B + £(k) -\ f B U (x(k, q): q 6 a).
Proof. Case 1. T = n°+1. Let |(x) be s.t.
P r- $(£) - Vy({2°+1}(Ï(F) , y) - *(*, y)).
Then (i) follows at once from Lemma l(ii). To prove (ii), suppose \p is 2°+,, B -\ A,
and 5 + £(rV)r~ #■ By Lemma l(iii), there is a <?s.t.
P+AtKIT),?)!-^.
Hence, by Lemma l(i),
¿, + Yy<iX(*..>')+ -.*!-$(*).
But then, since B + £(&) I- $>it followsthat 5 + Vy < ¿7x(£ >0 •" '/'•
Case 2. T = 2°+,. Let {(*) be s.t.
P V-Z(k) - 3y(^{n«+1}(i(iy,
y) A Vz < y x(k, *)).
The proof that £(x) is as claimed is almost the same as in Case 1.
Lemma 3. Suppose Xo(x> y)is f and X\(x, y) is T. Then there is a T formula £(x)
s.t. for i = 0, 1
(i) A + i\k) hVv< mXj(k, y) -» Xx-AK m),
(ii) if,4 is tandA+
£'(k) h ^', í/¡en^ U {Xi-,(^ <?):? e w) h ¥•
Proof. We need only prove this for T = 2°+1. Let |(x) be s.t.
P h S(£) - 3y((^{no+I}(i(iy,
y) V ^Xo(fc, jO) A
Vz<y({2°+1}(^(¿),z)AXl(¿,z))).
The proof of Lemma 3 is now very much the same as the proof of Lemma 2.
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438
PER LINDSTRÖM
Let us say that y is hereditarily T-conservative over A, y e HCons(T, A), if
<p6 Cons(I\ B) for every B -\A. The following result is due to Guaspari [2] (cf.
also [4, 14, 15]).
Theorem 1. Let X be any r.e. set. There is then a T formula £(x) s.t.
(i) ifk € X, thenA h -, £(£),
(ii) ifk £X, then i(k) £ HCons(f, 4).
Proof. Let p(x, y) be a PR binumeration of a relation /?(&, m) s.t. X = {&: 3m
R(k, m)) and let £(x) be as in Lemma 2 with x(*, y) = -i p(x, y). Then (i) follows
from Lemma 2(i) and (ii) follows from Lemma 2(ii).
A set X of sentences is mono-consistent with T if T + y is consistent for every
ipel(cf.
[9]).
Corollary
* 0.
1. If X is r.e. and mono-consistent with A, then T n HCons(f, A) —X
Proof. Let £(■*)be as in Theorem 1 and let y be s.t. P h <p«-*¿(9). If 9 e A',
then .4 H -1 l(^), whence ^4 I- -, <p.But this is impossible and so if Í X But then,
by Theorem l(ii), y is as desired.
Let Mi" = {y. -,<p£ X} and DCons(r, A) = Cons(r, A) n NCons(f, /I). Solovay
proved that T n DCons(f, A) =*=0 (cf. [2]). This can be improved as follows (cf.
[14, 15]).
Theorem 2. Suppose X is r.e. and mono-consistent with A. Then Y n DCons(f, A)
-(XV
NX)=* 0.
Proof. Let p¡(x, y) be PR binumerations of relations R¡(k, m) s.t. X = {k: 3m
R0(k, m)) and NX = (jfe:3m P,(rc, m)}. Let £(x) be as in Lemma 3 with x,(x, y) =
-, p¡(x, y) and let 6 be s.t. P I- 0 ** £(§). Suppose ieXU
NX. Let « be the least
number s.t. Ro(0, «)orP,(0,
«). Suppose Rt(0, n). Then not P,_,(ö, m) form < n.
(We may assume that R0(k, m) implies not Rx(k, m).) Hence, by Lemma 3(i),
A r- -, ¿'(a), whence y4 I- -, 6', which is impossible since $' £ X Thus 6 £ XV NX.
But then, by Lemma 3(h), 0 £ DCons(f, ¿).
The result of Solovay mentioned above together with Corollary 1 with X = Th(A)
led to the question if Y = n{DCons(r, B): B H A) ■* 0 (cf. [2, 14]). It is easily
seen, however, that this is not true in general. Let y be a T sentence and f af
sentence s.t. P If <p'V \p', / = 0, 1, and let A = P + y A ip. Suppose SeK Then,
since P + 6-*y + 6r-y, we get P + 0 -> <)?I- <pand so P + -, 6 \- y. Similarly
P + 8 \- \p. But then P \- y V \p, contrary to hypothesis.
Our next result will be applied in §2 to answer a question of Orey [12].
Theorem
3. There are sentences y¡ s.t.
9,, -,(<p0 A <¡p,)£ Cons(T, A) -
NCons(n°, A), i - 0, 1.
p
We write A H XB ("/?" for "proper") to mean that A H XB -h XA.
p
Lemma 4. Suppose A ~\ B-r\r\oA. Then there is a sentence x s.t. A -\ noA +
X' H rP, i = 0, 1.
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PARTIALLY CONSERVATIVE SENTENCES
439
Proof. Let 0 be a n? sentence s.t. B \- 0 and A If 0 and let 0, be s.t.
P h- 0. « Vz(Prfa(0V^,
z) * 3h <;zPrfa(0V0,_,,M)),
where < 0 is < and < , is < (cf. [6]). Then, by a standard argument,
(l)Ph0oV0„
(2)a y-e v 0,,
Suppose r U f ç T*. By Lemma 3, there is a sentence x s.t.
(3M+x'r-(0V0,.)^(0V0w),
(4) if $ is T* and A + x'' H <//',then A + (0 V 0,_,) h ^'.
By (1), (2), (3), A ^ noA+ x'. Finally, by (4), A + xrH r*
Proof of Theorem 3. Let tj and 0 be s.t. if <£Cons(n5\ ^) and 0' £ Cons(r,,4),
; = 0, 1 (cf. [17] and Theorem 2). Let \p = n A 0. Then
(1H e NCons(r,/I) - Cons(n?, ,4),
(2) ^ V -, t) g Cons(r, ^).
Moreover v4 H no/l + -, tj. So, by Lemma 4, there is a sentence x s.t.
p
(3) A H no^l +x'Hr^
+-1Î?-
Finally let <p,be ^ V x'. Then, by (2) and (3), <p,e Cons(r, ^) - NCons(n?, A).
Moreover I- (y0 A <p,) «->\p. So, by (1), y0 and <p,are as desired.
The next two results are refinements of the following simple and certainly
well-known observation. Suppose X is r.e., bounded, i.e. X ç T for some T, and
A v X is consistent. Then there is a sentence 0s.t. A V X -\ A + 6 and ^ + 0 is
consistent. This can be improved as follows. Let us say that S is a Y-conservative
extension of T if T -\ S -\ rT.
Theorem 4. Let X be an r.e. set of Y sentences. Then there is a Y sentence 0 s.t.
A + 0 is a Y-conservative extension of A V X.
Proof. By Craig's theorem, we may assume that Xis primitive recursive. Let £(x)
be a PR binumeration of X. Then
(1) P U X\ m h Vz < m(Í(z) -* r-true(z)).
By Lemma 2, there is a Y sentence 0 s.t.
(2M + 0 H |(ip) ^ r-true(cp),
(3) A + 0 H f A V {£(#) -» r-true(^): ? E «}.
From (2) it follows that A + 0 \- X. Suppose \p is f and A + 0 I- 4>.Then, by (1)
and (3),A VXhè. Thus A + 0-\ fA V X.
A closely related result is the following
Theorem
5. Let X be an r.e. set of Y sentences and let Y be any r.e. set s.t.
A V X If \pfor every if/£ Y. Then there is a Y sentence 0 s.t. AVX-\A
every \¡/ e Y.
+ 0W-4>for
Proof. We may assume that X and Y are primitive recursive. Let £(x) an¿ y(x)
be PR binumerations of X and Y, respectively.
Io Let 0 be s.t.
Case1.r = Il2+,.
P h 0 ~ Vy(è(y)
A Vzm *y{r,(z)
- -Wa(x)Wx.f(z,
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«)) - n°+1-true(y)).
440
PER LINDSTRÖM
Suppose \p g Y and A + 0 V-\L. Then P U X \- 0, whence A V X \- ¡p, contrary to
hypothesis. Thus A + 0 If <|*for »//g y But then it follows that ¿ + 0 h X
Case 2. r = 2°+1. Let 0 be s.t.
P h 0 ~ 3y(3z« <y(î,(z)
A Prfa(;c) Vx=,-(z, «)) A
Vz<y(ê(z)-2°+1-true(z))).
The verification that 0 is as claimed is now straightforward.
Guaspari [2] observed that Cons(r, A) is n° and that Theorem 1 implies that
T n Cons(f, A) is not r.e. and suggested the problem of classifying these sets. A
complete solution (and more) follows from our next theorem. Partial results have
been obtained by Hájek [4], Quinsey [13] and Solovay [16] using more complicated
methods (cf. also [14]). Let
Cons(T, y, A) = {y: for every ^ g r, if A + y \- \p, then^ g Y).
Theorem 6. Suppose Y * Yl° and Y is r.e. and mono-consistent with P. Then to any
Ylj set X, there is a Y formula |(x) s.t.
(i) ifk g X, then£(£) g HCons(f, A),
(ii) ifkr£X,r*i
p, then V'r<p£(kr) £ Cons(n°, Y, A) V Cons(2°, Y, A).
To prove this we need the following
Lemma 5. Suppose X and Y are r.e. and Y is mono-consistent with P. Then there is a
2° formula £0(x) anda Yl\ formula £x(x) s.t.
(l)Pr-Uk)^Uk\
(ii)if k £ X, thenP t- !0(£),
(iii) ifkr<£X,r^
Proof.
p, then Vr<p ¿,(*r) £ Y.
Let p(x, y) and o(x, y) be PR binumerations of relations R(k, m) and
S(k, m) s.t. X = {k: 3m R(k, m)) and Y = {k: 3m S(k, m)) and let ¿,(x) be s.t.
P V-Uk) *+V«(x(«) -» 3z < u p(k, z)),
where x(") is the formula
3u < uio(v,
u) A "v is of the form V £,(&,.)" A Vw < uVq 4: p -,p(kq,w)\.
Finally let £0(;t) be s.t.
P h è0(k) « 3z(p(£, z) A V« < z -,x(«)).
Then (i) is immediate. Next we show that for every m, x(m) is false. Suppose x(m) is
true. Then
P I- ¿,()t) -» 3z < mp(k,z).
Moreover there are Acr,r < p, s.t. Vrs¡/) £,(&,.) g Y and P h -, p(/cr, s) for r < /?
and s < m. But then P h- -, Vr</) £,(&,.), a contradiction. Thus x(^0 is false. But
then (ii) and (iii) follow at once, the latter, since we may assume that S(m, n) only if
m < n.
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PARTIALLY CONSERVATIVE SENTENCES
Proof of Theorem 6. We may assume that if y g y and Phip^f
441
then
\f»g y Let R(k, m) be an r.e. relation s.t. X = {k: Vm R(k, m)). By Lemma 5, there
is a 2° formula p0(x, y) and a n° formula px(x, y) s.t.
(l)PrPo(k,m)^
p,(rc, m)L
(2) if P(A:, m), then P h- p0(/c, m),
(3) if not R(k, m), m^p, then VOT<ppx(k, in) £ y.
By Lemma 2, there is a T formula £(x) s.t.
(4)^ + ¿(*) H p0(£ m),
(5) if B -{A, then P + £()t) H f P U (p0(rc,$: <?g w}.
Now (i) follows from (2) and (5) and (ii) follows from (1), (3) and (4).
Corollary
2. The sets Y n Cons(f, A), where Y * II?, Yl°2n Cons(2°, A), and
2° n Cons(2°, A) are complete YY\sets.
If A is 2°-sound, then 11° n Cons(2°, /I) is il°. However, Quinsey [13] has shown
that if A is not 2°-sound, then this set is a complete n" set.
There are results similar to Theorem 6 and Corollary 2 for the sets Y n
DCons(f, A) but at this point they are quite easy and are therefore omitted.
The following corollary is applied in [10].
Corollary 3.IfT*
n°, then Z = {y. 3xpg r n Cons(f, A) (A + \¡>h y)) is a
complete 2° set.
Proof. Clearly Z is 2°. Let X be any 2° set and let R(k, m) be a YY\relation s.t.
X = {k: 3m R(k, m)}. By Theorem 6, there is a Y formula p(x, y) s.t.
if R(k, m), then p(k, m) g Cons(f, A),
if not R(k, m), m < p, then Vm^pp(k, in) £ Cons(f, /I).
Let y¿ = {-ip(Íc, in): m g u). Then, by (the proof of) Theorem 4, there is a formula
H(x) s.t. ^ U Yk-\ A + n(k) -i fA V Yk.It follows that X = (k: -,y(k) g Z> and
so the proof is complete.
Suppose X is r.e. and let Y = {y: A + y \- X). Clearly Y is r.e. unless X is infinite
over A in the sense that A U X\ k If X for every /c.
Theorem
7. Suppose X is r.e., bounded, and infinite over A. Then Y = {y:
A + y \- X) is a complete Yl° set.
This was proved independently by Christian Bennet. In fact, the proof below is
due to him and is presented here with his permission.
Proof. As usual we may assume that X is primitive recursive. Let £,(x) be a PR
binumeration of X. Let n be s.t. X ç n°+1. Next let Z be any Yl° set and let
R(k, m) be an r.e. relation s.t. Z = {k: Vm R(k, m)). By Theorem 1, there is a
formula p(x, y) s.t.
(1) if R(k, m), then ,4 I- p(k, in),
(2) if not R(k, m), then -,p(/V, in) g Cons(2°+2, A).
Let t)(x) be the formula
Vz{£(z) AV«<zp(x,u)
-* n°+1-true(z)).
It sufficesto showthat Z = {k: v(k) g Y). If fc£ Z, then, by (1), /I + i}(*)H f(f)
-» n°+,-true(9)
for every y, whence r}(k) g Y. Next suppose k € Z. Let m be s.t.
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442
PER LINDSTROM
not R(k, m). Then A V Xf m + -, p(k, in) h v(k). Let 9 £ X. If A + v(k) h 9,
then .4 + -, p(iV,m) \- -X[ m -* 9, whence, by (2), ^4 U Xf m h 9. But, X being
infinite over A, this fails for some y g X It follows that t}(íV)£ y
2. Applications to interpretability. We write 5 < T to signify that S is interpretable
in T. Most applications of partially conservative sentences to interpretability are
based on the following (cf. [2, 3, 9])
Lemma 6. If S, T are r.e. reflexive extensions of P, then S < T iff S H noT.
For brevity we assume from now on that A and B are essentially reflexive.
From Corollary 1 and Lemma 6 we get at once the following result essentially due
to Hájek [3] (cf. also [5, 9, 14, 15, 16]).
Theorem
8. If X is r.e. and mono-consistent with A, then there is a 2° sentence y
s.t. A + y < A and y £ X
Similarly, Theorem 2 yields the following (cf. [9])
Theorem 9. If X is r.e. and mono-consistent with A, then there is a sentence 6 s.t.
A + 0' ü A and 0' £X, / = 0, 1.
Theorem 8 was originally proved to give an example of a sentence y s.t.
ZF + y < ZF and GB + y £ GB and this follows at once, since {y: GB + y < GB}
is r.e., GB being finite and mono-consistent with ZF. Similarly, Theorem 9 yields a
sentence 0 s.t. ZF + 0' «s ZF and GB + 0' =£GB, i = 0, 1.
Our next result, which is an immediate consequence of Theorem 3, answers a
question raised by Orey [12].
Theorem 10. There are sentences y¡ s.t. A + 9,< A, A + y0 A 9, < A, A +
-, 9, ^ A, and A + -, y0 V -, 9, sg A, i = 0, 1.
Let A = P mean that /I < P < /Í. Then from Theorem 4 we get ((i) is proved in
[9])
Theorem 11. (i) If A H B, then there is a sentence 0 s.t. A + 0 = B.
(ii) // X is an r.e. set of 2° sentences, then there is a 2° sentence a s.t.
A + a = AV X.
The following corollary answers a question suggested by Svejdar [17].
Corollary
4. There is a 2° sentence o s.t. A + o * A + $ for every Yl° sentence
Proof. There is an r.e. set X of 2° sentences s.t. A V X ^ A V X\ m for every m
(cf. [9]). By Theorem 11(ii), there is a 2° sentence a s.t. A + a * A V X. Let \p be
any n° sentence and suppose A + o = A + \¡/. Then, by Lemma 6, there is an m s.t.
A V X\ m \- \p. But then /lUX<,4
+ a</l+i/'-|,4UXrm,
whence A V X <
j4 U Xf m, a contradiction.
The following result which was first proved by Solovay [16] (cf. also [2, 9, 14])
answers a question of Hájek [3]. It is an immediate consequence of Lemma 6 and
Theorem 6 with Y = Th(P).
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PARTIALLY CONSERVATIVE SENTENCES
443
Theorem 12. If A < B, then 2° n (9: A + y < B) is a complete n° set.
If there is a n° sentence 0 s.t. B \- 8 and P < A + 0, then, by Lemma 6, {9:
P < A + 9) is r.e. In contrast to this we have (cf. [9])
Theorem
13. If there is no IIo sentence 0 s.t. B f- 0 ¿md P < ^ + 0, i/zen {9:
P < ^4 + 9} is a complete Y[°2set.
Proof. By assumption, the set of n0 sentences provable in B is infinite over A.
Moreover P < A + 9 iff A + 9 h- X Now apply Theorem 7.
Suppose e.g. a is as in Corollary 4. Then, by Theorem 13, {9: A + a < A + 9} is a
complete n° set.
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Department of Philosophy, University of Göteborg, Göteborg, Sweden
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