Provability of the Pigeonhole Principle and the Existence of Infinitely Many Primes
Author(s): J. B. Paris, A. J. Wilkie and A. R. Woods
Source: The Journal of Symbolic Logic, Vol. 53, No. 4 (Dec., 1988), pp. 1235-1244
Published by: Association for Symbolic Logic
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THE JOURNAL OF SYMBOLIC
LoGic
Volume 53, Number 4, Dec. 1988
PROVABILITY OF THE PIGEONHOLE PRINCIPLE
AND THE EXISTENCE OF INFINITELY MANY PRIMES
J. B. PARIS, A. J. WILKIE, AND A. R. WOODS
In this note we shall be interested in the following problems.
Can IlO H-Vx3y > x (y is prime)?
Here IzJOis Peano arithmetic with the induction axiom restricted to bounded
(i.e. z1O)formulae.
PROBLEM 2. Can IlO H-z10PHP?
Here /1OPHP(,AOpigeonhole principle) is the schema
PROBLEM 1.
Vx < z3y < zO(x,y) -+ 3x1, x2 < z3y < z(x1 $ x2
A O(x1,y) A 0(x2,y))
for 0 E JO, or equivalently in IJlO,for a /O formula F(x, y)
- [Vx < z + 13!y < zF(x,y)
A VX1, X2 < ZVY
< z[F(x1,y)
A F(x2,y)-
X1 = X2)]
written -i3zF:z + 1 "-- z.
By obtaining partial solutions to Problem 2 we shall show that Problem 1 has a
positive solution if IJlOis replaced by IJlO+ Vx xlo9(x)exists.
Our notation will be entirely standard (see for example [3] and [4]). In particular
all logarithms will be to the base 2 and in expressions like log(x), (1 + ?)x, etc. we
shall always mean the integer part of these quantities.
Concerning Problem 2 we remark that it is shown in [5] that for k e N and F e J0,
IJlOH 3z F: log(z)k + 1 "-- log(z)k.
As far as we know this is the best result of this form, in that we do not know how to
replace log(z)k by anything larger. However, as we shall show in Theorem 1, we can
do much better if we increase the difference between the sizes of the domain and
range of F.
In what follows let M be a countable nonstandard model of IJO, and let A'/ be
those subsets of M defined by JOformulae with parameters from M.
THEOREM 1. For k e N and F c-0,
I/1O H-VX(X,,gk(x)
Here log?(x) = x, logk +I (x)
exists and x > 1 -+ -iF:
x2 I_+ x).
= log(logk(x)).
Received June 11, 1986; revised August 10, 1987.
? 1988, Association for Symbolic Logic
0022-4812/88/5304-001 8/$02.00
1235
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1236
J. B. PARIS, A. J. WILKIE, AND A. R. WOODS
PROOF. To simplify matters, consider first the case k = 1. So assume M k=a Wg(a)
exists and F: a22l-4 a with F E za1 and a > 1. The idea of the proof is the following.
Suppose ab E M and G: 2b -+ a with G E AJ'. Think of G as a binary tree of height b
with each tip a associated with G(cr)e.g. for b = 3
000 001 010 011 101 100 110 111
00
A
G(000) G(001) ...
\
/
1
0
g/\
G(110) G(111)
/~~
;
/
0
Now associate 00 with F(G(OOO)
+ aG(001)) < a, 01 with F(G(010) + aG(011))
< a, 0 with F(F(G(OOO)
+ aG(001)) + aF(G(010) + aG(011))), etc. and so on down
the tree so that finally 0 gets associated with some value c < a. But clearly, since F is
one-to-one, G uniquely determines c and conversely. Of course this was all clear
for the case b = 3, and we now make this argument precise for general b such that
ab E M.
Fix b such that abE M, and for c < a and d < b define H d C 2d
-+
a by
Hj(of)= e t 3 sequence eo, el,..., ed< a
such that eo = c & ed = e
& VO< i < d[a(i - 1) = 0 -* 3x < a F(e, + ax) = ei-1]
& [a(i - 1) = 1 3-+ x < a F(x + aej) = ej- 1]
where we identify a < 2" with the map from d to 2 which sends x < d to 0 iff the
coefficient of 2x in the binary expansion of a is 0. Recall that since the graph of
exponentiation is JO (in 1L/O,see [1]) and since the sequence eo, el,...,ed can be
coded by a number to the base a of size <a d+l < ab+l we see that H d has a AO
definition in M, uniformly in a, c, d, and F.
Claim. If Gei mand G:2b +athen3c < aH = G.
Proof of Claim. We prove by induction on d < b that
Va < 2b3c
< aVT< 2 Hd(r) = G(c v)
where ant is the (code for the) map from b to 2 formed by concatenating the maps
coded by a and v. Notice that this formula is JO, so IzLO
suffices.The proof is entirely
straightforward since if
Hd7'(v)
=
G(&O
0t),
H7l'(r)
=
G(
1 r) for
<2b-d
and 2v<
then H~d(v)= G(ufv) for v < 2", where c = F(co + acj).
Having proved the claim, it is now easy to obtain the required contradiction, for
on taking b = 1 + log(a) we see that c < a H-*Hb gives us a map from a onto the set
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1237
PROVABILITYOF THE PIGEONHOLEPRINCIPLE
of all '1Mmaps from 2log(a) +1 to a. But the usual diagonal argument now works, since
the map sending x < 2log(a)?+1 to 1 Hmnin(x,a-1)(X)is in 4[ but cannot equal HI for
any c < a.
This proves the result for k = 1. Now suppose k = 2, i.e. only (a1g2(a) exists) holds
in M. Then we can again prove the above claim with b = 2 + log2(a). This means
that we have a map H" E JO such that for c < a, the maps HI enumerate all z1Mmaps
from 22+ log2(a) to a and in particular from log(a) + 1 (< 22 + log2(a)) to a. We can now
apply the same method of proof as above to obtain a contradiction as above, with
the only change that instead of coding the sequence eo, e1,..., ed < a by a single
number less than a' +Ig(a) we can code it using the enumeration
H2+Iog2(a)
Similarly for the case k = 3 we repeat this process twice, and so on.
Under the assumption that Vx, xlog(x) exists we can now prove a strengthened
version of the pigeonhole principle of Theorem 1. Precisely,
COROLLARY 2. For F E z0 and rational ? > 0,
IAo + Vx(xlog(x)exists) H m 3x, F: (1 + ?)x
PROOF.
1H+
x and (1 + e)x
>
x.
Assume that Vx(xlog(x)exists) holds in M and that
Fe Am,F: (1 + e)a
H+
a and (1 + e)a > a, a c M.
Then a must be nonstandard. Clearly F will give us a Jm map from (1 + e)a
+ ea H+ (1 + e)a, etc., so without loss of generality we may assume ? = 3.
Now let G E-,Am G: a2 lo a2 be defined by
G(4ab + c) = ab + F(c),
where 0 < c < 4a.
Then G(x) < x/2 for 4a < x < a2 and G(x) < a for x < 4a. Hence
Glog(a)+l1
a21bbe max(4a, a2/2log(a)+ 1) < 4a,
so GI(a)+:
a2 F-*a. Also, since (a2)Iog(a)+2 E M, GIo(a)+2 is definable in M by a /1
M
formula, and the result now follows by Theorem 1.
We do not know whether the assumption Vx(xlog(x)exists) in Corollary 2 can be
weakened to, say, Vx(xIOg2(x) exists).
We now use these methods to prove
THEOREM 3. If IJo is finitely axiomatizable, then for F E oA,
IA0 --i3x > 1 F:X2 IX.
Of course the finite axiomatizability of IAo is open and may well be false.
However, the method of proof of Theorem 3 is perhaps more important than the
theorem itself. Theorem 3 will be a consequence of a series of other results which we
will prove first.
THEOREM,4. Supposeat E M, F E JAmand F: al -+ a. Then there is an end extension
K of M (K DeM) such that Kk IAOand aY2 E K.
PROOF. We may obviously assume that a 2 0 M.
As in Theorem 1 we can find a function H E AJmsuch that {Hc Ic < a} enumerates
all AO maps from 2" into a. The idea now is to treat c < a as the "number"
Ei < 2y HCi)-* a'.
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1238
J. B. PARIS, A. J. WILKIE, AND A. R. WOODS
With this in mind we define for q, e, c, d < a and j < 2"
c
Gd= e
qtc
=
t 3y < aVi < 2[HY(O)=
& (i + 1 < 2
0
HY(i)+ HI(i) + Hd(i) = He(i) + a* HY(i+ 1))
& (i + 1 = 2-+
HY(i) + HI(i) + Hd(i) = He(i))],
e [i.e. the "sum"of qaj copies of c is e]
3y < aVi < 27[(i < j
HY(i)= He(i) = 0)
& (j < i + 1 < 2HY(i) + q *Hj(i-j)
+ a HY(i +
= He(i)
))
= He(i))
Hy(i) + q *HFi(-j)
& (2Y-j < i < 2 >- Hj(i) = 0 v q = 0)],
& (j < i + 1 =2Y
c 0 d = e t 3y < aVi < 2"[HY(O)= Hc(O)*d & e = Hy(2y - 1)
& (1 + i < 2-+ HY(i+ 1) = HY(i)E (Hj(i)!'d))].
Now let y2 E I ce 2 with I closed under addition and let
K = {c < a
t e I Vt < i < 2 HI(i) = O}.
Then K is closed under G, 9 and satisfies c ? d E K = c E K where for c, d < a,
c ? d - 3e < a c G e = d.
Hence we can treat K as a structure for the language of arithmetic (with less than) by
restricting G, X, ? to K and identifying c, d E K if Hc = Hd. Then K satisfies
Peano's axioms less induction, and furthermoresince e, 0, 9 E z1mwe have that if
A c K is in A' then A = K n Bfor some B c a B EcAm.By the next theorem then it
follows that K l= I'o.
Also, up to isomorphism, M ce K. To see this let q E M, q = 0%qjai to base a.
Then m < y2, so m E I and there is c < a such that
HJ0i)=
qj for i < M,
CkJ0~
form~i<27.
Then identifying q with c gives the required embedding of M as an initial segment
of K.
It only remains to show that aY2 E K. Clearly we could try to show that, in K,
c = a 2, where c < a satisfies
H~i) = {
for i
7& y2.
However a more economical proof is to argue as follows: Certainly K is a proper end
extension of M and the hypotheses of the theorem still hold for K. Hence if a,2 K
we could repeat the construction to get K(2), etc. Repeating a) times would give
K40), an A)-like end extension of M. But then K("1) 1=Io + collection schema; so,
by results in [2], K("1) 1 Peano axioms and aY2E K("1), giving aY2E K(') for some
2
ot < (W.
L1
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PROVABILITYOF THE PIGEONHOLEPRINCIPLE
1239
It should be clear that with a little more effort we could have obtained a21 E K.
Repeating this would then give
'
COROLLARY 5. Suppose at E M, F E A and F: a H+ a. Then there is an end
extension K of M such that K IAo and {a Ia' E K} is closed underexponentiation.
The proof of Theorem 4 lacks the following fact.
Let a E M and suppose that ? is a partial orderingof a, ? E A'M and
c
$X
a, X EiA'. Then
THEOREM 6.
0
PROOF.
< x A y 0 x).
3xeXVyyeX-i(y
We show by induction on i < log(a) + 1 that
3j < 2i[Vy e X3x E X(ja/2' < x < (j + 1)a/2 & x ? y)]
& -' [Vy e X3x e X(x < ja/2 & x ? y)].
Now take i = log(a) + 1. Then there is at most one x between ja/2' and
O
(j + 1)a/2', and this must be a minimal element of X.
fact:
3
we
need
the
following
well-known
Before giving the proof of Theorem
THEOREM 7 [see for example Paris and Kirby [2]]. Let M ce K and K l= IA1.
Then M F BE1, whereBE1 is the Z1-collection schema,
Vx <y3zO(x,y,z) -+3tVx <y3z < tO(x,y,z), for 0 E .lp
Suppose M P- Vx < a3zO(x,z) with 0 E Xl, say 0 = 3ui4(x,z, ui) with 4
M < beK. Then, since M ceK, M-<40K; so K:= Vx < adz < b3uLet
ezAO.
< bj(x, z, ii). By IA/Oin K, let b' be the least b such that
PROOF.
l= Vlx< adz < b3u-< bo(x, z, U-).
~~~~K
(*)
If b' > M then by the previous reasoning (*) would hold for b' - 1, since b' - 1
would still exceed every element of M, contradicting the choice of b'. Hence b' E M
so, since M -<40K,
M
F-
Vlx < adz < b'3u- < b'o(x, z, U'),
and hence M 1= Vx < adz < b'0(x, z) as required.
Now at last we shall give the
PROOF OF THEOREM 3. Assume that IAo is finitely axiomatizable and that MO
A 1 < a E MO, F E Lof and F: a2 u-+ a. We shall derive a contradiction.
Ido,
By taking an ultrapower if necessary we may assume that a" E MO for some
nonstandard y and that y = 2' for some A. Also we may assume that the only parameter c appearing in the AOdefinition of F is less than ay.
Now let M be the substructure of MO with domain {x E MO Ix < any for some n
E N}. Then M l= IAol F E A', F: a2 - a in M, etc. Working in M, define G: atYl
aY
as follows. Given b < ay, let b = Yi<2;-, bia2i to base a2, and set
G(b) =
F(bi)a' < al'.
E
i<2A-
1
Set H = G'. Then H: aY w-4 a and H EcAM' (with parameters a, aY, y, c) since for b
the sequence b, G(b), G2(b),... ., G'(b) can be coded by a number < a2Y.
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< aY
1240
J. B. PARIS, A. J. WILKIE, AND A. R. WOODS
Now let {00, ,01 .., Ok- 31} IAo be the assumed finite axiomatization of IAo, and
write these together with the statements
- 3x(x
H: a" i-4 a,
=
ay2),
a" is a to the power y
in prenex normal form as
Vyo 3x1 < t1 Vy1 < sl 3X2 < t2 VY2 < S2 ... 3Xm < tmVYm < Sm oji
where
/j = OjIj(x, y,
j < k,
a, y, c, a') is quantifier free, and
ti = tj(x1,*
, xi- 1, Yo, ... . ,y - 1, a, y, c, ay)
and si = si(x,.
. .,xi, yO,. . .Yi - 1, a, V,c, a') are terms.
We now define a function Q on N as follows. For q e N if
q
= <0KKelul,...,um5w0,W
with j < k and 1 < e < m, set Q(q) to be the least
Ye
and
Xe<te
...,wm>>
xe
< t+..
<S3Xe+1
such that
3Xm < tnVYm < S4r
where
07=
IA(Q(ul),.
.
.,Q(Ue
I=
ti(Q(u1), * **,
Q(U-
i=
si(Q(UJ), * * * ?(Ue-,)
if such an
1), Xeo.. .,Xm, ?2(w) .. ., Q(We 1), Ye' .. Ymya,V, c, a ),
e -),
,
xe
Xe, ...,
Xi - 1, Q(wo),***,
Xe ...**
X? 2(wO),***, Q(We- 1), Ye* ... * yY1,
Q(We -.1), Ye **.
y j-,
a, y, c,
ar),
a, y, c, a )
exists, and 0 otherwise. If
q=<l,u>
set?Q(q)=0,
q = <2,u>
set Q(q) = 1,
q=<3,u>
set Q(q)=a,
q = <4,u>
set Q(q) = y,
q = <5,u>
set Q(q) = c,
q=<6,u>
setQ(q)=a",
q = <7,<UU2?>>
q = <8,<ui,u2>>
set Q(q) = Q(UJ ) Q(U2)
set Q(q) = ?(ul) + Q(u2),
and otherwise set Q(q) = 0.
Let J = Q"N. Then J is a substructure of M and an elementary substructure for
all subformulae of 00,..., Ok- 3 9 H: a" H+ a, 7 3x x = ay2, and a" is a to the power y.
Hence
J l= Iz1o + H: a" H+ a + 7 3x(x = ay2) + a" is a to the power y.
Hence J = IBo + H: a" "-+ a + - 3x x = ay2; so, by Theorem 4, J has a proper end
extension to a model of IAo. Hence, by Theorem 7, J l= BZ1.
We now obtain a contradiction by showing this to be false. First notice that there
is a Xl formula which, for n e N, expresses ?(n) = X. Furthermore, for n e N and
x e J, J l= ?(n) = x
M - ?(n) = X. Hence
J l= Vx < a3n3b[b = a"' & Q(n) = x].
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PROVABILITYOF THE PIGEONHOLEPRINCIPLE
1241
But clearly there are infinitelymany x in J less than a; so b cannot be bounded by any
element of J since {any I n E N} are cofinal in M and hence in J. [The fact that at is the
same in both J and M ensures that the any are also the same.] So BZ1 fails in J and
O
Theorem 3 is proved.
Theorems 1 and 3 give a partial answer to our Problem 2. Corollary 2 allows us to
give a partial answer to Problem 1; precisely,
E
COROLLARY 8. IJOH-Vx(xlo9(x)exists -+ 3y > x(y is prime)).
and
This corollary is immediate by earlier results
the following Theorem 11.
Before giving a proof of Theorem 11 we shall need two results on coding which are
of independent interest. The first seems to be quite well known in many disguises.
THEOREM 9. Let a E M, d < log(a)k, and b < 21o(a)' with a a standard rational, 0
<a < 1. Let G E Al', G: M2 -+ M. Then there is a function F E-Am (uniformly)such
that F(O) = b and for i + 1 < d
F(i + 1) = min(G(F(i), i), 2109(a)a)
PROOF. As usual we suppress mention of further parameters in G and F except
where they are necessary to the argument. Clearly we may assume that a is large and
that G:
, 210g(a)x
42
We first prove the result for all such G with k replaced by (1 - a)/2. In this case a
suitable definition of F(i) for i ? log(a)(1-a)/2 is
F(i)
=
z
3
sequencewo, w1,..., wi such that wo = b & wi = z
& O< j < i(wj + 1= G(wj,j)).
'
ThisgivesF E A sincethe sequencewo,wl,..., wican be coded,to base21og(a)',by a
number less than
(2log(a)c) 2 + log(a)(1- X)/2
a.
<
Now assume the result (for all such G) with k replaced by /3.We show how to replace
/3by /3+ (1 - a)/2. Clearly this will give the theorem.
So for such a G let Gy, be defined by
Gy w(x,u) = G(x, y + u)
and let Fy w be the Jm function (uniformly in y and w) guaranteed to satisfy, for
u < log(a)f,
Fy,(O) = w,
Fy w(u + 1) = Gy w(Fy w(u), u).
Then a suitable definition of F(i) for i < log(a)f log(a)(1-x)/2 is given by
.
F(i) = z
3 sequences
yo =
yo, Y1,. w.yj.
& yj = i &80
& j < log(a)(1 -)/2
, w1 ... ., wj such that
< t < j(O < y,+, - y < log(a)fl)
& Wo = b & wj = z
& VO < t < j, Fytwt(Yt+,)
= Wt+i.
This definition shows F E Am, since the sequence Yo, Yl'..., yj can be coded by a
< a and the sequence
number < (1 - log(a)fl+(i -a)/2)1 +log(a)(l0-)/2
wO,w1, . . .,wj can
E
be coded by a number < (1 + 210o(a)a)2+log(a)(1-a)/2 < a.
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J. B. PARIS,A. J. WILKIE,AND A. R. WOODS
1242
E
THEOREM
10. Let a,b
-Mg d < log(a)k and G Am, G: d -+ b. Then there is a
such that F(O) = G(O)and for all i < d, F(i + 1) =
function F E-A' (uniformly)
F(i) + G(i + 1).
PROOF.Here and hereafter we shall denote F(i) by the more descriptive notation
q< i G(q).
First notice that by Theorem 9 the result is true if b < d. We shall use this in what
follows without further mention.
Let G(i) = Es<log(b) Gs(i) *2' (in binary as usual). The intention now is to define
functions ps, HI E Am such that for 0 < j < d and t < log(b),
Gs(i)2)
E (E
i<j s<t
=
s<t
Ps()
* 2s
+ 2t * Ht(j)
with ps(j) E {0, 1}. To this end define for 0 < j < d
E Go(i)
i<j
Ho(j)=
Ht+1(i)
=
min ([Htj)1
+ E Gt+ 1(i), (t + 1)d)
=HtRj] +
Gt+1(i)
E
(since E<5jGt(i) < j < d).
Then H E A' ' by the previous theorem.
Now set
AU) =O
}I
if Ht(j) is even,
if Ht(j) is odd
for j < d and t < log(b). Then Pt(O)= Gt(O)and for 0 < j < d we can easily show by
induction on t < log(b) that
Z
Z
Gs(j + 1)2S = 2tHt(j + 1) + E ps(j + 1)2
ps(j)2S +
s<t
s<t
s<t
Putting t = log(b) then gives the required function F, i.e.
2tHt(j) +
F(j) = 2 O(b)Hjog(b)(j) +
Z
ps(j)
22*
s <log(b)
In what follows we shall use Theorem 10 without mention.
Let a e M and suppose that there is no primein M betweena and all.
Then there is F e Atm,F: 9alog(a) i-4 8alog(a).
THEOREM 11.
PROOF. Clearly we may assume a is large. Before proceeding with the main proof
we need to show that certain functions are inA1m. Precisely, notice that we can show,
by induction on y e M, that
3z < y [if p is the least prime dividing y then z = 0 mod p and
if p < q are primes dividing y and there is no prime r, p < r < q,
dividing y and z = s mod p then z
=
s + I mod q].
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1243
PROVABILITYOF THE PIGEONHOLEPRINCIPLE
Furthermore any two such z must be congruent mod p for each prime p, p Iy. Hence
we can unambiguously talk of the number of primes dividing y, v(y), and for i < v(y)
of the ith prime dividing y; and these relations are inA1'. Also using IAowe can show
that v(y) < log(y) for all y.
So now assume there are no primes in M between a and a". We shall describe a
function H E A'm
H: a log(a'0) o-4(1 + [a/2])log(a'0) + 2a log(a).
The requiredF then follows easily using well-known properties of log (which hold in
M) and the fact that a is large. Let x < a log(a'0). We describe H(x).
Let (i - )log(a'0) < x < ilog(a'0) and al' + i = lj<lPpj, where v = v(al' + i)
and p. is the jth prime dividing a10 + i. We now consider two cases.
Case 1. 0 < j < v such that for all 1 < t <apa + 1f a 10+ t and i is minimal such
that i ? 1 & pjj Ia'0 + i. In this case pick the least such j and map x to
2alog(a) + [(1 + pj)/2]log(al0) + x - (i - 1)log(al0).
In other words we map the block [(i
[2alog(a) +
-
1)log(al0), i log(al0)) linearly into the block
2 Pj]log(al0), 2alog(a) +
I + I + P])log(a10)].
Notice that sincepj < all, pj ? a; so [(1 + pj)/2] < [a/2] + 1.
Case 2. Not Case 1. By induction on y we can show that if y = Hj<t
qb,
where q
'is the ith prime dividing y, then
log(y) < E bi(1 + log(qi)) < 2 log(y).
(*)
i<:
So let j < v and k < ej be such that
do = (i - 1)log(al0) + E e(1 + log(p.)) + k(1 + log(pj))
m<j
< x < (i-)log(a'0)
+ E em(1+ log(pm))+ (k + 1)(1 + log(pj)) = d1.
m<j
Now let s, 1 < s<x, be such that, for some b, pI Ia'0 + s and VI < t < a,
pl+'ta'O
+ t and VI ? t < s, pj{a'0 + t. In other words for this prime pj Case 1
applies to a'0 + s. We now consider two subcases.
Subcase 2a. s < i. Then, since ej < b, pj a0 +i-(a0
+ s) =i-s.
Let i-s
-m
,<tqcm,where qmis the mth prime dividing i - s. Suppose qr = pj. Then by (*)
2(i - s - 1)log(a) + E Cm(1+ log(qm))+ (k + 1)(1 + log(q,))
m<r
< 2(i
-
s - 1)log(a) + 2 log(a) = 2(i - s)log(a)
and we linearly map the interval [dodj) (which includes x) onto the half-open
interval with end points
2(i
-
s - 1)log(a) + E cm(1 + log(qm))+ k(1 + log(q )),
m<r
2(i - s - 1)log(a) + E Cm(1+ log(qm))+ (k + 1)(1 + log(qr)).
m<r
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1244
J. B. PARIS, A. J. WILKIE, AND A. R. WOODS
Notice that this latter interval determines q, and k, hence s and i and hence [do, d1).
Subcase 2b. i < s. Notice first that the largest i for which Subcase 2a applied (for
the same k) was
S + [a]pi
In this case, then, we proceed as in Subcase 2a but with i - s replaced everywhere
by
S -
i + [~
PJ
>
(S
_
+
P -S)
Again the interval we map [do, d1) into will determine pjand k, hence s and hence i
and [dod).
Considering this, we see that the map H we have constructed is one-to-one, and
E
since it is clearly A ' the theorem follows.
REMARK. By using very delicate work of the last named author (see [6, Theorem
4.5]) this can be improved to
THEOREM 12. IAO+ Vx(x1?o(x) exists) F- Sylvester'stheorem,whereSylvester's theorem states that if 1 < x < y then some numberamongst y + 1, y + 2,.. .,y + x has
E
a primedivisor p > x.
REFERENCES
[1] H. GAIFMAN and C. DIMITRACOPOULOS, Fragments of Peano 's arithmetic and the MRDP theorem,
no. 30,
Logic and algorithmic (Zurich, 1980), Monographies
de l'Enseignement
Mathematique,
Universite de Geneve, Geneva, 1982, pp. 187-206.
[2] J. PARIS and L. KIRBY, Zn-collection schemes in arithmetic, Logic Colloquim '77, North-Holland,
Amsterdam, 1978, pp. 199-209.
[3] J. PARIS and A. WILKIE, Counting problems in bounded arithmetic, Methods in mathematical logic
(proceedings of the sixth Latin American symposium on mathematical logic, Caracas, 1983), Lecture
Notes in Mathematics, vol. 1130, Springer-Verlag, Berlin, 1985, pp. 317-340.
[4] A. WILKIE and J. PARIS, On the scheme of induction for bounded arithmetic formulas, Annals of Pure
and Applied Logic, vol. 35 (1987), pp. 261-302.
[5] J. PARIS and A. WILKIE, Counting AOsets, Fundamenta Mathematicae, vol. 127 (1987), pp. 67-76.
[6] A. WOODS, Some problems in logic and number theory and their connections, Ph.D. thesis,
University of Manchester, Manchester, 1981.
DEPARTMENT OF MATHEMATICS
UNIVERSITY OF MANCHESTER
MANCHESTER M13 9PL, ENGLAND
INSTITUTE OF MATHEMATICS
OXFORD OXI 3LB, ENGLAND
DEPARTMENT OF MATHEMATICS
YALE UNIVERSITY
NEW HAVEN, CONNECTICUT 06520
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All use subject to JSTOR Terms and Conditions