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INTERPRETATIONS, ACCORDING TO TARSKI
Harvey M. Friedman*
Nineteenth Annual Tarski Lectures
Interpretations of Set Theory in Discrete Mathematics and
Informal Thinking
Lecture 1
Delivered, April 9, 2007
Expanded May 24, 2007
1.
2.
3.
4.
5.
6.
7. Interpretations.
Tarski Degrees.
Adequate Sentences and Relative Consistency.
Predicative Extensions and Relative Consistency.
P degrees.
Infinite Theories.
Observed Linearity. Special thanks to Albert Visser for considerable input. 1. Interpretations.
The notion of interpretation was first carefully defined
and developed in the book [TMR53].
The notion of interpretation is absolutely fundamental to
mathematical logic and the foundations of mathematics. It
is also crucial for the foundations and philosophy of
science  although here some crucial conditions generally
need to be imposed; e.g., “the interpretation leaves the
mathematical concepts unchanged”.
The most obvious and direct use of interpretations is for
relative consistency.
PROPOSTION 1.1. Suppose S is interpretable in T. If T is
consistent then S is consistent.
But the notion was mostly used by Tarski to show that
various formal systems are undecidable in the sense that
there is no algorithm for determining whether a sentence in
its language is provable.
PROPOSITION 1.2. Suppose S is interpretable in T. If T has
a consistent decidable extension in its own language, then
S has a consistent decidable extension in its own language.
I.e., if S is essentially undecidable then T is essentially
undecidable. 2 Rafael M. Robinson in [Ro52] set up a crucial very weak
system of arithmetic, Q, and showed that Q is essentially
undecidable.
Proposition 1.2 was successively applied, starting with Q,
invoking transitivity: S interpretable in T, T
interpretable in W, therefore S is interpretable in W.
We now define interpretations in the setting of ordinary
first order predicate calculus with equality.
There are some choices along the road  the notion is still
not entirely standard. We will work with the most liberal
natural and convenient notion.
A theory T is a set of sentences together with a relational
type, L(T). Here we won’t demand that we close T under
logical consequence.
We begin semiformally and semantically, with several
details postponed.
p is an interpretation of S in T iff
i. p specifies a domain of objects from viewpoint of T.
Given by formula(s) in L(T).
ii. p specifies an interpretation of all constant, relation,
and function symbols of L(S), on the domain in i, by means
of formulas in L(T).
iii. This data induces a map p from formulas of L(S) into
formulas of L(T). We require that for all j Œ S, T  pj.
We have hidden details. We take the most liberal natural
view of what is allowed; i.e., “liberal” definability.
The specified domain D consists of tuples of any lengths ≥
1. Mixed lengths are allowed. There must be a bound on the
lengths used. D is to be given by a finite set of formulas
in L(T), one for each length allowed. In every model of T,
D is nonempty.
We have a binary relation E on D, given by a finite set of
formulas in L(T), one for each pair of lengths allowed. 3
In every model of T, E is an equivalence relation on D.
We have formulas in L(T) that interpret the constant,
relation, function symbols of S, on D/E. We don’t perform
factoring, but instead treat the interpreted constants,
relations, functions as relations on D that respect E, in
all models of T.
We have lied a little: we will allow parameters. I.e.,
extra variables in the above, where we merely say that in
any model of T, there exist choices for these variables
(parameters) so that all of the above hold.
We are asserting that there is a way of defining a model of
S inside any model of T, uniformly, multidimensionally, via
equivalence, and with parameters. The definition is
required to be uniform but the choice of parameters has no
uniformity requirement. There may be many choices of
parameters that work. We call this liberal definability.
This definition is particularly well behaved in the case
where S is finite. To begin with, since S is finite, the
data used and the requirements made in the definition are
finite before we existentially quantify over choices of
parameters. Therefore we are asserting that T logically
implies a finite set of sentences. Hence we can replace
these logical implications by provability in T. This gives
us the provability version of (liberal) interpretability.
The following result shows a great deal of robustness. On
the other hand, we can remove any requirements of
uniformity.
THEOREM 1.3. A finite S is interpretable in T if and only
if for all M = T, some M’ = S is liberally defined in M.
Proof: The forward direction is obvious. For the reverse
direction, we use compactness and completeness.
We claim that M’ can be liberally defined in M, uniformly,
up to a choice of parameters. I.e., there is a fixed
presentation of L(S) data in L(T) such that for all M = T,
there is some choice of parameters from dom(M) such that
the data constitutes a model M’ = S.
Suppose this is false. Then for any such presentation,
there exists M = T such that for all choices of parameters
from dom(M), the presentation fails to define a model of S. 4 We then conclude that for any finite set of such
presentations, there exists M = T such that for at least
one of these presentations, for all choices of parameters
from dom(M), the presentation fails to define a model of S.
For if we had a finite set of such presentations without
this property, then we could get a single such presentation
without this property through definition by finitely many
cases.
Let X be the infinite set of sentences asserting T, and
that for each of the countably many presentations, and all
choices of parameters, the presentation fails to define a
model of S. By hypothesis, we know that X is not
satisfiable. Hence some finite subset of X is not
satisfiable. This finite subset of X provides a
counterexample to the previous paragraph. QED
The case of finite S,T (in finite relational types) is
fundamental. We can create unimpeachable certificates of
interpretability: the required definitions in L(T) together
with the required proofs in T.
We will identify these certificates with their Gödel
numbers when convenient. Thus we speak of one certificate
being smaller than another if and only if the Gödel number
of the first certificate is less than the Gödel number of
the second certificate.
Unimpeachable certification is important for relative
consistency, where issues of impeachability are crucial.
After Tarski, many of the general investigations into
interpretability, such as in Feferman, concentrated on the
non finitely axiomatized cases (reflexive theories). We
touch on this case in section 6, and infinite theories also
are used in section 7.
In the infinite case (where S is infinite), it is natural
to disallow the use of parameters, and consider only
parameterless interpretations. Obviously the semantic
version of parameterless interpretations is equivalent, by
the completeness theorem, to the provability version of
parameterless interpretations. However, we will lose
Theorem 1.3. 5
We now focus on the finite case. By taking conjunctions, we
can assume that we have single sentences A,B., We will have
a limited discussion of the infinite case in section 6.
Also infinite theories are present in section 7. 2. Tarski Degrees.
S £I T means that the theory S is interpretable in the
theory T. Define
S =I T ´ S £I T Ÿ S £I T.
S <I T ´ S £I T Ÿ ÿT =I S.
For the purposes of this definition, we view theories as
sets of sentences without mentioning any underlying
relational type. Underlying relational types play no
significant role here.
Obviously £I is reflexive and transitive. The equivalence
classes of theories, under the equivalence relation =I, are
called the Tarski degrees.
It is easy to show that there is a proper class of Tarski
degrees, corresponding to the fact that there are a proper
class of theories.
If we restrict to countable theories, then there are 2w
countable theories, and 2w associated Tarski degrees. If we
restrict to finite theories in finite relational types,
there are w associated Tarski degrees. Obviously,
restricting to finite theories is the same as restricting
to single sentences.
The unary Tarski degrees are obtained by restricting £I to
single sentences. I.e., they are the equivalence relations
of single sentences under =I. There are w unary Tarski
degrees.
The unary Tarski degrees bear a rough resemblance to two
very well studied countable structures from recursion
theory – the Turing degrees (of subsets of w) and the r.e.
sets (subsets of w). We know from experience that the
latter two structures are very complicated in various ways.
It is not clear how complicated the unary Tarski degrees
are, and how these three countable structures are related. 6
THEOREM 2.1. The unary Tarski degrees form a (reflexive)
partial ordering with a minimum and maximum element. The
maximum element 1 is the equivalence class of all sentences
with no models. The minimum element 0 is the equivalence
class of all sentences with a finite model.
Proof: Obviously every A is interpretable in any B with no
models. We claim that in any M, we can define all models of
all nonzero finite cardinalities using a single parameter x
and the multidimensional apparatus. I.e., we can use x,
(x,x), (x,x,x), ..., (x,x,...,x) as the domain elements.
Thus any A with a finite model is interpretable in any B.
QED
THEOREM 2.2. The unary Tarski degrees form a distributive
lattice with a minimum and maximum element.
Proof: Let A,B be sentences. Define inf([A],[B]) = [A ⁄ B].
To see that this is the inf, we have to verify that C £I A,B
´ C £I A ⁄ B. The reverse direction is obvious. Now assume
C £I A,B. Let M = A ⁄ B. If M = A then there is a model of
C that is liberally defined in M. If M = B then there is a
model of C that is liberally defined in M.
sup([A],[B]) requires more care. Introduce a new monadic
predicate symbol R. Take
sup([A],[B]) = [($x,y)(Rx Ÿ ÿRy) Ÿ AR Ÿ BÿR].
To see that this is the sup, we have to verify that
A,B £I C ´ ($x,y)(Rx Ÿ ÿRy) Ÿ AR Ÿ BÿR £I C.
The reverse direction is obvious. Now assume A,B £I C. Let M
= C. Let M’,M’’ be liberally defined in M, where M’ = A,
M’’= B. Using multidimensionality, we can disjointify the
domains of A,B in order to liberally define a model of
($x,y)(Rx Ÿ ÿRy) Ÿ AR Ÿ BÿR in M.
For distributivity, because we are in a lattice, it
suffices to check
sup(x,inf(y,z)) = inf(sup(x,y), sup(x,z)).
Choose representatives A,B,C of x,y,z. Choose unary R not
in A,B,C. It suffices to show 7
($x,y)(Rx Ÿ ÿRy )Ÿ AR Ÿ (B ⁄ C)ÿR =I
(($x,y)(Rx Ÿ ÿRy) Ÿ AR Ÿ BÿR) ⁄ (($x,y)(Rx Ÿ ÿRy) Ÿ AR Ÿ
CÿR).
In fact, we have logical equivalence. QED
THEOREM 2.3. sup(x,y) = 1 ´ x = 1 Ÿ y = 1. inf(x,y) = 0 ´
x = 0 ⁄ y = 0. I.e., in the Tarski degrees, 1 is not a sup,
and 0 is not an inf.
Proof: Let [A],[B] <I 1. Then ($x,y)(Rx Ÿ ÿRy) Ÿ AR Ÿ BÿR
has a model, and so sup([A],[B]) <I 1. Let [A],[B] > 0.
Then A ⁄ B has no finite model, and so inf([A],[B]) > 0.
QED
It will be convenient to write A ŸŸ B for ($x,y)(Rx Ÿ ÿRy)
Ÿ AR Ÿ BÿR. Here R is not in A,B. We can live with the
ambiguity of which R is chosen.
THEOREM 2.4. [Fe60]. Every Tarski degree <I 1 lies strictly
below some Tarski degree <I 1.
Proof: Let A be consistent. Consider T Ÿ Con(A), where T is
the conjunction of a reasonably healthy finitely fragment
of PA = Peano Arithmetic. A £I T Ÿ Con(A) by formalizing
Gödel’s completeness theorem in T. Since T Ÿ Con(A) is
true, T Ÿ Con(A) <I 1.
Suppose T Ÿ Con(A) is interpretable in A. Then
T  Con(A) Æ Con(T Ÿ Con(A)).
T + Con(A)  Con(T + Con(A)).
T + Con(A) is inconsistent (by Gödel’s second
incompleteness theorem).
But T + Con(A) is true. QED
A particularly appropriate choice of T Õ PA is EFA =
exponential function arithmetic = IS0(exp). See [HP98], p.
3744, p. 495.
EFA is in 0,S,+,•,exp,< with the proper axioms
1.
2.
3.
4. Sx ≠ 0.
Sx = Sy Æ x = y.
x+0 = x.
x+Sy = S(x+y). 8
5. x•0 = 0.
6. x•Sy = x•y + x.
7. exp(x,0) = S0.
8. exp(x,Sy) = exp(x,y)•x.
9. x < y ´ ($z)(y = x+Sz).
10. Induction for all bounded formulas.
It is known that EFA is finitely axiomatizable. E.g., this
can be obtained from [HP98], Theorem 5.8, p. 366.
A P01 sentence is a sentence in 0,S,+,•,<, beginning with
universal quantifiers, followed by a formula with only
bounded quantifiers.
LEMMA 2.5. There is a polynomial time function a from P01
sentences to sentences in 0,S,+,•,exp,<, such that
if EFA  j then a(j) has a finite model.
if EFA  ÿj then a(j) has no model (A is inconsistent).
Proof: Let j be P01. Let a(j) be the conjunction of:
1. < is a linear ordering with minimum 0, and at least
three elements.
2. x is greatest Æ Sx = x+y = y+x = x•y = y•x = exp(x) = x.
3. x < y Æ Sx is the immediate successor of x in <.
4. x ≠ 0 Æ ($y)(x = Sy).
5. (x,y < z Ÿ Sx = Sy) Æ x = y.
6. x < y Æ x + 0 = x.
7. x < y Æ x • 0 = 0.
8. S(x+y) < z Æ x+Sy = S(x+y).
9. x•y + x < z Æ x•Sy = x•y + x.
10. x < y Æ exp(x,0) = S0.
11. exp(x,y)•x < z Æ exp(x,Sy) = exp(x,y)•x.
12. x is greatest Æ there is a proof with Gödel number < x
of EFA Æ j, and there is no proof with Gödel number < x of
EFA Æ ÿj.
13. There is no greatest element Æ EFA Ÿ Con(EFA + j).
In the above, we assume a fixed finite axiomatization of
EFA, and also that “x is the Gödel number of a proof of the
formula with Gödel number y” is formalized by a bounded
formula in the language of EFA, where y is formally
required to bound all of the values of the relevant terms.
Suppose EFA  j. Let n be the Gödel number of a proof of
EFA Æ j. We can build a model of a(j) on 0,1,...,n+1 as 9
follows. Take < to be as usual. Define S,+,•,exp on [0,n]
as usual provided the result is £ n; n+1 otherwise. Define
S,+,•,exp to be n+1 if at least one argument is n+1. Axioms
17,10 are immediate.
For 8, let S(x+y) < z in this model. Then val(x+y) < n.
Hence val(x+y) = val(x)+val(y) < n. Hence val(S(x+y)) =
val(x)+val(y)+1. Also val(x),val(y) < n. Hence val(Sy) =
val(y)+1 £ n. And val(x)+val(Sy) = val(x)+val(y)+1 £ n.
Hence val(x+Sy) = val(x)+val(y)+1 = val(S(x+y)).
For 9, let val(x•y+x) < z in this model. Then x,y £ n. Also
val(x•y+x) = val(x•y)+x £ n. Hence val(x•y) = x•y £ n. Hence
val(x•y+x) = x•y+x = x(y+1) £ n. Also val(Sy) = y+1. Hence
val(x•Sy) = x(y+1) since x(y+1) £ n.
For 11, assume exp(x,y)•x < z in
val(exp(x,y)) £ n, and so y £ n,
xy £ n, xy•x £ n, val(exp(x,y)•x)
val(exp(x,Sy)) = xy+1 since xy+1 £ this model. Then
S(y) £ n+1, val(exp(x,y)) =
= xy•x = xy+1. Hence
n. For 12, let x be greatest. Then x = n+1, and n is the Gödel
number of a proof of EFA Æ j. Hence this holds in the
model. Also, there cannot be a proof with Gödel number £ n
of EFA Æ ÿj, according to the model, since otherwise, there
would actually be a proof of EFA Æ ÿj. But this contradicts
that there is a proof of EFA Æ j.
Obviously 13 is vacuously true in this model.
Now suppose EFA  ÿj. Let M be a model of a(j). Suppose
there is a greatest element x in M. In M, < is a linear
ordering with least element 0, and < must be discrete by
axioms 3,4. Hence we can generate standard integers
0,1,2,... which either goes on forever, or stops when it
reaches x.
If it goes on forever, then this standard part is a copy of
the standard model with ordinary +,•,exp. Hence the second
part of axiom 12 is violated.
If it does not go on forever, then x is standard. We again
contradict the second part of axiom 12.
Now suppose there is no greatest element x in M. By axiom
13, M = EFA + Con(EFA + j). But this is again impossible
by EFA  ÿj. QED 10 THEOREM 2.6. The unary Tarski degrees form a dense
distributive lattice with 0 < 1. I.e., a <I b Æ ($c)(a <I c
<I b).
Proof: Let A,B be sentences, where A <I B. By self
reference, let j be a P01 sentence such that EFA proves j if
and only if
Every interpretation certificate of (A ŸŸ a(j)) ⁄ B in A
has a smaller interpretation certificate of B in (A ŸŸ
a(j)) ⁄ B.
Suppose (A ŸŸ a(j)) ⁄ B £I A, and let n be the least
interpretation certificate.
case 1. There is an interpretation certificate of B in (A
ŸŸ a(j)) ⁄ B that is smaller than n. Then EFA  j. By
Lemma 2.5, a(j) has a finite model. Now there is an
interpretation of B in A ŸŸ a(j). Since A ŸŸ a(j) £I A, we
have B £I A, which is a contradiction.
case 2. There is no interpretation certificate of B in (A
ŸŸ a(j)) ⁄ B that is smaller than n. Then EFA  ÿj. By
Lemma 2.5, a(j) has no model. Hence (A ŸŸ a(j)) ⁄ B is
logically equivalent to B. Therefore B £I A, which is a
contradiction.
We have refuted (A ŸŸ a(j)) ⁄ B £I A.
Now suppose B £I (A ŸŸ a(j)) ⁄ B. Let n be the least
interpretation certificate.
case 3. There is no interpretation certificate of (A ŸŸ
a(j)) ⁄ B in A smaller than or equaled to n. Then EFA  j.
By Lemma 2.5, a(j) has a finite model. Clearly B £I A ŸŸ
a(j). Hence B £I A, which is a contradiction.
case 4. There is an interpretation certificate of (A ŸŸ
a(j)) ⁄ B in A smaller than or equaled to n. Then EFA ÿj. By Lemam 2.5, a(j) has no model. Hence (A ŸŸ a(j)) ⁄ B
is logically equivalent to B. Therefore B £I A, which is a
contradiction.
We have refuted B £I (A ŸŸ a(j)) ⁄ B. 11
Since A £I (A ŸŸ a(j)) ⁄ B £I B, we have A <I (A ŸŸ a(j)) ⁄
B <I B, as required. QED
We can sharpen Theorem 2.6 as follows.
In any lattice, let a1,a2,... be a finite or finite sequence
of points. We say that this sequence is independent if and
only if in the lattice, for all n,m ≥ 1,
if inf(ai1,...,ain) £ sup(aj1,...,ajm) then
{i1,...,in} Õ {j1,...,jm}.
THEOREM 2.7. In the unary Tarski degrees, let a <I b. There
exists an independent infinite sequence c1,c2,... lying in
(a,b).
Proof: Let A Œ a, B Œ b. Then A <I B. By self reference,
let j(n) be a P01 formula with only the free variable n,
such that EFA proves j(n) if and only if
for every interpretation certificate of (A ŸŸ a(j(n*)) ⁄ B
in A, or one which ruins the independence of the finite
sequence
(A ŸŸ a(j(1*))) ⁄ B, ...,(A ŸŸ a(j(n*))) ⁄ B
using (A ŸŸ a(j(n*))) ⁄ B on the left (inside the inf),
there is a smaller interpretation certificate of B in (A ŸŸ
a(j(n*))) ⁄ B, or a smaller interpretation certificate
which ruins the independence of the finite sequence
(A ŸŸ a(j(1*))) ⁄ B,...,(A ŸŸ a(j((n)*)) ⁄ B
using (A ŸŸ a(j(n*))) ⁄ B on the right (inside the sup).
Here we use n* for the closed term S...S0, with n S’s.
We now show that the infinite sequence
(A ŸŸ a(j(1*))) ⁄ B,(A ŸŸ a(j(2*))) ⁄ B,...
is independent and lies in (a,b). Obviously, this infinite
list lies in [a,b].
Let n be least such that it is not the case that
(A ŸŸ a(j(1*))) ⁄ B,(A ŸŸ a(j(2*))) ⁄ B,..., 12
(A ŸŸ a(j(n*))) ⁄ B
is independent. We will obtain a contradiction.
Consider the following two cases.
case 1. (A ŸŸ a(j(n*))) ⁄ B £I A, or there is a ruining of
the independence of the finite sequence (A ŸŸ a(j(1*))) ⁄
B, ...,(A ŸŸ a(j(n*))) ⁄ B, using (A ŸŸ a(j(n*))) ⁄ B on
the left.
Suppose EFA  ÿj(n*). Then by Lemma 2.5, a(j(n*)) has no
model, and so (A ŸŸ a(j(n*))) ⁄ B is logically equivalent
to B. Hence B £I A or there is a ruining of the independence
of the finite sequence (A ŸŸ a(j(1*))) ⁄ B, ...,(A ŸŸ
a(j(n1*))) ⁄ B,B, using B on the left. Since (A ŸŸ
a(j(1*))) ⁄ B, ...,(A ŸŸ a(j(n1*))) ⁄ B £I B, the B, being
on the left side, which is an inf, gets absorbed. Hence B £I
A or there is a violation of the independence of the finite
sequence (A ŸŸ a(j(1*))) ⁄ B, ...,(A ŸŸ a(j(n1*))) ⁄ B.
The former is impossible, and the latter contradicts the
choice of n.
Since EFA does not refute j(n*), we see that EFA  j(n*).
Hence by Lemma 2.5, a(j(n*)) has a finite model, and so (A
ŸŸ a(j(n*))) ⁄ B =I A ⁄ B =I A.
Also since EFA does not refute j(n*), we see that B £I (A ŸŸ
a(j(n*))) ⁄ B, or there is a ruining of the independence of
the finite sequence (A ŸŸ a(j(1*))) ⁄ B,...,(A ŸŸ
a(j((n)*))) ⁄ B using (A ŸŸ a(j(n*))) ⁄ B on the right.
Hence B £I A, or there is a ruining of the independence of
the finite sequence (A ŸŸ a(j(1*))) ⁄ B,...,(A ŸŸ a(j((n1)*)) ⁄ B,A, using A on the right. Since A £I (A ŸŸ
a(j(1*))) ⁄ B,...,(A ŸŸ a(j((n1)*)) ⁄ B, the A, being on
the right side, which is a sup, gets absorbed. Hence B £I A,
or there is a violation of the independence of the finite
sequence (A ŸŸ a(j(1*))) ⁄ B,...,(A ŸŸ a(j((n1)*))) ⁄ B.
The former is impossible, and the latter contradicts the
choice of n.
case 2. B £I (A ŸŸ a(j(n*))) ⁄ B, or there is a ruining of
the independence of the finite sequence (A ŸŸ a(j(1*))) ⁄
B, ...,(A ŸŸ a(j(n*))) ⁄ B, using (A ŸŸ a(j(n*))) ⁄ B on
the right. Because case 1 is impossible, we have EFA j(n*). Hence by Lemma 2.5, a(j(n*)) has a finite model.
Hence (A ŸŸ a(j(n*))) ⁄ B =I A ⁄ B =I A. 13 Hence B £I A, or there is a ruining of the independence of
the finite sequence (A ŸŸ a(j(1*))) ⁄ B, ...,(A ŸŸ a(j((n1)*))) ⁄ B,A, using A on the right. The former is
impossible. For the latter, since A £I (A ŸŸ a(j(1*))) ⁄
B,...,(A ŸŸ a(j((n1)*)) ⁄ B, the A, being on the right
side, which is a sup, gets absorbed. Hence there is a
violation of the independence of the finite sequence (A ŸŸ
a(j(1*))) ⁄ B,...,(A ŸŸ a(j((n1)*))) ⁄ B. This contradicts
the choice of n.
Since both cases lead to a contradiction, we see that there
is no violation of the independence of the finite sequence
(A ŸŸ a(j(1*))) ⁄ B, ...,(A ŸŸ a(j(n*))) ⁄ B using (A ŸŸ
a(j(n*))) ⁄ B. Hence there is a violation of the
independence of the finite sequence (A ŸŸ a(j(1*))) ⁄ B,
...,(A ŸŸ a(j((n1)*))) ⁄ B. This contradicts the choice of
n.
Thus we have established the independence of the infinite
sequence
(A ŸŸ a(j(1*))) ⁄ B,(A ŸŸ a(j(2*))) ⁄ B,...
It remains to show that these sentences lie in (a,b). Let m
be least such that (A ŸŸ a(j(m*))) ⁄ B £I A. Since there is
no interpretation of B in (A ŸŸ a(j(m*))) ⁄ B, we see that
EFA  ÿj(m*). Hence a(j(m*)) has no model, and so B £I A,
which is a contradiction. Hence for all m, not (A ŸŸ
a(j(m*))) ⁄ B £I A.
Let m be least such that B £I (A ŸŸ a(j(m*))) ⁄ B. Since
there is no interpretation of (A ŸŸ a(j(m*))) ⁄ B in A, we
see that EFA  j(m*). Hence a(j(m*)) has a finite model,
and so B £I A ⁄ B and B £I A. This is impossible. Hence for
all m, not B £I (A ŸŸ a(j(m*))) ⁄ B. QED
Do all unary Tarski degrees other than 0,1 look alike? The
answer is strongly no.
We have already seen that 1 is not a sup. Therefore the sup
of any two incomparable Tarski degrees is a sup ≠ 0,1.
However, other unary Tarski degrees ≠ 0,1 are not sups.
Let M,M’ be two structures with disjoint domains. We write
M + M’ for the structure with domain dom(M) » dom(M’),
where the relations and constants of M and M’ remain 14
unchanged, the functions of M are extended to dom(M’) by
returning the first argument, and a new unary relation is
introduced caring out dom(M).
LEMMA 2.8. Let M,M’ be two structures with disjoint
domains. Suppose M + M’ liberally defines a linear ordering
with no greatest element. Then M or M’ liberally defines a
linear ordering with no greatest element.
LEMMA 2.9. Let M,M’ be two structures with disjoint
domains. Suppose M + M’ liberally defines a pairing
function with at least two inequivalent elements. Then M or
M’ alone liberally defines a pairing function with at least
two inequivalent elements.
THEOREM 2.10. In the unary Tarski degrees, “Linear ordering
with no greatest element” is not a sup. “Pairing function
with at least two elements” is not a sup.
Proof: Suppose [linear ordering with no greatest element] =
sup([A],[B]). Let M = A ŸŸ B. Let M’ be the part of M for
A and M’’ be the part of M for B. Since M liberally defines
a linear ordering with no greatest element, M’ or M’’
liberally defines a linear ordering with no greatest
element. Hence [A] or [B] is [linear ordering with no
greatest element]. The analogous argument proves the second
claim. QED
A sentence A is said to be complete if and only if for all
sentences B in L(A), A  B or A  ÿB.
THEOREM 2.11. Let A be a complete sentence. Then [A] is not
an inf in the unary Tarski degrees.
Proof: Let A be a complete sentence. Let [A] = inf([B],[C])
= [B ⁄ C]. Let M = A. Then M has a liberally defined model
of B ⁄ C. Hence M has a liberally defined model of B or M
has a liberally defined of C. We now use the completeness
of A. In the former case, every M = A has a liberally
defined model of B. In the latter case, every M = A has a
liberally defined model of C. QED 3. Adequate Sentences and Relative Consistency.
A very important condition is adequacy of a theory T. It is
convenient to first define adequacy for a model. 15
We say that M is adequate if and only if there is an M
definable binary relation whose cross sections are closed
under single point additions and single point deletions.
Another, more common, definition is that there is an M
definable binary relation with an empty cross section,
whose cross sections are closed under single point
additions.
THEOREM 3.1. These two definitions of adequacy are
equivalent.
Proof: Let M be a model with an M definable binary relation
R whose cross sections are closed under single point
additions and single point deletions. Fix c Œ dom(M).
Define S(x,y) ´ (R(x,y) ´ ÿR(c,y)). Clearly the cross
section of S at c is empty. Now let x,y be given. If
ÿR(c,y) then choose a cross section of R obtained by
inserting y in the cross section of R at x. If R(c,y) then
choose a cross section of R obtained by deleting y from the
cross section of R at x.
Let M be a model with an M definable binary relation R with
an empty cross section, whose cross sections are closed
under single point additions.
We first use the following construction for getting: an
empty cross section, and closure under single point
additions and pairwise intersection. Call a cross section
good if and only if its intersection with every cross
section is a cross section. Obviously, empty cross sections
are good. Let X,Y be good cross sections and x Œ X. To
check that X » {x} is good, clearly X » {x} » {y} and (X »
{x}) « Z are cross sections since X « Z is a cross section.
To check that X « Y is good, clearly (X « Y) » {x} and (X
« Y) « Z are cross sections, since X « Y, Y « Z are cross
sections.
Call a cross section X of R very good if and only if it is
good, and any single point deletions of any good cross
section that is a subset of X, is a good cross sections.
We claim that the very good cross sections of R are closed
under single point additions and single point deletions. To
see this, let X be a very good cross section and x Œ
dom(M). We must show that X » {x} and X\{x} are both very
good cross sections. Clearly there are both good cross 16
sections. We can assume without loss of generality that x œ
X.
Let Y Õ X » {x} be a good cross section of R. Any Y\{y} is
Y\{x}\{y} or (Y\{x}\{y}) » {x}. Since X is good, X « Y is
good, and X « Y = Y{x} Õ X. Since X is very good, Y\{x} is
a good subset of X, and so Y\{x}\{y} and (Y\{x}\{y}) » {x}
are good. Therefore X » {x} is very good.
Let Y Õ X\{x} be good. Since Y Õ X, single point deletions
to Y remain good. QED
We give four equivalent forms of adequacy for a theory T.
We begin with two syntactic definitions.
We say that a theory T is adequate if and only if there
exists a formula j(x,y,z1,...,zn) in L(T) such that the
following is provable in T. There exists 1 £ i £ k (as a
disjunction) and z1,...,zn (normal universal quantification)
such that
i. ($x)("y)(ÿji(x,y,z1,...,zn)).
ii. ("u,w)($x)("y)(ji(x,y,z1,...,zn) ´ ji(u,y,z1,...,zn) ⁄ y
= w).
i’. ("u,w)($x)("y)(ji(x,y,z1,...,zn) ´ ji(u,y,z1,...,zn) ⁄ y
= w).
ii’. ("u,w)($x)("y)(ji(x,y,z1,...,zn) ´ ji(u,y,z1,...,zn) Ÿ
y ≠ w).
In the case of finite T, it is easily seen that adequacy of
T is equivalent to the model theoretic condition that every
model of T is adequate (in any of the two senses shown
equivalent by Theorem 3.1). This uses a compactness
argument.
It will also be convenient to rephrase the adequacy of T in
terms of interpretations.
We call an interpretation regular if the domain used for
the interpretation is the full domain, and the equality
relation used is equality.
It is clear that the adequacy of T is equivalent to the
regular interpretability of the following weak system NW of
set theory, formulated in the usual language Œ,=. 17
N (Null axiom). ($x)("y)(y œ x).
W (With axiom). ($y)("z)(z Œ y ´ (z Œ x ⁄ z = x)).
Or we can equivalently use WD.
W (With axiom). ($y)("z)(z Œ y ´ (z Œ x ⁄ z = x)).
D (Delete axiom). ($y)("z)(z Œ y ´ (z Œ x Ÿ z ≠ x)).
The system NW is credited to Ed Nelson in [MM94], where it
is shown that Q is interpretable in NW. (They also credit
Jan Krajicek for an earlier and unpublished proof of this
result, not known to them at the time of their
publication).
The interpretability of NW in Q was already mentioned in
[TMR53], p. 34, where the authors report that the
interpretability of even NWE had been established by W.
Szmielew and A. Tarski in 1950. Here E is
E. Extensionality axiom. ("z)(z Œ x ´ z Œ y) Æ x = y.
The argument in Theorem 3.1 establishes the mutual regular
interpretability of WD and NW. However, WD, NW, WDE, NWE
are only mutually interpretable, not mutually regularly
interpretable. E.g., they have the same Tarski degree.
In fact, WD represents a very important Tarski degree with
many different kinds of representatives. A particularly
important representative of this Tarski degree is the well
known weak system of arithmetic called Q, or Robinson’s
arithmetic after R. Robinson [Ro52]. Also see [HP98], p.
28. The language is 0,S,+,•,£,=, and the nonlogical axioms
are
1.
2.
3.
4.
5.
6.
7.
8. S(x) ≠ 0.
S(x) = S(y) Æ x = y.
x ≠ 0 Æ ($y)(S(y) = x).
x + 0 = x.
x + S(y) = S(x + y).
x • 0 = 0.
x • S(y) = (x • y) + x.
x £ y ´ ($z)(z + x = y). Obviously the explicit definitional axiom 8 is redundant.
However, since for many purposes, Q is normally extended by
additional axioms that do involve £, it is convenient to
incorporate 8. 18 One such important extension of Q is the system IS0. This
also goes under the name “bounded arithmetic”. I prefer the
name PFA = polynomial function arithmetic. In this system,
we add
the induction scheme for bounded formulas in L(Q) to Q.
A bounded formula in L(Q) is a formula all of whose
quantifiers are bounded to terms that do not contain the
variable; i.e., ("x £ t), ($x £ t), where t is a term of
L(Q) in which x does not appear. Thus the nonlogical axioms
of PFA are 18 and
9. Bounded induction scheme. (j[x/0] Ÿ ("x)(j Æ j[x/Sx]))
Æ j, where j is a bounded formula of L(Q).
PFA is believed to be not finitely axiomatizable, although
this is not known. See [HP98], p. 350.
It is well known that PFA is interpretable in Q. See
[HP98], p. 366.
THEOREM 3.2. WD, NW, WDE, NWE, Q, PFA are mutually
interpretable.
Theories of a tame character such as Presburger arithmetic,
real closed fields, and algebraically closed fields, are
interpretable in Q, but cannot interpret Q.
Let T be adequate. An adequacy mechanism for T consists of
formulas of L(T) that
i. define the internal nonnegative integers by a unary
predicate, together with an equivalence relation
representing equality. T must prove that this is an
equivalence relation E.
ii. define <,0,S,+,• on the natural numbers. T must prove
everywhere definedness and univalence (with respect to E).
iii. define the internal finite sequences by a unary
predicate.
iv. define the length function and value function for
finite sequences (i.e., the ith term).
v. T proves that there is a finite sequence of length 0,
and that the finite sequences are closed under appending
any object on the right  and that appending raises the
length
by 1. 19
vi. T proves the axioms of PFA formulated using ii above,
with bounded induction extended to incorporate all of the
symbols of L(T).
THEOREM 3.3. Every adequate theory has an adequacy
mechanism.
We need a notion of restricted proof in first order
predicate calculus with equality. For definiteness, we use
the following logical axioms and rules in ",$,Ÿ,⁄,Æ,´,ÿ,=.
i. All tautologies.
ii. ("x)(j) Æ j[x/t], t substitutable for x in j.
iii. j[x/t] Æ ($x)(j), t substitutable for x in j.
iv. x = x.
v. x = y Æ (j Æ j’), where j is atomic and j’ is obtained
from j by replacing an occurrences of x in j by y.
vi. From j Æ y derive j Æ ("x)(y), where x is not free in
j.
vii. From j Æ y derive ($x)(j) Æ y, where x is not free in
y.
viii. From j, j Æ y, derive y.
A proof consists of a finite sequence of formulas, each of
which is either an axiom above, or follows from previous
entries in the sequence by one or more rules of inference
above. A proof of j is a proof whose last entry is j. We
write PROV(j).
For any formula j, we define lth(j) to be the total number
of symbols in j, including parentheses, connectives,
quantifiers, constants, relations, functions, and
variables. Officially, the constants, relations, functions,
and variables are given with a single symbol adorned with
subscripts in binary. Each occurrence of these symbols is
charged one plus the length of the subscript.
Note that there is a small fixed constant b such that the
number of formulas j with lth(j) £ n is at most bn. In fact,
we take b to be the length of
"$Ÿ⁄Æ´ÿ=RFc01(),
which is 16.
A restricted proof is a proof with the additional
requirements that 20 i. All nonlogical symbols used are present in the last
entry, j.
ii. Every entry is a propositional combination of formulas,
each of which have at most lth(j) quantifiers.
We write RPROV(j) for “j has a
of a proof is the total number
This is the same as the sum of
We write RPROV(j,n) for “j has
n”. restricted proof”.
of symbols that it
the lengths of its
a restricted proof The size
contains.
entries.
of size £ We write RCON(A) for ÿRPROV(ÿA).
We write 2[p](n) for an exponential stack of p 2’s with n on
top, where if p = 0, this is n.
THEOREM 3.4. Let ("n)($m)(j(n,m)) be a P02 sentence, where j
is bounded in L(Q). The following are equivalent.
i. EFA  ("n)($m)(j(n,m)).
ii. There exists p Œ N such that PFA  ("n,r)(r = 2[p](n)
Æ ($m £ r)(j(n,m))).
Proof: In ii, we are using an adequate formalization of the
exponentiation relation in PFA, which is well known to
exist. Let y = ("n)($m)(j(n,m)) be as given. Assume i.
Suppose ii is false. Let T be the theory in L(Q) with
constants d,e, consisting of
i. PFA.
ii. The scheme d = 2[p](e) Ÿ ("r £ d)(ÿj(d,r)), p ≥ 0.
Since ii is false, every finite fragment of T is
consistent. Hence by compactness, T is consistent. Let M =
T. Let M’ be the initial segment of M determined by the
iterated exponentials of d. Then M’ = EFA Ÿ
ÿ("n)($m)(j(n,m)). Hence ii holds.
The converse ii Æ i is trivial. QED
Let M be adequate (see Theorem 3.1), and let an adequacy
mechanism be given (see Theorem 3.3). A cut in M is taken
to be an initial segment of (the numerical part of) M, with
no greatest element. A proper cut in M is the cut in M
consisting of all nonnegative integers of M (under the
given adequacy mechanism). We will work with M definable
cuts in M, only. 21 The following result was essentially proved in [Fr76],
[Fr80], and exposited in [Sm85], [Vi90].
THEOREM 3.5. (Interpretability = relative restricted
consistency). Let A,B be sentences, where B is adequate.
The following are equivalent.
i. A £I B.
ii. EFA  RCON(B) Æ RCON(A).
i Æ ii does not require the adequacy of B. However, ii Æ i
does require the adequacy of B.
Proof: i Æ ii. Let p be an interpretation of A in B. Now p
specifies a parameterized family of structures M*. I.e.,
there are free variables for the unspecified parameters. We
have a proof from B that parameters can be chosen so that p
actually defines a specific model M* of A.
We now argue in EFA. Let a be a restricted proof of ÿA. So
p converts a to a proof from B of
there exists a choice of parameters such that the specific
model M* satisfies A as well as a1,...,ak = ÿA.
Obviously, this gives us a proof of a contradiction from B,
and hence a proof of ÿB.
But how complicated are the formulas involved in the proof
of ÿB obtained in this way? There are a number of sources
of complexity here, including the use of function symbols
in A that give rise to terms of uncontrolled complexity
that have to be interpreted in M*.
So we only claim that the formulas involved in the
refutation of B have a standard integer bound on their
quantifier depth. Quantifier depth counts blocks of like
quantifiers like a single quantifier.
Fortunately, cut elimination has been reworked in terms of
quantifier depth. See, e.g., [Zh91], [Zh94], [Ge05]. Since
we have only standard quantifier depth, we can reduce the
cut complexity to zero in a finite number of steps within
EFA. EFA will take care of the exponential blowup (no
worse) at each stage. 22
The implication ii Æ i is considerably more involved. Here
cuts in nonstandard models are used, combined with
formalized completeness proofs.
Let A,B be given, where B is adequate. Let an adequacy
mechanism for B be given (Theorem 3.3).
Assume EFA  RCON(B) Æ RCON(A). Obviously
EFA  ("n)($m)(RPROV(ÿA,n) Æ RPROV(ÿB,m)).
Hence by Theorem 3.4, let p be such that
IS0  ("n,r)(r = 2[p](n) Ÿ RPROV(ÿA,n) Æ RPROV(ÿB,r)).
Let M = B. We need to define a model M* = A in M.
We use an attempted Henkin construction of a model of A in
M. There are a number of problems that we encounter in
order to make this work. All of these problems stem from
the fact that we don’t have much induction in M. However,
we recover by making things work in M definable cuts.
Let L(A)’ be L(A) together with the Henkin constants
d0,d1,... . Let W be the set of all sentences j of L(A)’
which are propositional combinations of sentences whose
number of quantifiers is at most #(ÿA), and whose
nonlogical symbols appear in A.
We begin with listing certain elements of W. If EFA holds
in (the numerical part of) M, then there is no problem
listing all elements of W. Specifically, for each k, there
is a unique (coded) finite sequence consisting of the first
k elements of W, ordered first by length and second
lexicographically. These finite sequences are initial
segments of each other. Thus we can form the M definable
j0,j1,... which enumerates all elements of W in order.
If EFA fails in M, then there is an M definable cut. By cut
shortening techniques, we can construct an M definable cut
C which is closed under multiplication, and where there
exists t œ C such that 2^2^t exists. Then we can easily
enumerate the first t elements of W, lexicographically, in
order, as
j0,j1,...,jt. 23
We can restrict the indices to elements of C, obtaining {ji:
i Œ C} as a sequence.
It is easy to see that {ji: i Œ C} enumerates exactly the
elements of W whose length n has 2n Œ C. (Recall the
discussion above that {j: lth(j) £ n} has at most 16n
elements, and that C is closed under addition).
Note that {ji: i Œ C} is closed under propositional
combinations since C propositional combinations roughly add
lengths, and C is closed under multiplication. Also note
that the terms appearing in the elements of {ji: i Œ C} are
closed under the term building operations, since the term
building operations roughly add lengths, and C is closed
under multiplication.
So far, we can consolidate the cases where M satisfies EFA
and does not satisfy EFA, by allowing C to be the improper
cut of all n ≥ 0, or C to be a proper cut. In either case, C
is an M definable cut closed under multiplication.
We write log(C) for the cut {2i: i Œ C}. Then log(C) is an M
definable cut closed under addition, and {ji: i Œ C} = {j Œ
W: lth(j) Œ log(C)} = W[log(C)].
Recall that we are working entirely within M.
case 1. RCON(A). A finite sequence of sentences y0,...,y2k+1
from W[log(C)] is called good for A if and only if it
satisfies the following conditions for all 0 £ i £ k.
i. k Œ log(C).
ii. y2i is ji or ÿji.
iii. If y2i is ÿ("x)(j) and c is the first constant not
appearing in y0,...,y2i, then y2i+1 is ÿj[x/c].
iv. If y2i is not of the form ÿ("x)(j), then y2i+1 = y2i.
v. There is no proof of ÿ(A Ÿ y0 Ÿ ... Ÿ y2k+1) of any size,
whose entries lie in W[log(C)].
vi. For all 0 £ i £ k, if y2i = ÿji, then there is a proof
of ÿ(A Ÿ y1 Ÿ ... Ÿ y2i2 Ÿ ji) of any size, whose entries
lie in W[log(C)].
It is easy to see that there exists y0,y1 that is good for
A. Also, suppose y0,...,y2k+1 is good for A, k ≥ 0. Then
there is an extension y0,...,y2k+3 which is still good for A. 24
We now assume that there are good sequences for A of every
even length Œ log(C), and also for any two sequences good
for A, one is an extension of the other. Of course, these
two suppositions are obvious if we had enough induction in
M  which we don’t.
The previous paragraph defines a particular “path” if we
think of this set of M sequences as forming an internal
tree.
We can define a structure on the closed terms in the
obvious way from the sequences good for A. Note that the
atomic sentences in L(A)’ among the ji are exactly the
atomic sentences whose length is in log(C), which are the
atomic sentences whose closed terms have length in log(C).
These closed terms are closed under the term building
operations since log(C) is closed under addition.
The domain of the prospective model of A is the set of
closed terms of length in log(C). We make this into a
structure interpreting L(A)’ in the obvious way. The
interpretation of equality is: s ≡ t ´ s = t lies on the
path.
We have defined an interpretation M* of L(A)’. Obviously M*
is M definable. We have to show that as a structure
external to M, M* = A.
Let q be the number of quantifiers in A. We show by
external induction on j from 0 through q, that for all
standard formulas r in L(A)’ with at most j quantifiers,
r holds in M at any assignment comprised of internal closed
terms of M of length Œ log(C), in the external sense, if
and only if the result of making the substitution, creating
a sentence in W[log(C)] internally in M, lies on the
internal path.
This follows from the usual basic facts about what is
sitting on the path. Since A is a sentence on the path, we
see that M* is in fact a model of A. Both M and the outside
world agree on this.
case 2. ÿRCON(A). Since M = PFA, we have
M = ("n,r)(r = 2[p](n) Ÿ RPROV(ÿA,n) Æ RPROV(ÿB,2[p](n))) 25
where p was chosen above in advance of the choice of M.
We now produce an M definable cut C such that ("n Œ
C)(RCON(A,n)).
If there is no least n such that RPROV(ÿA,n), then we
immediately have our desired cut. Namely the cut of all n
such that RCON(A,n).
Now let n be least such that RPROV(ÿA,n).
case 2a. 2[p](n) does not exist. By cut shortening
techniques, we can explicitly construct an M definable cut
C below n.
case 2b. 2[p](n) exists. Then RPROV(ÿB,2[p](n)). Let p be a
witness to this statement. We now construct an M definable
cut below 2[p](n) as follows.
We would like to give an M definable valuation of all terms
in L(B) at all assignments of objects. However, we may not
have enough induction in M. in M, let C be the set of all m
such that for every finite list t1,...,tn of terms in L(B),
with a total of at most m symbols, and every assignment of
objects to the variables in t1,...,tn, there is a unique
assignment of objects to all subterms of t1,...,tn obeying
the obvious valuation clauses. It is easy to see that C is
an M definable cut (not necessarily proper).
Let C’ be the set of all m Œ C such that there is a
satisfaction relation for all propositional combinations of
formulas in L(B) with at most lth(ÿB) quantifiers each,
whose total number of symbols is £ m, based on the above
definition of valuation. Again, C’ Õ C is a cut.
Finally, let C’’ be the set of all m Œ C’ such that there
is a satisfaction relation for all finite sequences of
propositional combinations of formulas in L(B) with at most
lth(ÿB) quantifiers each, whose total number of symbols (in
the whole finite sequence) is £ m, based on the above
definition of satisfaction. Once again, C’’ Õ C’ Õ C is a
cut.
We cannot have 2[p](n) Œ C’’ because the satisfaction
relation provides a sequence of truth values for the
universal closures of the formulas in this Hilbert style
proof, p. We can then do an induction on the resulting bit 26
sequence, taking the first entry which is false, and
obtaining a contradiction.
In particular, C’’ is a proper cut below 2[p](n). We can
then do cut shortening to obtain an M definable cut below
n.
The cut shortening procedure can be made to yield an M
definable cut C* below n in which PFA holds. We can also
obtain an M definable cut C** Õ C* several exponentials
lower, where C** is closed under multiplication.
We are now in the same position that we were at the
beginning of case 1. Here the full numerical part of M
corresponds to the cut C* here, and the cut C there
corresponds to the cut C** here. The identical argument
will produce an M definable model of A. QED
The following result is used crucially in section 5.
THEOREM 3.6. Let A be a P01 sentence. There exists an
adequate sentence B in L(Q) such that EFA  A ´ RCON(B). B
can be taken to be the conjunction of any sufficiently
large finite fragment of PFA with some S01 sentence in L(Q).
Proof: Firstly, we may assume A is in L(Q), since every P01
sentence is provably equivalent, in EFA, to a P01 sentence
in L(Q).
Choose a sufficiently large finite fragment PFA’ of PFA.
Then PFA’ is adequate. By self reference, let C be a S01
sentence in L(Q) provably equivalent over PFA’ to
C*. there exists a restricted proof a of ÿ(PFA’ Ÿ C) such
that A is true (with outermost universal quantifiers
restricted to natural numbers) £ lth(a).
The adequate sentence B that we want is the sentence B =
PFA’ Ÿ C*. Note that C* is S01. To see that this works, we
argue in EFA.
Suppose A. If ÿRCON(B) then C*. Hence C. Since C is a true
S01 sentence, we can prove PFA’ Æ C using formulas in L(Q)
which are propositional combinations of formulas each of
which have at most a number of quantifiers that depends
only on the choice of PFA’ and C. Also, we have RCON(PFA’).
Hence we conclude that RCON(PFA’ Ÿ C), because we can 27
perform cut elimination several times (EFA is our
background theory). I.e., RCON(B).
Now suppose RCON(B). Let n be a counterexample to A. Note
that PFA’ sees that there is no restricted proof a of ÿB
with lth(a) £ n, by RCON(B). Hence PFA’  ÿC* using
formulas in L(Q) which are propositional combinations of
formulas each of which have at most a number of quantifiers
that depends only on the choice of PFA’ and C. Hence PFA’
 ÿC using formulas in L(Q) which are propositional
combinations of formulas each of which have at most a
number of quantifiers that depends only on the choice of
PFA’ and C. The same must be true of a proof of ÿB = PFA’ Ÿ
C. By successive cut elimination, we obtain ÿRCON(B). Hence
the counterexample n to A must not exist. Therefore A. QED
Here is a sharpening of Theorem 3.6.
THEOREM 3.7. Let A be a P01 sentence and B be adequate, with
an adequacy mechanism. There exists a S01 sentence C such
that EFA  (A Ÿ RCON(B)) ´ RCON(B Ÿ C).
Proof: We again assume that A is in L(Q), and choose a
sufficiently large finite fragment PFA’ of PFA. Then PFA’
is adequate. By self reference, let C be a S01 sentence in
L(Q) provably equivalent over PFA’ to
C*. there exists a restricted proof a of ÿ(B Ÿ C) such that
A is true (with outermost universal quantifiers restricted
to natural numbers) £ lth(a).
We argue in EFA. Suppose A Ÿ RCON(B). If ÿRCON(B Ÿ C) then
C*. Therefore C. Hence B proves C with a restricted proof.
Therefore RCON(B Ÿ C), which is a contradiction. (For these
last two statements, we use cut elimination for a standard
number of steps in the background theory EFA). Hence RCON(B
Ÿ C).
Now suppose RCON(B Ÿ C). Let n be a counterexample to A.
Note that B sees that there is no restricted proof a of ÿ(B
Ÿ C) with lth(a) £ n, by RCON(B Ÿ C). Hence B  ÿC* with
restricted proof. Therefore B  ÿC with restricted proof.
(We have again used cut elimination for a standard number
of stpes in the background theory EFA). This contradicts
RCON(B Ÿ C). QED 28
COROLLARY 3.8. Let A,B be consistent sentences, where B is
adequate. There exists consistent adequate B Ÿ C I A. In
fact, under any adequacy mechanism for B, C can be taken to
be S01.
Proof: Let A,B be as given. By Theorem 3.7, let C be such
that EFA  (RCON(A) Ÿ RCON(B)) ´ RCON(B Ÿ C). In
particular, EFA  RCON(B Ÿ C) Æ RCON(A), and B Ÿ C is
adequate. By Theorem 3.5, A £I B Ÿ C. QED
An interpretation p from S to T is said to be faithful iff
it does not use any parameters, and for all sentences j Œ
L(S), S  j ´ T  pj.
THEOREM 3.9. (Interpretability = faithful
interpretability). Let A,B be sentences, where B is
adequate without parameters. The following are equivalent.
i. A is interpretable in B.
ii. A is faithfully interpretable in B.
For a proof of Theorem 3.9, see [Vi05].
Theorem 3.9 is easier if we assume that there is an
adequacy mechanism for B, without parameters, which
satisfies all true P01 sentences. For this result, we need a
Lemma.
LEMMA 3.10. Let B be consistent, with an adequacy
mechanism. There is a formula j(n) in L(B) such that for
all n ≥ 0, B Ÿ j(0) Ÿ ... Ÿ j(n1) Ÿ ÿj(n) is conservative
over B for all S01 sentences. In fact, we can replace S01 by
S0k, for any k given in advance of j.
Proof: By a well known Rosser construction. QED
Here is the weakened form of Theorem 3.9 that was promised.
THEOREM 3.11. (Interpretability = faithful
interpretability). Let A,B be sentences, where B is given
an adequate mechanism without parameters. Assume that B is
consistent with the true P01 sentences. The following are
equivalent.
i. A is interpretable in B.
ii. A is faithfully interpretable in B. 29
Proof: Let p be an interpretation from A into B, where B is
adequate without parameters. Let j(n) be as given by Lemma
3.10. We now define a faithful interpretation of A to B.
Let M = B. We define a model of A within M.
Let n be least such that ÿj(n). If this does not exist, use
the default value n = 0.
Let E be the nth sentence in L(A). If M satisfies the
consistency of A Ÿ E then return the model of A Ÿ E
obtained using the formalized completeness proof.
Otherwise, return the model of A given by p.
We claim that for any E Œ L(A) such that A Ÿ E is
consistent, some M = B returns a model of A Ÿ E.
Let E be as given, and let n be its Gödel number. Let B* =
B Ÿ j(0) Ÿ ... Ÿ j(n1) Ÿ j(n). Then B* remains consistent
with the true P01 sentences, according to Lemma 3.10.
Let M = B*. Then the calculation in M of n, and therefore
E, is correct. Hence M = Con(A Ÿ E) since Con(A Ÿ E) is a
true P01 sentence.
Now under this interpretation p’ of A into B, which
sentences r in L(A) are satisfied in all models arising in
the interpretation? I.e., in all p’M, M = B?
Such a sentence r must be consistent with every consistent
sentence A Ÿ E. So if A does not prove r, then A + ÿr is
consistent, and r must be consistent with A + ÿr. This is
patently absurd, and so r must be provable from A, as
required. QED 4. Predicative Set Extensions and Relative
Consistency.
The predicative set extension of a theory T is defined as
follows. First we introduce a second sort for “sets”, with
the binary relation Œ between objects of the original sort
and objects of the new sort. Then we add the predicative
comprehension axiom
($A)("x)(x Œ A ´ j) 30
where j is a formula in the extended language without any
quantifiers over the second sort, in which A is not free.
Since we are officially working in single sorted logic,
convert this two sorted system to a single sorted system in
the obvious way by introducing a unary predicate.
We write this extension as PSE(T). It is well known, by a
simple model theoretic argument, that PSE(T) is a
conservative extension of T.
THEOREM 4.1. Let A be an adequate sentence. PSE(A) is
adequate, and finitely axiomatized. EFA proves RCON(PSE(A))
´ CON(A). EFA does not prove RCON(A) Æ CON(A). EFA does
not prove CON(A) Æ CON(PSE(A)). PSE(A) is not interpretable
in A.
Proof: Well known. See Theorems 5.5 and 7.1 of [Vi06]. QED
THEOREM 4.2. Let A,B be adequate sentences. The following
are equivalent.
i. PSE(A) is interpretable in PSE(B).
ii. EFA proves CON(B) Æ CON(A).
Proof: Apply Theorem 3.5 to the adequate theories PSE(A),
PSE(B). We obtain that i ´ (EFA proves RCON(PSE(B)) Æ
RCON(PSE(A))) Æ ii, using Theorem 4.1. QED 5. Adequate Tarski Degrees and P Degrees.
The adequate Tarski degrees are the Tarski degrees of
adequate sentences. There is another kind of degree which
is intimately connected with Tarski degrees.
It is most convenient to work in the language of EFA, 0,S,
+,•,exp,<. For P01 sentences A,B, we write
A £P B ´ B logically implies A.
£P is reflexive and transitive. Define =P, ≥P in the obvious
way. The P degrees are the equivalence classes under =P.
EFA represents a particular P degree.
Recall Theorem 3.6. 31
THEOREM 3.6. Let A be a P01 sentence. There exists an
adequate sentence B in L(Q) such that EFA  A ´ RCON(B). B
can be taken to be the conjunction of any sufficiently
large finite fragment of PFA with some S01 sentence in L(Q).
THEOREM 5.1. The adequate Tarski degrees are isomorphic to
the P degrees ≥ EFA.
Proof: Let A be adequate. Map it to A* = EFA Ÿ RCON(A). By
Theorem 3.5, for adequate A,B, we have
A
A
A
A £I
£I
£I
£I B
B
B
B ´
´
´
´ EFA  RCON(B) Æ RCON(A).
EFA Ÿ RCON(B) logically implies EFA Ÿ RCON(A).
EFA Ÿ RCON(A) £P EFA Ÿ RCON(B).
A* £P B*. This defines an embedding from the adequate Tarski degrees
into the P degrees ≥P EFA.
To see that the embedding is onto, let EFA Ÿ A be given,
where A is a P01 sentence. By Theorem 3.6, Let B be an
adequate sentence such that EFA  A ´ RCON(B). Then
B* = EFA Ÿ RCON(B) =P EFA Ÿ A.
QED
THEOREM 5.2. The adequate Tarski degrees, and the P degrees
≥ EFA, are isomorphic distributive lattices with 0,1. The P
degrees are a distribute lattice with 0,1.
Proof: By Theorem 5.1, we need only deal with the P
degrees. The P degrees ≥ EFA have EFA as 0 and 1 = 0 as 1.
Obviously ⁄ and Ÿ are the inf and sup. The P degrees are
also a distributive lattice 0 = 0 as the 0, and 1 = 0 as
the 1. QED
THEOREM 5.3. In the adequate Tarski degrees, every degree <
1 is an inf. 1 is not an inf in the adequate Tarski degrees
(obvious).
Proof: It suffices to work within the P degrees ≥ EFA, where
it is basically well known. Let A be a consistent P01
sentence that logically implies EFA. By self reference, let
j be provably equivalent, over EFA, to
*) every proof of A Æ j has a smaller proof of A Æ ÿj. 32 Let y be
every proof of A Æ ÿj has a smaller proof of A Æ j.
Note that j is the usual Rosser sentence construction over
A, and y is its dual.
We claim that EFA  j ⁄ y, so that A is the inf of A Ÿ j
and A Ÿ y. To see this, argue in EFA. Suppose ÿj Ÿ ÿy. Then
ÿ* Ÿ ÿy. Then we get two distinct Gödel numbers, neither of
which is > the other. This is impossible.
It remains to refute A  j and A  y.
Assume A  j. Let n be the least proof. If there is a
smaller proof of A Æ ÿj then EFA  ÿA. This is
impossible. Hence there is no smaller proof of A Æ ÿj. So
ÿj is true. Since ÿj is S01, we have EFA  ÿA. This is a
contradiction.
Assume A  ÿj. Let n be the least proof. By the previous
paragraph, there is no proof of A Æ j. Hence EFA  j, and
so EFA  ÿA. This is a contradiction.
Note that the previous two paragraphs were conducted within
EFA + Con(A). Thus we obtain
EFA + Con(A)  Con(A + j) Ÿ Con(A + ÿj).
Note that EFA + j Ÿ y  Con(A).
Assume A  y. Then A + j  Con(A). Hence A + j  Con(A +
j). By Gödel’s second incompleteness theorem, A + j is
inconsistent. I.e., A  ÿj. This is impossible. QED
THEOREM 5.4. In the adequate Tarski degrees, every degree
other than 0,1, is a sup. 0,1 are not sups in the adequate
Tarski degrees (obvious for 0).
Proof: (repaired by Albert Visser, private communication).
We work in the P degrees ≥ EFA. Let A be a consistent P01
sentence that proves EFA, but is not provable in EFA. We
show that A is a sup in the P degrees ≥ EFA.
By self reference, let j be provably equivalent, over EFA,
to 33 j*. Every proof of EFA Ÿ ÿj Æ A has a smaller proof of EFA
Æ ÿA or of EFA Ÿ j Æ A.
Let y be
every proof of EFA Æ ÿA or of EFA Ÿ j Æ A has a smaller
proof of EFA Ÿ ÿj Æ A.
We claim that A is the sup of EFA Ÿ (A ⁄ j) and EFA Ÿ (A ⁄
y). Obviously A logically implies these two sentences.
Note that EFA Ÿ j Ÿ y  Con(A)  A. To see this, argue in
EFA, and assume j,y. There cannot be a proof of EFA Ÿ ÿj Æ
A, and so Con(A).
Assume EFA Ÿ ÿj  A. There is no proof of EFA Æ ÿA and no
proof of EFA Ÿ j Æ A. Hence EFA  ÿj, EFA  A. This is a
contradiction.
Assume EFA Ÿ j  A. Let n be the least proof. If there is
a smaller proof of EFA Ÿ ÿj Æ A then, since there is no
proof of EFA Æ ÿA, we have EFA  ÿj, and hence EFA A.
This is impossible. If there is no smaller proof of EFA Ÿ
ÿj Æ A then EFA  j. Hence EFA  A, which is again a
contradiction.
Assume EFA Ÿ y  A. Since EFA  j ⁄ y, we have EFA Ÿ ÿj
 A. But this is impossible. QED
Theorems 5.3 and 5.4 may suggest that in the adequate
Tarski degrees, or alternatively, in the P degrees ≥ EFA,
the degrees ≠ 0,1 look alike. This is not the case.
THEOREM 5.5. In the P degrees ≥ EFA, ($d < 1)("e
0)(inf(d,e) > 0). The d’s with this property are
upwards. This property holds of EFA Ÿ RCON(EFA),
EFA Ÿ j, where j is the Rosser sentence over EFA
proof of Theorem 5.3.). >
closed
but not of
(see the Proof: Obviously the property is closed upwards. Let B be a
P01 sentence. Assume EFA  RCON(EFA) ⁄ B. Then
EFA + ÿB EFA + ÿB EFA + ÿB is
theorem for RCON(EFA).
RCON(EFA + ÿB).
inconsistent (Gödel’s second incompleteness
restricted consistency). 34
EFA  B.
By the proof of Theorem 5.3, inf(j,y) = 0, and y > 0. QED
COROLLARY 5.6. In the adequate Tarski degrees, ($d < 1)("e
> 0)(inf(d,e) > 0). The d’s with this property are closed
upwards. This property holds of EFA, but not of all
adequate Tarski degrees.
Proof: Immediate from Theorem 5.5, using the isomorphism
from the adequate Tarski degrees onto the P degrees ≥ EFA,
defined in the proof of Theorem 5.1. The isomorphism maps
EFA to EFA Ÿ RCON(A). QED 6. Infinite Theories.
We scratch the surface concerning interpretability between
infinite theories. The nature of this subject is quite
different than the finite case, as is clear from Theorems
6.2 – 6.4. Here we will assume that we are dealing with
parameterless interpretations only.
We begin with a well known sufficient condition for
interpretability of infinite theories.
THEOREM 6.1. Every r.e. set of sentences S is interpretable
in Q + {Con(S’): S’ is a finite subset of S}. As a
consequence, if S,T are theories, S is r.e., T is adequate
with an adequacy mechanism, and T proves the consistency of
each finite fragment of S, then S is interpretable in T.
Proof: Let M = Q + {Con(S’): S’ is a finite subset of S}.
Pass to an M definable cut C in M satisfying PFA. Then C =
{Con(S’): S’ is a finite subset of S}.
If C = Con(S) then we can use formalized completeness to
return a model of S. Suppose C = ÿCon(S). In C, let n be
largest such that in C, the first n axioms of S are
consistent. Since in C, for each standard k, the first k
axioms of S are consistent, this n must be a nonstandard
integer. Again use formalized completeness to return a
model M* which, according to C, is a model of the first n
axioms of S. Hence M* really is a model of S. The stated
consequence is immediate. QED
Theorem 6.1 applies to such pairs of theories as ZFC and
ZFC + ÿCH. 35 THEOREM 6.2. (No certificates). There is a S02 sentence A
such that
i. PA + A is interpretable in PA.
ii. ZFC doesn’t prove that PA + A is interpretable in PA,
provided ZFC is consistent.
iii. Moreover, “PA + A interpretable in PA” is provably
equivalent to Con(ZFC) over PA.
iv. ZFC does not prove (R)Con(PA) Æ (R)Con(PA+A), provided
ZFC is consistent.
Proof: Let A = “the greatest n such that ISn is consistent
is less than any inconsistency in ZFC”.
We claim that PA proves the consistency of every ISn + A. To
see this, we argue in PA, and fix n. We know that ISn +
Con(ISn) + ÿCon(ISn+1) is consistent. By exhaustive search,
we see that n is less than the Gödel number of any
inconsistency in ZFC. Hence ISn + A is consistent.
By Theorem 6.1, PA + A is interpretable in PA. Note that
this argument is verifiable in PA + Con(ZF).
Now suppose PA + A is interpretable in PA. Suppose ZFC is
inconsistent, and let n be the least Gödel number of an
inconsistency in ZFC. Since PA  Con(ISn+1), we see that PA
+ A is inconsistent, and therefore not interpretable in PA.
Hence ZFC is consistent. This establishes ii,iii.
In particular, we have shown that if PA + A is consistent
then ZFC is consistent. Hence iv. QED
For additional results along these lines, see [Sh97].
THEOREM 6.3. Let A be a P01 sentence. PA + A is
interpretable in PA if and only if PA proves A.
THEOREM 6.4. There is a P01 sentence A such that
i. PA + A is not interpretable in PA.
ii. EFA  Con(PA) ´ Con(PA + A) ´ RCON(PA).
Proof: Set A to be a usual Rosser sentence over PA. Since
PA does not prove A, we have i by Theorem 6.3. Also ii is
clear as in the proof of Theorem 5.2. QED
On the other hand, interpretations from r.e. theories into
single sentences behave like interpretations from single 36
sentences into single sentences, as the following result
indicates.
THEOREM 6.5. Let T be r.e. and B be adequate. If EFA RCON(B) Æ RCON(T) then T is interpretable in B. T is
interpretable in B if and only if T is faithfully
interpretable in B.
Let T be r.e. and adequate, with finite L(T). In PSE(T) we
can construct a truth definition for L(T) in the standard
way. We can replace T by “all axioms of T are true” in
PSE(T), resulting in the finitely axiomatized system
PSE*(T), which is a conservative extension of T.
Note that PSE*(T) depends on the r.e. presentation of T.
Depending on some details about how to formulate PSE*(T),
one can prove the following result.
THEOREM 6.6. Let S,T be adequate r.e. theories with
adequacy mechanisms. If EFA proves CON(T) Æ CON(S) then
PSE*(S) is interpretable in PSE*(T).
[Sh97] tells us, for example, that the set of S01 sentences
j such that ZF + j is interpretable in NBG is complete S03. Section 7. Observed Linearity.
We restrict attention to the adequate theories from the
literature which constitute a coherent system of axioms for
mathematical reasoning. Here we don’t require
philosophically or foundationally coherent, as this would
arguably be quite restrictive. We mean coherent only in
that there is a modicum amount of naturalness. But we do
require that there be no metamathematical ingredients
present beyond a very standard base theory.
In particular, we include any theory of the form
EFA + {A1,...,Ak}
PA + {B1,...,Bn}
RCA0 + {C1,...,Cm}
Z + {D1,...,Dr}
ZF + {E1,...,Es}
and much more, where A1,...,Ak are robust formulations of
published mathematical theorems in the language of EFA,
B1,...,Bn are such in the language of PA, C1,...,Cm are such 37
in the language of RCA0, and D1,...,Dr,E1,...,Es are such in
the language of set theory.
The striking observation is that one finds a remarkable
linearity. This linearity is found not only with finitely
axiomatized systems – which we have emphasized here – but
with the non finitely axiomatized systems such as PA and
ZFC.
This is perhaps the most intriguing, thought provoking,
fundamental, and deep phenomenon in the whole of the
foundations of mathematics.
In practice, the non finitely axiomatized systems
encountered are almost finitely axiomatized, in the sense
that the axiomatizations consist of finitely many axioms
and finitely many axiom schemes. E.g., PA and ZFC meet this
criteria.
Of course, for any adequate system T based on finitely many
axioms and axiom schemes, there is a canonical version of
PSE(T), which is finitely axiomatizable. Under
interpretability, PSE(T) is a little bit higher than T.
However, in the other orderings that we consider, with one
exception, £S, PSE(T) and T are equivalent. Examples: PA and
ACA0. ZF and NBG.
The formal systems from the literature range from very weak
axioms of arithmetic, to the much stronger axioms of
infinite set theory such as ZFC  and beyond, with the
extensions of ZFC by the so called large cardinal axioms.
We consider the following relations between theories. For
each of these notions, we allow systems with finitely many
axioms and finitely many axiom schemes.
£I. T £I T’ ´ T is interpretable in T’.
£S. Logical strength. T £S T’ ´ EFA  Con(T’) Æ Con(T).
£S*. Liberal logical strength. T £S* T’ ´ PA  Con(T’) Æ
Con(T).
£R. Provably recursive functions. We say that f:N Æ N is a
provably recursive function of T if and only if there is a
Turing machine code e computing f such that T  “e
computes a function from N into N”. We define T £R T’ ´ 38
every provably recursive function of T is a provably
recursive function of T’.
£1. T £1 T’ ´ every P01 sentence provable in T is provable
in T’.
£2. T £2 T’ ´ every P02 sentence provable in T is provable
in T’.
£•. T £• T’ ´ every arithmetic sentence provable in T is
provable in T’.
£O. Provable ordinals. For all systems below discussing
subsets of w (or general sets), we say that a is a provable
ordinal of T if and only if there is a Turing machine code
e computing a well ordering of N of order type a, such that
T  “e computes a well ordering of N”.
Here are some some observed linearity phenomena.
Note that comparability under £O is entirely automatic. This
kicks in when we are extending RCA0.
A. Any two natural systems interpreting Q are comparable
under £I.
B. Any two natural systems extending EFA are comparable
under £I, £S, £S*, £R, £1, £2. They agree under £S*, £R, £1,
£2. If they are finitely axiomatized, then they agree under
£I, £S, £S*, £R, £1, £2.
C. Any two natural systems extending PA are comparable
under £I, £S, £S*, £R, £1, £2, £•. They agree under £S*, £R,
£1, £2, £•. If they are finitely axiomatized, then they
agree under £I, £S, £S*, £R, £1, £2, £•.
D. Any two natural systems extending RCA0 are comparable
under £I, £S, £S*, £R, £1, £2, £•, £O. They agree under £S*,
£R, £1, £2, £•, £O. If they are finitely axiomatized, then
they agree under £I, £S, £S*, £R, £1, £2, £•, £O.
E. If T and T’ are natural systems extending PA, and T < T’
under any of <S*, <R, <1, <2, <•, <O, then T’ proves the
consistency of T.
Here is a table that lists the finite number of levels that
have figured prominently in f.o.m. by means of preferred 39
representatives. We have linearity under £I (with axiom
schemes allowed here). We also exemplify B – E above.
The relevant interpretations here can all be taken to be
parameterless.
PFA.
EFA.
SEFA.
PRA.
RCA0.
IS2.
IS3.
PA.
ACA0.
ACA0 + ("n,x)(TJ(n,x)Ø).
ACA.
RCA0 + TJ(w)Ø.
ACA0 + TJ(w)Ø.
ACA + TJ(w)Ø.
ACA0 + ("x)(TJ(w,x)Ø).
ACA0 + {("a,x)(TJ(a,x)Ø):a < ww}.
ACA0 + {("a < ww)("x) (TJ(a,x)Ø)}.
RCA0 + TJ(ww)Ø.
ACA0 + TJ(ww)Ø.
ACA0 + {("x)(TJ(ww,x)Ø)}.
ACA0 + {("x)(TJ(a,x)Ø):a<Œ0}.
D11CA.
RCA0 + TJ(Œ0).
ACA0 + TJ(Œ0).
ACA + TJ(Œ0).
ACA0 + ("x)(TJ(Œ0,x)Ø).
{ATI(a):a < G0}.
ATR0.
ATI(<G0).
ATR.
P12TI0.
P12TI.
TI.
ID2.
ID<w.
P11CA0.
P11CA.
P11CA + TI.
P11TR0.
P11TR.
P12CA0. 40
P12CA.
P12CA + TI.
Z2.
Z3.
Type Theory.
Weak Zermelo.
ZC.
ZC + ("a < w1)(V(a)Ø).
KP(℘).
ZFC.
ZFC + strongly inaccessibleØ.
ZFC + strongly MahloØ.
ZFC + {strongly nMahloØ: n < w}.
ZFC + ("n < w)(strongly nMahloØ).
ZFC + (weakly compactØ).
ZFC + (indescribableØ).
ZFC + (subtleØ).
ZFC + (almost ineffableØ).
ZFC + (ineffableØ).
ZFC + {nsubtleØ: n < w}.
ZFC + ("n < w)(nsubtleØ).
ZFC + k Æ w Ø.
ZFC + ("a < w1)(k Æ a Ø).
ZFC + 0# Ø.
ZFC + ("x Õ w)(x# Ø).
ZFC + k Æ w1 Ø.
ZFC + RamseyØ.
ZFC + MeasurableØ.
ZFC + Concentrating MeasurableØ.
ZFC + StrongØ.
ZFC + WoodinØ.
ZFC + SuperstrongØ.
ZFC + SupercompactØ.
ZFC + ExtendibleØ.
ZFC + VopenkaØ.
ZFC + Almost HugeØ.
ZFC + HugeØ.
ZFC + SuperhugeØ.
ZFC + ("n < w)(nhugeØ).
ZFC + Rank into ItselfØ.
ZFC + Rank + 1 into ItselfØ.
VB + V into VØ.
There are extremely natural theories for which linearity
fails, in which Q is not interpretable. For example, let 41
T = axioms for discrete linear orderings without endpoints
(every point has an immediate predecessor and an immediate
successor).
T’ = axioms for dense linear orderings without endpoints
(between any two elements there is a third, and no
endpoints).
Then T £I T’ and T’ £I T both fail. Also note that T,T’ are
finitely axiomatized and complete.
Interpretability is very interesting in this kind of tame
world, below Q – where it takes on an entirely different
character than the theory ≥I Q that we have focused on.
In particular, it would be interesting to understand
interpretability among subsystems of RCF (real closed
fields) and ACF (algebraically closed fields) and
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* This research was partially supported by NSF DMS 0245349. ...
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 JOSHUA
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