1
5.6.
Π
0
1
correct internal arithmetic,
simplification.
We fix M* = (A,<,0,1,+,,•,
↑
,log,E,c
1
,c
2
,...,X
1
,X
2
,...) from
Lemma 5.5.8. Let M = (A,<,0,1,+,,•,
↑
,log,E,c
1
,c
2
,...). We
can view the main point of this section as the derivation
of a suitable form of the axiom of infinity.
Note that we have yet to use that the c’s lie outside
α
(E;2,<
∞
), from Lemma 5.5.8 iii). In this section, we use
this in an essential way. It is needed in order to obtain
any useable form of the axiom of infinity.
The one and only use of the fact that the c’s lie outside
α
(E;2,<
∞
), in this Chapter, is in the proof of Lemma 5.6.6.
There we use that c
5
∉
α
(E;2,<
∞
).
We first prove the existence of a least internal set I
containing 1, and closed under + 2c
1
(see Lemma 5.6.7 and
Definition 5.6.3). We then define natural arithmetic
operations on I (see Lemma 5.6.10), resulting in the
structure M(I) which satisfies PA(L) (see Lemma 5.6.11).
Then we define a natural external isomorphism h from M(I)
into M. We then show that M(I) Mrng(h) satisfies PA(L) +
TR(
Π
0
1
,L) using the solution to Hilbert's 10th Problem (see
Lemma 5.6.14).
At this point, we only care that M(I) satisfies TR(
Π
0
1
,L),
and that I is internally well ordered. The external h is
used only to take advantage of the fact that M satisfies
TR(
Π
0
1
,L). We think of h as external because its range is
not be a subset of E.
M(I) will provide us with the arithmetic part of the
structure M# in Lemma 5.6.18.
We remind the reader that for x,y
∈
A, xy always means
xy if x
≥
y; 0 if x < y.
Recall that
α
(E) is the set of all values of terms in L at
arguments from E (Definition 5.3.3).
LEMMA 5.6.1.
α
(E) = EE.
Proof: According to Lemma 5.4.12,
α
(E) = EE holds in the
structure M given by Lemma 5.3.18, which is the same as the
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structure M given by Lemma 5.4.17. Therefore
α
(E) = EE
also holds in the structure M* given by Lemma 5.5.8, which
is an expansion of M. QED
DEFINITION 5.6.1. We say that x is critical if and only if
x
∈
EE
∧
2xc
1
+1
∈
E.
LEMMA 5.6.2. Let p,x > 0, where p
∈
N and x is critical.
Then p,2x+p are critical.
Proof: Let p,x be as given. By Lemma 5.6.1, p,2x+p
∈
EE.
Note that pc
1
∈
α
(E;1,<
∞
). Hence by Lemma 5.5.8 v), 2pc
1
+1
∈
2
α
(E;1,<
∞
)+1
⊆
E. Hence p is critical.
By 2xc
1
+1
∈
E and Lemma 5.5.8 v),
2xc
1
+1,c
1

≤
(2xc
1
+1)+(pc
1
1)
≤
4p2xc
1
,c
1
.
(2xc
1
+1)+(pc
1
1) = 2xc
1
+pc
1
= (2x+p)c
1
∈
α
(E;1,<
∞
).
2(2x+p)c
1
+1
∈
2
α
(E;1,<
∞
)+1
⊆
E.
Hence 2x+p is critical. QED
LEMMA 5.6.3. Let x
≥
1 be critical. Suppose that for all
critical y
∈
[2,x], there is a critical z such that y
∈
{2z,2z+1}. Then x+1 is critical and there is a critical z
such that x+1
∈
{2z,2z+1}.
Proof: Let x
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 Fall '08
 JOSHUA
 Math, Order theory, Lemma, M*

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