1
5.5. Comprehension, indiscernibles.
We fix M = (A,<,0,1,+,-,•,
↑
,log,E,c
1
,c
2
,...) and terms
t
1
,t
2
,... of L(E) be given as in Lemma 5.4.17.
We now consider unbounded quantifiers. Below, Q indicates
either
∀
or
∃
. All formulas of L(E) are interpreted in M.
LEMMA 5.5.1. Let n,m
≥
0, r
≥
1, and
ϕ
(v
1
,...,v
n+m
) be a
quantifier free formula of L(E). Let x
n+1
,...,x
n+m
∈
E
∩
[0,c
r
]. Then
(Q
n
x
n
∈
E)...(Q
1
x
1
∈
E)(
ϕ
(x
1
,...,x
n+m
))
↔
(Q
n
x
n
∈
E
∩
[0,c
r+1
])...(Q
1
x
1
∈
E
∩
[0,c
r+n
])(
ϕ
(x
1
,...,x
n+m
)).
Proof: We prove the following statement by induction on n
≥
0.
Let m
≥
0, r
≥
1,
ϕ
(x
1
,...,x
n+m
) be a quantifier free formula
in L(E), Q
1
,...,Q
n
be quantifiers, and x
n+1
,...,x
n+m
∈
E
∩
[0,c
r
]. Then
(Q
n
x
n
∈
E)...(Q
1
x
1
∈
E)(
ϕ
(x
1
,...,x
n+m
))
↔
(Q
n
x
n
∈
E
∩
[0,c
r+1
])...(Q
1
x
1
∈
E
∩
[0,c
r+n
])(
ϕ
(x
1
,...,x
n+m
)).
The basis case n = 0 is trivial. Assume this is true for a
given n
≥
0. Let m
≥
0, r
≥
1, and
ϕ
(x
1
,...,x
n+1+m
) be a
quantifier free formula in L(E). Let x
n+2
,...,x
n+1+m
∈
E
∩
[0,c
r
]. We wish to verify that
(Q
n+1
x
n+1
∈
E)...(Q
1
x
1
∈
E)(
ϕ
(x
1
,...,x
n+1+m
))
↔
(Q
n+1
x
n+1
∈
E
∩
[0,c
r+1
])...(Q
1
x
1
∈
E
∩
[0,c
r+n+1
])
(
ϕ
(x
1
,...,x
n+1+m
)).
By duality, we may assume that Q
n+1
is
∃
. Thus we wish to
verify that
1) (
∃
x
n+1
∈
E)(Q
n
x
n
∈
E)...(Q
1
x
1
∈
E)(
ϕ
(x
1
,...,x
n+1+m
))
↔
(
∃
x
n+1
∈
E
∩
[0,c
r+1
])(Q
n
x
n
∈
E
∩
[0,c
r+2
])...
(Q
1
x
1
∈
E
∩
[0,c
r+n+1
])(
ϕ
(x
1
,...,x
n+1+m
)).
Let x
n+1
∈
E
∩
[0,c
r+1
] witness the right side of 1). I.e.,
2) (Q
n
x
n
∈
E
∩
[0,c
r+2
])...(Q
1
x
1
∈
E
∩
[0,c
r+n+1
])(
ϕ
(x
1
,...,x
n+1+m
)).