2
Extension Axioms Revisited
•
Lemma
Forall t, s0:
m
(
t
t
,
s0
) = lim
n
m
n
(
t
t,s0
) = 1
•
This is if and only if
•
Lemma
Forall s0:
m
(
s
s0
) = 1
•
Proof
:
Let t
1
, t
2
, …, t
m
be all k-types
•
Then:
s
s0
=
t
t
1
,s0
Æ
…
Æ
t
t
m
,s0
t
t,s
=
8
x
. (t(
x
)
)
9
z.s(
x
, z))
s
s0
=
8
x
.
9
z. s0(
x
,z)
Extension
axiom
“Loose”
extension
axiom
Where s(
x
, z) = t(
x
)
Æ
s0(
x
, z)