3
Extension Axioms Revisited
•
Moreover, this is iff
•
Lemma
Forall s0:
m
(
s
s0
¹
) = 1
•
Proof
:
Every
s
s0
is the conjunction of several
expressions
s
s1
¹
, where s1 states s0 over fewer
variables, by repeating some of them.
s
s0
¹
=
8
x
.
¹
(
x
)
)9
z. s0(
x
,z)
“not-so-loose”
extension
axiom
s
s0
=
8
x
.
9
z. s0(
x
,z)
“Loose”
extension
axiom