26
Fagin’s 0/1 Law for FO
•
Theorem
For every
f
2
FO, either
m
(
f
) = 0 or
m
(
f
) = 1.
•
•
Proof
.
•
Case 1:
R
˛
f
.
Then there exists m extension
axioms s.t.
t
1
, …,
t
m
˛
f
.
Then
m
n
(
f
)
¸
m
n
(
t
1
Ć
…
Ć
t
m
)
!
1
•
Case 2:
R
2
f
.
Then
R
˛
:
f
, hence
m
(
:
f
) = 1, and
m
(
f
) = 0