2
Extension Axioms Revisited
Lemma Forall t, s0: m(t t,s0) = limn mn(t t,s0) =
1
This is if and only if
•Lemma Forall s0: m(ss0) = 1
•Proof: Let t1, t2, …, tm be all
k-types
Then: ss0 = tt1,s0 Æ … Æ ttm,s0
tt,s = 8 x. (t(x) ) 9 z.s(x, z))
ss0 = 8 x.9 z. s0(x,z)
Extension
axiom
axiom
“Loose”
extension
axiom
extension
axiom
Where s(x, z) = t(x) Æ s0(x, z)