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0/1 Law for FO
•Lemma 
For every extension axiom t,  m(t) = limn mn(t) = 1
•Proof: later
•
•Corollary For any m extension axioms t1, …, tm:         m(t1 Æ … Æ tm) = 1
•Proof
mn(:(t1 Æ … Æ tm))
=   mn(: t1 Ç … Ç : tm)
·   mn(: t1) + … + mn(: tm)   ! 0