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By Wolfram Pohlers (author), Thomas Glaß (editor)

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Thus :F1 2 M which implies F 2= M because otherwise f:F1 F g would be an S -inconsistent nite subset of M: The remaining cases are similar (or may be reduced to the previous ones because f: ^g is a complete set of connectives). 10 we obtain the following properties of maximally nitely sententially consistent sets. 11. Let M be a maximally nitely sententially consistent set of Lformulas. 4. Propositional Properties of First Order Logic 31 a) (:F) 2 M , F 2= M b) (F _ G) 2 M , F 2 M or G 2 M c) (F ^ G) 2 M , F 2 M and G 2 M d) (F !

We are going to study the propositional structure and properties of rst order formulas in the next section. 6. Thus all we have to check is g). Assume S j= :9xF ] for some L-structure S and an S -assignment . e. 1. a) Prove: j= 8x(F ^ G) ! 8xF ^ 8xG: b) Let L be a rst order language including a constant symbol 0: Determine Lformulas F and G with 6j= 8x(F _ G) ! 2. Let L be a rst order language and P a predicate symbol of L: Which of the following formulas are valid? a) (F ! G) ! ((F ! :G) ! :F) 26 I.

B) We call M a Henkin set if M contains all witnesses and also all formulas Fx (t) ! 9xF where t is an arbitrary term. 36 I. Pure Logic The term c9xF is supposed to witness the element x whose existence is claimed in 9xF: In general there is a di erence between c9xF and c9yF . But there isn't any semantical di ernce for the Henkin constants for 9xF and 9yFx (y): This will be proved in the following proposition. 3. Let M be a sententially consistent Henkin set and B a boolean assignment such that GB = t for all G 2 M: If F and F~ are obtained by renaming bounded variables, then it is F B = F~ B : Proof.

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