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Unformatted text preview: Gdels Metatheorem (45.17) and the Strong Completeness Theorem for FOTs (46.2) Branden Fitelson 04/12/07 Before getting to the salient proofs, its important to understand Hunters terminology consistent set of WFFs of a first order theory K . For Hunter, is a consistent set of WFFs of K iff there is no WFF A of K such that K A and K A . As a result, this definition of consistent set of WFFs of K implies that K is itself a consistent first-order theory! That is, an inconsistent first order theory K does not have any consistent sets of WFFs on this definition. This sounds a bit odd, but its crucial for the proofs below. In this handout, I will go through the proper proofs of 45.17 and 46.2. To this end, I will begin with the background ingredients of the proof of 45.17: metatheorem 45.16, and Lemmas 1 and 2. 45.16 . If is a consistent set of closed WFFs of a first order theory K , then has a denumerable model. Proof. Assume is a consistent set of closed WFFs of a first order theory K . Then, by Hunters definition (above), there is no WFF A of K such that K A and K A . Therefore, it follows that the first order theory K + is a consistent first order theory. If K + were inconsistent, then there would have to be a WFF A of K + such that both A and A were theorems of K + . That would imply the existence of an A such that K A and K A , which contradicts Hunters definition of consistent set of WFFs of K . Since K + is a consistent first order theory, it must have a denumerable model [this is implied by theorems 45.1045.14]. Thus, itself has a denumerable model ( is a subset of the set of theorems of K + ). Lemma 1 for 45.17 . If is a consistent set of WFFs of a first order theory K , then is also a consistent set of WFFs of of the first order theory K , where K is the first order theory one gets when one adds denumerably many new constant symbols with an effective enumeration h c 1 ,...c n ,... i to K . Proof. Assume is a consistent set of WFFs of a first order theory K , and assume that K is K with the new constant symbols h c 1 ,...c n ,... i added to it. Now, assume, for reductio , that is an inconsistent set of K . Then, by definition, this means that there is a WFF B of K such that K B and K B . So, since derivations are finite, there is a finite subset such that K B and K B . These derivations in K can be converted into derivations in K , as follows. Let X = Xv i /c i , where v i is the i th variable in our enumeration that does not occur in either of the derivations K B or K B , and c i is the i th constant symbol in our enumeration of new symbols added to K to yield K . Then, = , since and is a set of WFFs of K (and so do not contain any c i s). Moreover, K B and K B . Why? Think about...
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- Spring '07