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ONT Re: Model Theory




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Note 13

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| 1.  Introduction
|
| 1.2.  Model Theory for Sentential Logic (cont.)
|
| 1.2.10.  Lemma.  Suppose !C! is a maximal
|          consistent set of sentences in $S$.
|
|          Then:
|
|          1.  For each sentence p, exactly one of
|              the sentences p, ~p belongs to !C!.
|
|          2.  For each pair of sentences {p, q},
|              we have that p & q belongs to !C!
|              if and only if both p and q
|              belong to !C!.
|
| We leave the proof as an exercise.
|
| Now consider a set !S! of sentences of $S$.
|
| We shall say that A is a 'model' of !S!,
| in symbols,
|
|       A |= !S!,
|
| iff every sentence p in !S! is true in A.
|
| !S! is said to be 'satisfiable' iff it has at least one model.
|
| We now prove the most important theorem of sentential logic,
| which is a criterion for a set !S! to be satisfiable.
|
| 1.2.11.  Theorem.  (Extended Completeness Theorem).
|
|          A set !S! of sentences of $S$ is consistent
|          if and only if !S! is satisfiable.
|
| Proof.   Assume first that !S! is satisfiable, and let A |= !S!.
|
|          We show that every sentence deducible from !S! holds in A.
|
|          Let q_0, q_1, ..., q_n be a deduction of q_n from !S!.
|
|          Let m =< n.
|
|          If   q_m is in !S! or
|          if   q_m is a tautology,
|          then q_m holds in A.
|
|          If q_m is inferred from two sentences
|          q_k and q_k => q_m which hold in A,
|          then q_m must clearly hold in A.
|
|          It follows by induction on m that each of
|          the sentences q_0, q_1, ..., q_n holds in A.
|          Since S & ~S does not hold in A, it is not
|          deducible from !S!, so !S! is consistent.
|
|          Now assume that !S! is consistent.
|
|          By Lindenbaum's theorem we enlarge !S!
|          to a maximal consistent set !C!.
|
|          We now construct a model of !S!.
|
|          Let A be the set of all sentence symbols S in $S$
|          such that S is in !C!.  We show by induction that,
|          for each sentence p, we have:
|
|          1.  p in !C!  if and only if  A |= p.
|
|          By definition, (1) holds when p is a sentence symbol S_n.
|
|          Lemma 1.2.10.1 guarantees that,
|          if   (1) holds when p =  q,
|          then (1) holds when p = ~q.
|
|          Lemma 1.2.10.2 guarantees that,
|          if   (1) holds when p = q and when p = r,
|          then (1) holds when p = q & r.
|
|          From (1), it follows that A |= !C!,
|          and since !S! c !C!, that A |= !S!.  -|
|          
| We can obtain a purely semantical corollary.
|
| !S! is said to be 'finitely satisfiable' iff
| every finite subset of !S! is satisfiable.
|
| 1.2.12.  Corollary.  (Compactness Theorem).
|
|          If   !S! is finitely satisfiable,
|          then !S! is satisfiable.
|
| Proof.   Suppose !S! is not satisfiable.
|
|          Then by the extended completeness theorem !S! is inconsistent.
|
|          Hence,  !S! |- S & ~S.
|
|          In the deduction of the sentence S & ~S from !S!
|          only a finite set !S!_0 of sentences of !S! is used.
|
|          It follows that !S!_0 |- S & ~S,  so !S!_0 is inconsistent.
|
|          Then !S!_0 is not satisfiable, so !S! is not finitely satisfiable.  -|
|
| Notice that the converse of the compactness theorem is trivially true,
| that is, every satisfiable set of sentences is finitely satisfiable.
|
| Chang & Keisler, 'Model Theory', pages 10-11.
|
| C.C. Chang and H.J. Keisler, 'Model Theory',
| North-Holland, Amsterdam, Netherlands, 1973.

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