Logic: Lecture 15, 09 October 2012
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Aim of the next few lectures: 
  To understand Godel's Incompleteness Theorem

Reference:  
  "A Mathematical Introduction to Logic"
  Herbert B Enderton
  Chapter 3

Basic definitions
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* Logical consequence:

  Let T be a set of sentences, phi a sentence.  

  phi is a logical consequence of T, written T |= phi, if for
  every model M of T, M |= phi (i.e. M |= T implies M |= phi).

  Consequences of T:  Con(T) = { phi | T |= phi }

  Recall that for a model M, Th(M) is the set of sentences true
  in M: Th(M) = { phi | M |= phi).  

  Thus, Con(T) = Intersection_{M | M |= T} Th(M)

* Theory

  A set of sentences T is a theory if it is closed with respect
  to logical consequence: If T |= phi, then phi is in T.  In
  other words, T is a theory iff T = Con(T).

  A complete theory T is one where for every sentence phi, either
  phi or ~phi is in T.

  Recall that for any model M, Th(M) is a complete theory.  Thus,
  one way to show that T is a complete theory is to show that any
  two models M and M' of T are elementary equivalent --- i.e., 
  Th(M) = Th(M').

  A complete theory with an effective axiomatization is
  decidable.  We can enumerate all deriviations and try them out
  systematically.  For every sentence, either a proof of phi or
  ~phi will eventually emerge.

  Note: Effective axiomatization means the set of axioms is
        "decidable" --- given a sentence, we can unambiguously
        compute whether it is or is not an axiom.  A special case
        is a finite axiomatization

* Categoricity

  A theory T is aleph_0-categorical if all countable models of T
  are isomorphic.  In general, for a cardinal kappa, T is
  kappa-categorical if all models of T of cardinality kappa are
  isomorphic.

* Los-Vaught test:

  Let T be a theory in a countable language.  Assume that T has
  no finite models.

  a. If T is aleph_0-categorical then T is complete.
  b. If T is kappa-categorical for some infinite cardinal kappa,
     then T is complete

  Proof:

  a. Let A and B be any two infinite models of T.  Consider
     Th(A).  By Lowenheim-Skolem, there is a countable model A'
     of Th(A).  Likewise, B' for Th(B).  A' and B' are isomorphic
     (by assumption), so Th(A) = Th(B).  So all infinite models
     of T are elementary equivalent and T is complete.

  b. Likewise.  Upward+downward Lowenheim-Skolem imply there are
     infinite models of all cardinalities, so get to a model of
     cardinality kappa and use the same argument as part a.


Number Theory
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Consider the language NT = (R,F,C) where R = {<}, F = {S,+,*,E}
and C = {0}.  The intended interpretation is the structure
MN = (Nat,0,S,<,+,*,E), where Nat = {0,1,2,...}, S is successor, < is
less than, + is plus, * is times and E is exponentiation.  By
"number theory" we mean the first order theory of this structure.

We will look at the following sublanguages of NT:

  NT_S = ({},  {S},     {0})
  NT_< = ({<}, {S},     {0})
  NT_+ = ({<}, {S,+},   {0})
  NT_* = ({<}, {S,+,*}, {0})

We use similar names for the corresponding restrictions of the
model MN = (Nat,0,S,<,+,*,E), so MN_S is the structure (Nat,0,S),
MN_< is the structure (Nat,0,S,<) etc.

Typical questions:

a. Is the corresponding theory decidable?  If so, is there a nice
   set of axioms?  Is there a finite set of axioms?

b. What subsets of Nat are definable in the theory?

c. What do nonstandard models look like?

Notation: 
   S^i(0) = S(S(..S(0)...)), i.e., S applied to 0 i times, is a
   term denoting the number i.

NT_S: Natural numbers with successor

Axioms

S1.   Ax S(x) =/= 0
S2.   Ax Ay (S(x) = S(y) -> x = y)
S3.   Ay (y =/= 0 -> Ex y = S(x))
S4.1  Ax S(x) =/= x
S4.2  Ax S^2(x) =/= x
..
S4.n  Ax S^n(x) =/= x

Let A_S be this (infinite) set of axioms.

Proposition:  Con(A_S) is included in Th(MN_S)

- Verify that all axioms from A_S hold in MN_S

To show that Con(A_S) = Th(MN_S) we will use the Los-Vaught test.

We first examine the structure of models of A_S.

  Every model of A_S looks like a "standard" copy of Nat plus
  zero or more Z-chains (sequences isomorphic to the set of
  integers Z).  The number of Z chains may be countable or
  uncountable.

  Each Z-chain is countable.  If there are kappa Z-chains, it
  follows from basic set theory that the overall cardinality of
  the model is max(aleph_0,kappa).

  Conversely, any model that consists of a standard copy of Nat
  plus a collection of Z-chains is a model of A_S.

  Any two such models with same cardinality kappa, kappa strictly
  above aleph_0, are isomorphic:

  - the standard copies of Nat are isomorphic to each other

  - any Z-chain is isomorphic to any other Z-chain

  - since the models have the same cardinality, there is a
    bijection between the two sets of Z-chains

  - combine the bijection between the sets of Z-chains with the
    internal isomorphism between Z-chains to get an overall
    isomorphism of the models

  By Los-Vaught test, Con(A_S) is complete.

Proposition:  
  If A and B are theories such that A is a subset of B, and both
  theories are complete, then A = B.

  - Prove this for yourself.

From the proposition, since Con(A_S) is a subset of Th(MN_S) and
both are complete theories, Con(A_S) = Th(MN_S).

Hence, Th(MN_S) is complete and has an effective axiomatization,
so it is decidable.

Next time we will give a more direct proof using "quantifier
elimination" --- we will show that every formula in NT_S is
equivalent to one without quantifiers and argue that this yields
decidability.

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