Logic: Lecture 17, 16 October 2012
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NT_< = ({<}, {S}, {0})
-----------------------

Not categorical for any infinite cardinality, but finitely
axiomatizable

(S3)  Ay     (y =/= 0 -> Ex y = S(x))
(L1)  AxAy   (x < S(y) <-> x <= y)
(L2)  Ax     ~(x < 0)
(L3)  AxAy   (x < y or x = y or y < x)
(L4)  AxAy   (x < y -> ~(y < x))
(L5)  AxAyAz (x < y -> y < z -> x < z)


It can be verified that axioms (S1), (S2), (S4.n) for all n from
the axioms A_S can be derived from (S3), (L1)-(L5) (see
Enderton), so any model of A_< is also a model of A_S. Hence, any
model of A_< looks like a copy of Nat plus some Z-chains.

Unlike NT_S, here all Z-chains must be ordered with respect to
each other.  Since we don't specify anything, the order among
Z-chains may be discrete or dense or a mixture.  Hence, at any
fixed cardinality we can construct models that are not order
isomorphic, so we don't have categoricity.

Quantifier elimination:

Again suffices to consider

  Ex (a_1 and ... and a_n), each a_i atomic or negation of atomic

  Atomic formulas are either S^i(u) = S^j(v) or S^i(u) < S^j(v)

1. Eliminate negation

   ~(t < t') is (t' < t or t = t')
   ~(t = t') is (t' < t or t < t')

2. Assume x occurs in each atomic formula

3. Eliminate quantifiers

   Case 1: There is at least one equality.  Use same technique as
           NT_S.

   Case 2: Each a_i is an inequality.

     Separate into conjunctions of upper and lower bounds: i.e, 
     formula is of the form

      OR   t_i < S^{m_i}(x) or   OR    S^{n_j}(x) < u_j
     i < l                        j < k
            
     |---------------------|    |-----------------------|    
           Lower bounds               Upper bounds
 
     No upper bounds => formula reduces 0 = 0.

     No lower bounds => x has a valid assignment iff 0 satisfies
                        the upper bounds, so replace x by 0.

     Otherwise, pair up lower and upper bounds.

      OR   t_i < S^{m_i}(x) and  S^{n_j}(x) < u_j
     i < l 
     j < k

     Convert into intervals

      OR   S^{n_j}(t_i) < S^{m_i+n_j}(x) and
     i < l                S^{m_i+n_j}(x) < S^{m_i}(u_j)
     j < k

     i.e., for some i < l, j < k
           S^{n_j}(t_i) < S^{m_i+n_j}(x) < S^{m_i}(u_j)

     Need to check that some interval is feasible.  Replace by
     two checks:

     S_{n_j+1}(t_i) < S^{m_i}(u_j)   [Interval is nonempty]
     S^{m_i+n_j}(0) < S^{m_i}(u_j)   [Upper bound is feasible]

Corollary:
 
a. Con(A_<) is complete  (by quantifier elimination)

b. Con(A_<) = Th(MN_<)   (inclusion + completeness of Con(A_<))

c. Th(MN_<) is decidable

   - Completeness plus axiomatization
   - More directly, by quantifier elimination

Definability

  Subsets of Nat:    finite or cofinite, like NT_S

  Binary relations:  + is not definable (else we can define
     evenness, which is neither finite nor cofinite)


NT_+ = ({<}, {S,+}, {0})
-----------------------

Presburger arithmetic

Models are what we saw earlier: Copy of Nat plus countable
  Z-chains densely ordered.

Will not axiomatize, just prove quantifier elimination.

Cannot eliminate quantifiers directly

   Ex y = x + x  has no equivalent without quantifiers

Instead, replace by (mod m) relations for all m and show
quantifier elimination for that.   For instance, the earlier
formula is equivalent to

   y = 0 (mod 2)


As usual, suffices to consider formulast

   Ey (a_1 and ... and a_n), each a_i atomic or negation of atomic

Terms are of the form S^k(0) + n_1 x_1 + ... + n_k x_k, where nx
   abbreviates x + ... + x, n times.

Atomic formulas are t = t', t < t', t = t' mod m

1. Eliminate negation

   ~(t < t') is (t' < t or t = t')
   ~(t = t') is (t' < t or t < t')
   ~(t = t' mod m) is t = t'+ 1 mod m or ... or t = t'+ (m-1) mod m

   Now atomic formulas are of the form

        ny + t = u
        ny + t = u mod m
        ny + t < u
             u < ny + t

   Rewrite, for convenience, as

           ny < t - u
           ny = t - u mod m
           ny < t - u
        u - t < ny

2. Take LCM to make coefficients of y uniform.  Use the fact that

       a = b mod m  iff ka = kb mod km

3. All y terms now have a common coefficient p.  Eliminate the
   coefficent.  Replace all py by x and add conjunct "x = 0 mod p".

4. Special case:

   If one of the atomic formulas is  x + t = u,

   replace x by (u - t) everywhere and add conjunct t <= u.

5. No occurrence of =

   Formulas consist of lower bounds, upper bounds and
   congruences.

   a. If no congruences, proceed as for NT_<

   b. If there are congruences modulo m1, m2, ... mn, let M = lcm
      of (m1,...,mn).  Observe that residues (r0,...,rn) modulo
      (m1,...,mn) are preserved modulo (M,....,M).

      Let lower bounds be L1 to Ll.  Suffices to search for valid
      solutions within

      L1, L1+1, ... , L1+(M-1)
      L2, L2+1, ... , L2+(M-1)
      ..
      Ll, Ll+1, ... , Ll+(M-1)
   
      In case all Lj's are negative add a constraint x >= 0.
      Express this as an additional lower bound 0 < x + 1.

      Rewrite the constraints as quantifier free formula by
      taking disjuntion over (all lower bounds, all q in 1 to M)

      See Enderton for details.

Decidability:

  Quantifier elimination translates sentences to sentences.  For
  sentences, no variables, so the resulting formula can be
  checked one atomic formula at a time.

Note that quantifier elimination blows up the formula: the
complexity is double exponential, so this procedure is not
computationally feasible in general, but it still used in modern
verification tools!

Theorem:
  A set of numbers is definable in MN_+ iff it is ultimately
  periodic.

  Periodic:  Ep such that for all n, n in X iff n+p in X
  Ultimately periodic:  EN such for all n > N, n in x iff n+p in D.

Corollary:
  Set of perfect squares is not definable.

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