CS 641 Lecture -*- Outline -*-

* selection and individuals (7.5)

  To allow formulation of arithmetic, we need two more postulates
  for higher-order logic.

  Need a time with an infinite number of elements,
  and the ability to choose one element from any set of elements.

** selection
------------------------------------------
       SELECTION (AXIOM OF CHOICE)

A new constant, the selection operator

   \choose : (S -> Bool) -> S

Notation:

   (\choose v :: t) == \choose.(\ v . t)

            INFERENCE RULES

\choose introduction:

   ____________________________________
   Phi |- (\exists v :: t)
           ==> t[v := (\choose v :: t)]
   * if (choose v :: t) is free for v in t

\choose elimination:

          Phi |- t ==> s
   _____________________________________
   Phi |- t[v := (\choose v :: t)] ==> s
   * if v is not free in Phi or s, and
     if (choose v :: t) is free for v in t
------------------------------------------

  Here is a basic lemma we would like,
  to confirm our intuition about the \choose operator.

  Lemma. (One point rule for \choose).
         Suppose x does not occur free in t.
         Then |- (\choose x :: x == t) == t.

  Proof: Suppose x does not occur free in t.  Since we have to
  "eliminate" the \choose operator, and since the inference rules for
  \choose are Boolean formulas, we start with the whole formula,
  as otherwise we have no Booleans.  Since we thus have to simplify the
  formula to T, and since the \choose rules are implications, the
  overall shape will be (\choose x :: x == t) == t  <==  T.
  Hence we will actually be using the \choose introduction rule.

  |-  (\choose x :: x == t) == t.
  <==> { substitution, x does not occur free in t }
      (x == t)[x := (\choose x :: x == t)]
  <==  { \choose introduction, since we have to conclude from T,
         (\choose x :: x == t) is free for x in (x == t),
          since x does not occur free in t, and so can't be captured }
      (\exists x :: x == t)
  <==  { \exists introduction, t is free for x in x == t,
          since x does not occur free in t, and so can't be captured }
      (x == t)[x := t]
  <==> { substitution }
      t == t
  <==> { == is reflexive }
      T
  QED

  Another way to view the use of select is as an equivalent way
  to write a formula that is universally quantified, by selecting a witness.

  Lemma. (\exists \choose correspondence)
  Let t be a Boolean term in which v may appear free
  and such that t is free for v in t.
  Then |- (\exists v :: t) <==> t[v := (\choose v :: t)].

  Proof: 
  |-  (\exists v :: t) 
  <==> { ==> antisymmtric }
         *  (\exists v :: t) 
         ==> { \choose introduction, (\choose v :: t) is free for v in t,
               because t is free for v in t }
            t[v := (\choose v :: t)]
         *  t[v := (\choose v :: t)]
         ==> { \exists introduction,  (\choose v :: t) is free for v in t,
               because t is free for v in t }
            (\exists v :: t) 
      t[v := (\choose v :: t)].
  QED

------------------------------------------
         ARBITRARY ELEMENT CONSTANT

For any type S

        ^
    arb = (\choose x :: x \in S)
------------------------------------------

** type of individuals
------------------------------------------
          TYPE OF INDIVIDUALS

Ind, a type with infinitely many elements

infinity axiom:
  |- (\exists f \in Ind -> Ind ::
        oneone.f /\ !(onto.f))

Pictorial example:

     infinite    finite
       set        set
      .     .
      .     .
      .     .
      3 --> 4    3
      2 --> 3    2 --> 3
      1 --> 2    1 --> 2
      0 --> 1    0 --> 1
            0          0

definitions:
         ^
oneone.f =
 (\forall x x' :: f.x == f.x' ==> x == x')

       ^
onto.f =
 (\forall y :: (\exists x :: y == f.x))

            ^
bijective.f = oneone.f /\ onto.f
------------------------------------------