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C00002 00002	@BEGIN[Comment]
C00005 00003	@BEGIN[Comment]  Outtakes
C00010 00004	@BEGIN[Comment] Clauses in Predicate Calculus, before Propositional Case
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Outline

*** Intro ***
  - Need for Learning & COMMUNICATION
  - Not obvious (esp at first)
  - Problem of def'n -- like Church's thesis
  <Reading guide>

*** BODY ***
  - Loose (intuitive) definition
  - Several sketches examples

Propositional case
  - Simplifying assumptions

  - Def'n 1: NU - new wrt to Theory
  - Guess: Independence  [proof based definition]
  - Thm: elimination of interp follows from proof procedures]

  - Def'n 2: New(Th,s,A)
	[need symbols for things like analogy, communication]
  - Guess#0: Syntactic approach (embedded in sentence)
  - Thm: G#0 neither necessary nor sufficient (exs)

  - Guess#1: I[Th'](A) ⊂ I[Th](A)
  - G#1 matches Guess for Def'n 1
  - Thm: G#1 Insufficient (counterexample: A=B)

  - Analysis of Guess#1's problem
	[no mention of other symbols, dependency]
  - Guess#2: Use [[Ij]] equivalence class
  - Thm: G#2 better than G#1
	[If s new (using G#1) then s new (using G#2)]
  - Thm: G#2 insufficient (counterexample: B=C, given A=B)

  - Guess#3: All n-tuples of assignments
  - Thm: G#3 better than G#2
	[If s new (using G#1) then s new (using G#2)]

  - Justification for G#3

  - Comments wrt Predicate Calculus case

Conclusion
  - Rehash of examples
  - comparison w/Statistical (info theory)
  - Summary
	* Def'n again
	* Uses (communication/learning)
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@END[Comment]
@BEGIN[Comment]  Outtakes

@****** Which of the below two cases should I use --
the one with =, or the one which takes so many dumb symbols? *****@*
@SubHeading[Case 1: Father(Arnold)]
Suppose
@BEGIN[Equation]
Th1 @K[Gets] @K[LeftBrace] @↑Father(Arnold Bill)
@\Ancestor(Bill Charles)
@\@K[ForAll] x,y,z. (Ancestor x y)@K[And](Ancestor y z) @K[Implies] @~
(Ancestor x z) @K[RightBrace]@FOOT{
The notation @T<T @K[Gets] @g[x]> means the theory @T<T> is
the deductive closure of the set, @g[x].}.
@END[Equation]
Then hearing @g[t],
@BEGIN[Equation]
@g[t] @K[Identical] @K[ForAll] x,y. (Father x y) @K[Implies] (Ancestor x y),
@END[Equation]
would indeed say something about @T[Arnold],
even though it did not include that symbol.
(E.g. we can then deduce facts like
@T<Ancestor(Arnold Bill)>, and @T<Ancestor(Arnold Charles)>;
and perhaps use these facts to help specify the referent of the @T[Arnold] symbol.)
That is, we would want
@T<New(Th1,Arnold,@g[t])>.

and the (pseudo-)sentence
@T<"A@K[Eq]x"> refers to the sentence in which the referent for @T<x> is
substituted in the underlined part of the sentence @T<x@K[Eq]@u[  ]>.

<<this is irrelevant, yes?>>
Following our earlier notational convention, we can define
@BEGIN[Equation]
@K[LeftDoubleBracket]I@+[j]@K[RightDoubleBracket][A] @K[Eq]
@K[LeftBrace]I@-[j,i][A] @K[Modulo] @~
I@-[j,i]@K[MemberOf]@K[LeftDoubleBracket]I@+[j]@K[RightDoubleBracket]@K[RightBrace]
@END[Equation]

Notice that in each case, even here,
we can produce a symbol whose "interpretive range" is restricted.
Namely,
we can relation-izing of the @T<@g[f][x]> formula to form
the @T[R] relation,
which was only true its argument was equal to the value of the constant @T[C].

What we want is a definition of the form
@BEGIN[Equation]
Nw( T, d, @g[s] ) @K[And] Related( d, c )
	@K[Implies] New( T, c, @g[s] ).
@END[Equation]
In the example above, @T<Related(R,A)>;
and, in general, @T<Related(x,x)>.

The problem here is to formally specify this
@T<Related(x,y)> relation.
Realize that we do NOT have liberty to arbitrarily "cons up" a new relation
to serve this purpse -- recall the @T<Floogle> example from above.

Unfortunately we need something else,
to cover for Case 3 ("@T<A@K[Eq]C>") above.
One (misguided) attempt would be to combine this semantic approach
with the syntactic one shown above  -- i.e.
@BEGIN[Equation]
[Nw( T, c, @g[s] ) @K[Or] SynInclude( c, @g[s] )]
    @K[Implies] New( T, c, @g[s] ).
@END[Equation]
where 
@T<SynInclude> is as defined above, in Section @Ref[SynMethod].
While this does handle every case so far,
it fails miserably for Case 3',

@END[Comment]
@BEGIN[Comment] Clauses in Predicate Calculus, before Propositional Case

Unfortunately this syntactic condition is neither necessary nor sufficient.
To show it is not necessary, consider Case 1:
@SubHeading<Case 1: B, when A@K[IFF]B>
@Label[Bis1]
Suppose
@BEGIN[Equation]
Th1 @K[Gets] @K[LeftBrace] @↑A @K[MemberOf] @K[LeftBrace]0,1@K[RightBrace],
@\B @K[MemberOf] @K[LeftBrace]0,1@K[RightBrace],
@\A=B@K[RightBrace]@FOOT{
The notation @T<T @K[Gets] @g[x]> means the theory @T<T> is
the deductive closure of the set, @g[x].}.
@END[Equation]
@****** Hmmm.  Do I really mean that Th3 is empty,
and the universe consists only of the two elements, {0,1} ?
The problem really emerges at the "tagged" passage below. *****@*
Then hearing @T<B@K[Eq]1>,
would indeed say something about @T<A>,
even though it did not include that symbol.
(E.g. we can then deduce the fact that
@T<A@K[Eq]1>,
and use this fact to specify the referent of the @T<A> symbol.)
That is, we would want
@T<New(Th1,A,"B@K[Eq]1">.

@SubHeading<Case 2: @K[Not]((A@K[Eq]0)@K[And]B@K[Eq]1))>
@Label[AorNotB]
Suppose
@BEGIN[Equation]
Th2 @K[Gets] @K[LeftBrace] @↑A @K[MemberOf] @K[LeftBrace]0,1@K[RightBrace]@↑,
@\B @K[MemberOf] @K[LeftBrace]0,1@K[RightBrace],
@\@K[Not]((A@K[Eq]1)@K[And]B@K[Eq]1))@\@K[RightBrace].
@END[Equation]
Now imagine hearing that @g[t], where
@BEGIN[Equation]
@g[t] @K[Identical] @K[Not]((A@K[Eq]0)@K[And]B@K[Eq]1)).
@END[Equation]
At this point,
we see that @T<A> can be either @T<0> or @T<1>;
and that @T<B> must equal @T<0>.
Realize first that this is precisely the same as hearing that
@T<B@K[Eq]0>,
which we know says nothing about @T<A>!
Hence I would claim that @g[t] says nothing about @T<A>,
even though it contains that symbol.
As this statement conveys no new usable information about @T<A>, we want
@T<@K[Not]New(T,A,@g<t>)>.

@SubHeading<Case 3: A@K[Eq]B>
@Label[AisB]
Suppose 
@BEGIN[Equation]
Th3 @K[Gets] @K[LeftBrace] @↑A @K[MemberOf] @K[LeftBrace] 0,1 @K[RightBrace],
@\B @K[MemberOf] @K[LeftBrace] 0,1 @K[RightBrace] @K[RightBrace]
@END[Equation]
There are four obvious interpretations,
depending on the values of @T<A> and @T<B> --
@BEGIN[Verbatim]
@tabclear
    @↑  A   @↑B
@\@&@ux[ ]@\
I@-[0]@\| @u[0@\0]
I@-[1]@\| @u[0@\1]
I@-[2]@\| @u[1@\0]
I@-[3]@\| @u[1@\1]@Foot{
The underbar notation denotes the referents of the corresponding
linguistic symbols.
We are, for now, assuming an oracle fixes this correspondence; and it is
therefore not subject to ambiguity.}
@END[Verbatim]
Hence @T<@F1[I]@-[Th3][A] @K[Eq] @K[LeftBrace]@un[0,1]@K[RightBrace]>.
Now add to @T<Th3> the independent statement
@BEGIN[Equation]
@g[f][A] @K[Equivalent]   A @K[Eq] B,
@END[Equation]
forming @T<Th3'>.
While this only leaves two of the four interpretations, 
@T<I@-[0]> and @T<I@-[3]>, 
@T<@F1[I]@-[Th3'][A]> remains @T<@K[LeftBrace]@un[0,1]@K[RightBrace]>;
indicating that @T<@g[f][A]> says nothing @T<Nw> about @T<A>.

@SubHeading<Case 4: B@K[Eq]C, when A@K[Eq]B>
Take
@BEGIN[Equation]
Th4 @K[Gets] @K[LeftBrace] @↑A @K[MemberOf] @K[LeftBrace]0,1@K[RightBrace],@↑
@\B @K[MemberOf] @K[LeftBrace]0,1@K[RightBrace],
@\C @K[MemberOf] @K[LeftBrace]0,1@K[RightBrace],
@\A=B@\@K[RightBrace]
@END[Equation]

The tableau below describes this situation
@BEGIN[Verbatim]
@tabclear
                A
          @↑  @un[0   @↑1]
       @\@ux[@& @\ ]
        @un[0,0]@\| I@-[0]@\--
         @\|
        @un[0,1]@\| I@-[1]@\--
<B,C>    @\|
        @un[1,0]@\| --    @\I@-[6]
         @\|
        @un[1,1]@\| --    @\I@-[7]
@END[Verbatim]

The proportedly new sentence is @T<A=C>.  Certainly this should be new --
see Case #?, where we agreed that sentences of this form should count.
Only here it appears not to: 
The second and third rows vanish, and nothing else changes!
Something is wrong!

@END[Comment]