# [FOM] Re: Formal treatment of expressions that refer to each other

Sandy Hodges SandyHodges at attbi.com
Sat Mar 29 15:38:08 EST 2003

```XI. THE DEPENDENCY FUNCTION DETERMINED BY A FORMULA

Consider the expression:

Three times the sum of all integers designated by the utterance
(if any) on occasion U1, plus six times the sum of all integers
designated by the utterance (if any) on occasion U4, plus the sum of all
integers designated by the utterance (if any) on occasion U6.

This expression, expressed in the formalism, is this formula:

3 * ValRef(ZZ,U1) + 6 * ValRef(ZZ,U4) + ValRef(ZZ,U6)

Call this formula, formula A.     Without knowing what, if anything, was
uttered on occasions U1, U4, and U6, we can't say what any particular
utterance of formula A will designate.   But we can say that the very
form of formula A indicates the following function:

f(a,b,c) = 3 * a + 6 * b + c

That formula A indicates this function f, rather than some other
function, is not a claim about how the value designated by an utterance
of A depends on values designated by other utterances.   It is just a
claim about what symbols occur in formula A.   We can call this function
f the dependency function indicated by formula A.   If we have the Gödel
number of a formula such as A, we can work out what the indicated
function is, and we can express the claim that this function is the
indicated function for that Gödel number, in an object language
expression.

An axiom will say that the formula:

ValRef(d,u)

always indicates the dependency function:

f(x) = x

and the dependency function indicated by a more complex formula can be
built up from the dependency functions of its parts.  But the details
are tedious.  In the next post I'll review what all this is supposed to
accomplish.  Here are some of the tedious details:

XII.   SOME TEDIOUS DETAILS RE. DEPENDENCY: RefSelect

Looking at formula A, one can see it refers to three utterance
occasions.  But we may be interested in more utterance occasions than
just these three.   So given a list of n utterance occasions which
includes at least all those mentioned by a formula, we can say that the
"reference selection vector" indicated by the formula, is list of n
values from the set {0,1}, with 1 indicating that the occasion is
mentioned.  Suppose we were interested in six occasions, U1 through
U6.   Formula A mentions only U1, U4, and U6.   So relative to the list
(U1 ... U6), the reference selection vector for by A is (1,0,0,1,0,1).

The relation

RefSelect(g,w,s)

shall mean that for the formula of which g is the Gödel number, w is a
list of utterance occasion references (actually the Gns. of them)
including at least all those references used in the formula, and s is
the selection vector saying which references from w are used in g.

Here's an axiom to say that if g is the Gn of the formula:

ValRef(d,u)

and then RefSelect(g, w, s) will hold just if w is a list of occasions
including u, and s is the selection vector which selects u only out of
w, will hold.   For example if g is the Gn. of "ValRes(ZZ,U2)" and w is
the list (U1,...,U6), and s is the list (0,1,0,0,0,0), then RefSelect(g,
w, s) holds.   Recall that the Gn. of "ValRef(d,u)" is BinR(8209,p,v),
for p and v, the Gns. of d and u.    Here's the axiom:

(\/ g, n, b, p e NN)(\/ r e Pow(NN))(\/ w,s e {functions from r to NN})
(
( r = {1,2,...,n}   &    b e r   &   g = BinR(8209,p,v)  ) => (
(  w e { 1-to-1functions from r to NN } & (\/ <x, y> e w) ( y = v
<=> x = b )   &
s e { functions from r to {0,1}  & (\/ <x, y> e s) (  y = 1 <=> x =
b )   )
<=>
RefSelect(g, w, s)
) )

XIII.   TEDIOUS DETAILS: Domain

In the expression "ValRef(d,U2)," which means the sum of the set of
members of d designated on occasion U2, d is the domain.    If g is the
Gn. of an expression containing one or more uses of "ValRef" then

DomainRef(g,p,d)

shall mean that all the ValRef's use d as the domain, and p is the Gn.
of the expression used to Denote d.    For example if g is the Gn. of
"ValRef(ZZ,U2)", then DomainRef(g,8210,ZZ) holds.  (8210 is the Gn. of

"ZZ", which Denotes ZZ, the integers.)

The axiom is

(\/ g,p,v e NN)(\/ d) (
g =  BinR(8209,p,v) => ( Denotes(p,d) <=> DomainRef(g,p,d) )
)

XIII.   TEDIOUS DETAILS: DependFunc

If we are interested in n utterance occasions, and a formula mentions
only m of them, then it's convenient to think of the dependency function
as a function of all n inputs, rather than just the m mentioned
inputs.   So the dependency function for formula A, relative to the
list, (U1,...U6) is

f(a,b,c,d,e,f) = 3 * a + 6 * d + f

The relation

DependFunc(g,w,d,f)

shall mean that for the formula of which g is the Gödel number, w is a
list of utterance occasion references (actually the Gns. of them)
including at least all those references used in the formula, and f is
the function on the domain d, the dependency function indicated by g.
This axiom says that if g is the Gn. of "ValRef(d,u)", the dependency
function is f(x1,x2,...xM) = xN.

(\/ p, n, b, v e NN)(\/ d, w, s, h, r, f, s) (

(\/ g, n, b, p e NN)(\/ r e Pow(NN))(\/ d)(\/ w e {functions from r to
NN})
(\/ h = {functions from r to d) (\/ f e {functions from h to d})  (

r = {1,2,...,n}  &  b e r  &  a = BinR(8209,p,v) => (
(  w e { 1-to-1functions from r to NN } &
(\/ <x, y> e w) ( y = v <=> x = b )  &
( \/ <y, t> e f )( <b, t> e y )   )
<=>
DependFunc(g, w, d, f)
) )

XIV.   TEDIOUS DETAILS: Dependency

Dependency puts it all together.   Axiom is

(\/ g,p,n e NN)(\/ d,s,w,f) (
Dependency(g,p,d,w,s,f)
<=>
(  RefSelect(g,w,s) & DomainRef(g,p,d) & DependFunc(g,w,d,f) )
)

XIV.   TEDIOUS DETAILS: Dependency of sums

If you have g the Gn of  "A+B" for expressions A and B with Gns. a and
b, and you have:

Dependency(a,p,d,w,s,f) and Dependency(a,p,d,w,s,h) , that is, you know
the dependency functions for A and B relative to the same list of
utterance occasions and the same domain, then you can deduce that
Dependency(g,p,d,w,s,k), where k is the function

k(x1,...,xN) = f(x1,...,xN) + h(x1,...,xN)

But I'm to bored with this to work out the axiom.

XV.   TEDIOUS DETAILS: Uniqueness

We need axioms to say that a function in only the indicated function of
some formula, if it can be shown to be by the axioms.  That this could
be done seems too obvious to take the trouble to do it.

------- -- ---- - --- -- --------- -----
Sandy Hodges / Alameda,  California,   USA
mail to SandyHodges at attbi.com will reach me.

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