# [FOM] Tokens and Predicates

Sandy Hodges SandyHodges at attbi.com
Sun Nov 17 00:05:21 EST 2002

```I've been trying to work out how the concept of token-relativism, which
has some interesting properties when applied to the semantic paradoxes,
might extend to the paradoxes of membership.   Here's how I'm thinking

Assume a "Pred" operator, so that, for example:

(Pred x) ( Mortal(x) & Bipedal(x) )

names the predicate that says of something that it is mortal and
bipedal.    We can use this in definitions, so that

g =def (Pred x) ( Mortal(x) & Bipedal(x) )

makes "g" a name for that predicate.    A relation:

Appl(n, a, b)

is defined to say that token n is an instance of the application of
predicate a to some noun phrase that designates b.   Thus of these
tokens:

1.   Mortal(Tully) & Bipedal(Tully)
2.   Appl(1, g, Cicero)
3.   Appl(1, g, Cicero) & True(1)
4.   (E token n) ( Appl(n, g, Cicero) & True(n) )

Token 2 is true because token 1 is the application of predicate g to the
noun phrase "Tully" which designates Cicero.    Token 1 is true because
Cicero had two legs and died.  Hence token 3 is true, and 4 follows from
3.
- - -
The token:

5.   ~ (E token n) ( Appl(n, y, y) & True(n) )

says of y, that when applied as a predicate to itself, the resulting
formula is not anywhere instanced as a true token.     We now have what
we need to construct a membership paradox.    Define:

h =def (Pred y) [ ~ (E token n) ( Appl(n, y, y) & True(n) ) ]

h is the predicate which token 5 applies to y.    So the paradox will
arise in applying h to itself.

Consider these tokens:

6.  ~ (E token n) ( Appl(n, h, h) & True(n) )
7.  Appl(6, h, h)
8.  Appl(6, h, h) & True(6)
9.  (E token n) ( Appl(n, h, h) & True(n) )
10.  ~ (E token n) ( Appl(n, h, h) & True(n) )
11.    Appl(10, h, h) & True(10)

Token 7 is true, as can be seen by comparing 6 with the definition of
h.   Suppose 6 were true.  Then 8 would be true, and thus 6 would be
false.   Suppose 6 were false, and let m be any token for which Appl(m,
h, h) is true, for example, token 6 or token 10.   Any such token will
be equiform with token 6, so if token 6 is false, any such token m will
not be true.    So there can be no token m such that (Appl(m, h, h) &
True(m)).   Thus token 6 is true.

So we have a situation in which token 6, if true, is false, and if
false, is true.    But of course this situation is not in the least
surprising or unusual - it is merely a paradox.    The result will be
that token 6, at least, is declared GAP.    But which tokens are GAP?
My system in
http://sandyhodges.topcities.com/logic/sybil/forhtm.htm
although devised for semantic paradoxes, applies to this membership
paradox as well; it calls tokens 6 and 10 GAP, 7 true, and 8, 9, and 11,
false.    Gaifman's system would produce the same results, given a
suitable definition of "refers."
------- -- ---- - --- -- --------- -----
Sandy Hodges / Alameda,  California,   USA
mail to SandyHodges at attbi.com will reach me.

```