[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
about it:

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

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
- - -
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
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.

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