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