[FOM] Frege's error

Hartley Slater slaterbh at cyllene.uwa.edu.au
Mon Jul 11 21:10:41 EDT 2005


Carnap's notes on Frege's lecture course on the Begriffschrift, 
recently published, show precisely, I believe, where he went wrong. 
At least they do this unless someone can show how the last sentence 
in the following passage makes sense.  He says: 'A function of two 
arguments, e.g. x-y, can be transformed into a function of one 
argument in two different ways, either by saturation (x-2), or by 
identifying the two argument places (x-x).  Functions of two 
arguments that always have a truth value as value are relations. 
Therefore we can transform the relation x>y into a concept, e.g. x>0 
(the concept of a positive number).  Or we can form the concept x>x.' 
(Frege's Lectures on Logic, translated and edited by Erich H. Reck 
and Steve Awodey, Open Court, Chicago 2004, p155).

The question is whether (to put it into more modern language) from 
LyLx(x > y), one can obtain Lx(x > 0), and Lx(x > x), where 'L' is 
the lambda of lambda abstraction.  Remembering the general format of 
lambda abstraction reduction goes LyFy[a] = Fa, the first reduction 
is OK, since applying the two term
relation to 0 one gets the concept of being greater than 0:
LyLx(x > y)[0] = Lx(x > 0).
But the second reduction hits a problem.  Proceeding as before one might try
LyLx(x > y)[x] = Lx(x > x),
but in 'LyLx(x > y)' the 'x' is a bound variable, and so the whole is 
equivalent to 'LyLz(z > y)', and using that form one merely gets
LyLz(z > y)[x] = Lz(z > x),
i.e. the concept of being greater than x.

Frege talks about getting his second concept by identifying the two 
variables, but he cannot mean LxLx(x>x), since that is as malformed 
as (Ex)(x)Rxx, etc.; also LyLx(x>y.x=y) runs into the same problems 
as before.  Nor can he be thinking one can produce his second concept 
from LyLz(z>y)[x][x], since while that produces the statement 'x>x', 
it still does not identify the concept he mentions, because that 
statement is still analysed as a relation between two arguments, not 
as involving a single subject with a predicate expressing the concept 
Lx(x>x).  Unless someone shows me otherwise, it seems to me one 
cannot get to the latter concept from the relation; it is a quite 
separate idea.

And it is that idea which got him into trouble with
x is in {y | y is not in y} iff x is not in x,
since there is no trouble with (please check if you do not believe me!)
<x,x> is in {<y,z> | z is not in y} iff x is not in x.
Directly in terms of relations and concepts, the trouble with the 
paradox of predication:
(EP)(x=LyPy.-LyPy[x]) iff LyQy[x],
is precisely its taking the two argument relational expression on the 
left (with the two arguments identified the same) to be equivalent to 
a single-subject and predicate expression, as on the right.  Instead 
it is equivalent to: LyLz(EP)(z=LtPt.-LtPt[y])[x][x].

If I am right, then it was clearly Frege's training in Mathematics 
which got him into trouble - things are different in Logic.  If there 
is no derivation of the supposed concepts from the given relations 
(both in his case, and in that just now), then he must have been 
working entirely on the basis of his understanding of mathematical 
functions - which is also illustrated in the passage above.  For 
there is no doubt that, given a two-valued numerical function f(y,z), 
one can obtain a function of one variable f(x,x)=g(x).  But the 
analogy with the 'truth value' case limps just at this point.  First, 
if anything like 'Pa=T', or 'Rab=F' holds it is with '=' as material 
equivalence, 'T' a tautology, and 'F' a contradiction (thus sentences 
are not referential terms with the same reference as 'The true' or 
'The false').  But then one has that 'Pa=T' and 'Rab=F' are 
equivalent to 'Pa' and '-Rab' respectively, making 'Pa' and 'Rab' 
quite unlike mathematical functions, and 'T' and 'F' nothing like 
their values.  The former sentence indeed says that a is a member of 
a set - the set of individuals which fall under the concept LxPx - 
but the latter says instead that <a,b> is a member of the set of 
ordered pairs which fall under the relation LxLyRxy, which means that 
'Raa' is not about a set that *a* belongs to, but merely about a set 
that <a,a> belongs to.
-- 
Barry Hartley Slater
Honorary Senior Research Fellow
Philosophy, M207 School of Humanities
University of Western Australia
35 Stirling Highway
Crawley WA 6009, Australia
Ph: (08) 6488 1246 (W), 9386 4812 (H)
Fax: (08) 6488 1057
Url: http://www.philosophy.uwa.edu.au/staff/slater



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