The Paradox of the Preface and the Law of Non-Contradiction

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2 responses to “The Paradox of the Preface and the Law of Non-Contradiction”

1. 

   Jan

   January 19, 2012

   There are two separate questions here. The first is whether this example shows that there exists a proposition p such that the following are both true:
   a. It is rational for S to believe that p.
   b. It is rational for S to believe that ~p.
   In what follows, Rp denotes *the writer is rationally justified in believing that p*. The second question is if the following is true:
   (*) Rp & Rq ⇒ R(p&q)
   which given a proposition p as in (a) – (b) would imply that it is rational to believe some necessarily false proposition.
   I think Clark fails to provide a proposition p as in (a) – (b) and his argument rests on a confusion concerning quantificators. The following true:
   1. For every proposition p expressed in the book, the writer is rationally justified in believing that p.
   2. The writer is rationally justified in believing that there exists a proposition p expressed in the book such that p is false.
   This pair is not inconsistent. Indeed, to form a pair of the type (a) – (b), (1) must be replaced with
   1′. The writer is rationally justified in believing that every proposition p expressed in the book is true.
   (1′) is however simply not true. The argument that is given for it is in fact an argument for (1) with which (1′) is confused. The difference between (1) and (1′) is clearly seen by using symbolic logic:
   1. ∀p R(p)
   2. R ( ∃p ¬p )
   and
   1′. R( ∀p p )
   Expressing the point in a somewhat pompous way, the operator R does not commute with the universal quantifier. This example does however constitute a counterexample to (*). (1) is equivalent to
   1*. Rp &…& Rq ( p, …, q being the propositions from the book).
   By (*) (and induction) we have
   2*. R(p &…& q)
   (2*) is easily seen to be ¬(2). Therefore, an aporetic polyad cannot be formed unless (*) is given strong independent support, which I don’t think can be done. This may be to the chagrin of the proprietor of the blog, who as we know loves them aporetic polyads.
   In Bayesian terms, (*) says that an intersection of big sets is big (whatever big exactly means). This need not be true under more reasonable understandings of ‘big’ (and not very pathological underlying spaces).

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2. 

   Matthew

   January 23, 2012

   I’m not a philosopher, I’m an economist, so you may have to excuse any abuse of nomenclature or other mistakes, but it seems to me that this example is easily resolved by introducing a very basic form of uncertainty. That is, I believe that every proposition is true with near-certainty. So proposition q=1 if true and q=0 if false, and I believe that each q=1 with probability p=.99. So the “expected truth value” of any given proposition q is 1*.99=.99. Thus I believe each proposition is true with something very close to certainty, but if I amass a collection of hundreds of such propositions, then it is also nearly certain that at least one of those statements is false (e.g., 500 statements have a probability of containing at least one false statement with a probability around 99.3%).
   In this case then, applying the same concept to both statements, it is rational for me to believe with near-certainty that each statement, considered individually, is true, and that that at least one statement in the book is false.

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