Rational Inconsistent Beliefs

Suppose rational belief is closed under competent deduction:

(RBC) Necessarily, if S has a rational belief that p, and p entails q, then S is a competent deduction away from a rational belief that q, maintaining her belief that p throughout.

Given this moderate closure schema, which seems somewhat plausible, considering what seems to be a close link between rationality and deduction, it is possible to come to rationally believe a set of inconsistent propositions, by deduction alone.

Consider the preface paradox: An author publishes a book. It is rational for the author to believe that the conjunction of the propositions in this book are false, given reflection on her fallible nature, etc.:

(1) RB(~(p₁&p₂&…&p_n))

Now, further suppose that the author then begins looking through the book, or reflecting on each proposition individually. Now for each p₁,…,p_n it is not a stretch to consider that the author rationally believe the proposition, as there is a good chance that the individual proposition countenanced is not that which makes the conjunction false, and the author stands behind her work. But we can take this further. Suppose for each p₁,…,p_n the possibility of error were raised. For example, someone–perhaps the author herself–were to point out that the editor is known for transposition errors, spelling mistakes, etc., and it would be beneficial to double check whether the proposition in question were true. In each of these cases, the author could, by disjunctive addition, come to the deductive conclusion that there is no editing mistake present ¹:

(2) p₁ v ~EM(p₁)

(3) ~~(p₁ v ~EM(p₁))

(4) ~(~p₁ & ~~EM(p₁))

(5) ~(~p₁ & EM(p₁))

Thus, by (2)-(5), we have the conclusion that it is not the case that not p₁ and there is an editing mistake regarding p₁. This argument can be repeated for each p₂,…,p_n, with the result that, not only does it seem initially plausible that each of p₁,…,p_n are rational to accept individually, they are in fact robustly rational beliefs, immune to the possibility of error. At this point we now have:

(6) RB(p₁) & … & RB(p_n) & RB(~(p₁ & … & p_n))

If rational belief is aggregative, or closed under conjunction, then we have rational inconsistent beliefs:

(7) RB((p₁ & … & p_n) & ~(p₁ & … & p_n))

We can make this a bit uncomfortable for the author previous to the commitment to belief aggregation, however. For suppose that once the author had iterated through all of her beliefs, someone pointed out to her that she rationally believed the negation of the conjunction of her beliefs, and the iteration she just preformed committed her to rationally believing each proposition within the conjunction individually, which is inconsistent–one of her beliefs must not be rational. Given this, she could first prove that it is rational to believe the negation of the conjunction in the face of a series of rational beliefs in each conjunct:

(8) RB(~(p₁&p₂&…&p_n)) v ~(RB(p₁) & … & RB(p_n))

…

(9) ~(~RB(~(p₁&p₂&…&p_n)) & (RB(p₁) & … & RB(p_n)))

Then she could prove that it is rational to believe each conjunct in the face of the negation of the conjunction:

(10_(1-n)) RB(p_(1-n)) v ~RB(~(p₁&p₂&…&p_n))

…

(11_(1-n)) ~(~RB(p_(1-n)) & RB(~(p₁&p₂&…&p_n)))

If we wanted to take this a step further, and assume aggregation, by stopping one step short of the full-out contradiction:

(12) RB(p₁&p₂&…&p_n) & RB(~(p₁&p₂&…&p_n))

She could prove that it is rational to have a rational belief in both the conjunction and the negation of the conjunction:

(13)  RB(p₁&p₂&…&p_n) v ~RB(~(p₁&p₂&…&p_n))

…

(14) ~(~RB(p₁&p₂&…&p_n) & RB(~(p₁&p₂&…&p_n)))

——-

(15) RB(~(p₁&p₂&…&p_n)) v ~RB(p₁&p₂&…&p_n)

…

(16) ~(~RB(~(p₁&p₂&…&p_n)) & RB(p₁&p₂&…&p_n))

So, not only can you have inconsistent rational beliefs, but you can have rational beliefs that your inconsistent beliefs are rational.