# The set of Logical Inductors is not Convex

Sam Eisenstat asked the following interesting question: Given two logical inductors over the same deductive process, is every (rational) convex combination of them also a logical inductor? Surprisingly, the answer is no! Here is my counterexample.

We construct two logical inductors over PA, , and .

Let be any logical inductor over PA.

We consider an infinite sequence of sentences .

Let be computable in time.

We construct as in the paper, but instead of using all traders computable in time polynomial in , we use all traders computable in time polynomial in time. Since this also includes all polynomial time traders, is a logical inductor.

However, since the truth value of is computable in time, if the difference between and the indicator of did not converge to 0, a trader running in time polynomial in f(n) can easily exploit . Thus,

and

Now, consider the market

Observe that

so

Now, consider the trader who exploits by repeatedly either buying a share of when the price is near 34, or selling a share when the price is near 14, waiting for that sentence to be resolved, and then repeating. Eventually, in each cycle, this trader will make roughly 14 of a share, because eventually the price will always be close enough to either 14 or 34, and all shares that this trader buys will be true, and all shares that this trader sells will be false.

Thus is not a logical inductor.