Russell introduces a new axiom which he calls axiom of reducibility. As I’m not sure I fully understood his thought, I will let him speak. “We assume, that every function is équivalent, for ail its value to some predicative function of the same argument.“ But to understand this assertion, we must go back to the definitions given at the beginning of the paper. What is a function, and what is a predicative function? If a proposition is asserted of a given object to a, it is a particular proposition; if it is said of an indeterminate object x, it is a propositional function of x. The proposition will be of a certain order in the hierarchy of types, and this order will not be the same regardless of x, since it will depend on the order of x. The function will then be said to be predicative, if it is of order K + 1, when x is of order K.
After these definitions the meaning of the axiom is not yet very clear and some examples would not be superfluous. Mr. Russell has not given it, and I hesitate to give it for myself, because I am afraid of betraying his thought, that I am not certain of having entirely seized it. But, without having grasped it, there is one thing I can not doubt, is that it is a new axiom. Thanks to this axiom, we hope to be able to demonstrate the principle of mathematical induction; that it is possible, I would like to deny it all the less because I suspect this axiom to be another form of the same principle.
And then I can not help but think of all the people who claim to demonstrate Euclid’s postulate, relying on one of its consequences, and looking at this consequence as an obvious truth by itself. What did they win? This truth, however obvious it may be, will it be more than the postulatum itself?
We thus gain nothing on the number of postulates; do we at least gain on quality?
In what does the new axiom prevail over the principle of induction?
1° Is it susceptible to a simpler and clearer statement? It is possible, because the one Mr. Russell gives us can probably be improved; but it’s not likely.
2° Is the axiom of reducibility more general than the principle of induction? so that we can not demonstrate this axiom on the basis of this principle?
3° Or, on the contrary, is the axiom less general in appearance than the principle, so that we do not immediately perceive that the second is contained in the first, though it is?
4° Is the use of this axiom more in keeping with the natural tendencies of our minds; can we justify it psychologically?
I am merely asking these questions; I miss the elements to solve them since I have not been able to even fully understand the meaning of this axiom.
But if I can not, with the too summary indications given by Mr. Russell, hope to penetrate entirely this meaning, I am at least permitted to make some conjectures. Here is a proposal such as the definition of the whole number; a finite integer is a number that belongs to all recursive classes; this proposition does not make sense, by itself; it would only be one if the order of the recurrent classes in question were specified. But happily this happens; the whole of the second order is a fortiori an integer of the first order, since it will belong to all the recurrent classes of the first two orders, and consequently to all those of the first order; likewise the whole of the K order will be a fortiori an integer of K – 1 order. We are thus led to define a series of classes more and more restricted, integers of the 1st, 2nd, …, of the n order, each of which will be contained in that which precedes. I will call an integer of order ω any number that will belong to all these classes at once; and this definition of the integer of the order ω will have a meaning and may be regarded as equivalent to the definition first proposed for the integer and which did not. Is this a correct application of the axiom of reducibility, as understood by Mr. Russell? I offer this example only timidly.
Let’s admit it, however, and take up the theorem to be proved about the sum of two integers. We have established that the sum of two integers of the K order is an integer of order K – 1, and we want to conclude that if x and n are two integers of order ω, the sum n + x is also an integer of order ω. And indeed it suffices to establish that it is an integer of order K, however large K. But if n and x are integers of order ω, they will be a fortiori integers of order K + 1, so by virtue of the already established theorem, n + x is an integer of order K …
Is that how you can use Mr. Russell’s axiom? I feel that it is not quite that and that Mr. Russell would give the reasoning a whole other form, but the substance would remain the same.
I do not want to discuss here the validity of this demonstration mode.
I will confine myself for the moment to the following observations. We have been led to introduce next to the notion of the objects of the n order, that of the objects of order ω and we believe to have succeeded with regard to the integers, to define this new notion. But that would not always succeed; for Epimenides for example, it would not work at all. What ensured success is the following circumstance. The classification studied was not predictive, and the addition of new elements made it necessary to modify the classification of the elements previously introduced and classified. However, this modification could only be done in one direction; we could be forced to transfer objects of the class A into the class B (that is, from integers to non-integers), but never to transfer them from class B in class A. It would take a new convention to define the objects of order ω in cases where the modification should be done sometimes in one direction, sometimes in the other.
Secondly, the definition of integers of order ω is not the same as that of integers of order K, K being finite. We define the integers of order K by recurrence by deducing the notion of integer of order K from the notion of integer of order K – 1. We define the integers of order ω, by passing to the limit, by making this new notion depend on an infinity of previous notions, those of the integers of all the finite orders. The two definitions are therefore incomprehensible to someone who does not already know what a finite number is; they presuppose the distinction between finite numbers and infinite numbers. It is not on them that we can hope to base this distinction.