In Mysticism and Logic, Russell says that :

"Pure mathematics consists entirely of assertions to the effect that, if such and such a proposition is true of anything, then such and such another proposition is true of that thing. It is essential not to discuss whether the first proposition is really true, and not to mention what the anything is, of which it is supposed to be true. [...] Thus mathematics may be defined as the subject in which we never know what we are talking about, nor whether what we are saying is true."

I understand in which sense the theorems of a mathematical system can be considered as conditionally true statements: they are true if axioms and definitions are assumed as hypothesis.

But, is Russell's claim correct regarding axioms?

If Russell's claim were correct, then the axioms of a mathematical system could not be considered as true in this system, for, apparently, they do not play the role of consequent in any conditional statement belonging to this system.

Does not this consequence contradict the standard thesis according to which the axioms of a deductive system are true in this system?

Remark. can one say that axiom A is true because : A --> A is true?

But in a mathematical system, an axiom has not the form : A --> A, but the categorical form : A.

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    $\begingroup$ I don't think Russell was talking about axioms. As I understand Russell's position, mathematics doesn't really study axioms, so much as it studies their consequences. The axioms aren't "really true", but if in a certain situation they were true, so would all of their consequences. $\endgroup$ – Theo Bendit May 18 '19 at 16:08
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    $\begingroup$ But $A$ is not equivalent to $A \to A$; the second one is a tautology, while the first one is not. $\endgroup$ – Mauro ALLEGRANZA May 18 '19 at 16:55
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    $\begingroup$ Some systems of axioms have turned out to be very useful in applications as they successfully model certain aspects of physical reality, e.g. axioms about numbers. So useful have they been that they have been promoted to universal truths, even though they are, in a some sense, provisional or conditional. $\endgroup$ – Dan Christensen May 19 '19 at 17:11

Mathematics, logicians and concomitant philosophers distinguish between explicit and implicit definitions. If you've seen plenty of mathematical definitions but haven't thought or read much about foundations, you're probably only familiar with the former. The latter is what axioms often provide. Let me explain with some examples.

What is a group? If you don't know the definition, feel free to Google it. It's a list of "group axioms". A group is anything satisfying those axioms; that's an explicit definition. Any theorem saying "all groups are like this" is provable as-is, which sounds contra Russell. (The proof might use stronger rules of inference than someone else is comparable with, but let's assume we fix these rules so we know which theorems count. Besides, Russell doesn't seem to be talking about that subtlety anyway.)

On the other hand, what is a set? Ooh, that's harder. I can copy-paste my preferred axiomatic set theory, let's say ZFC or whatever. Then we have an explicit definition of the models of ZFC, just as the group axioms explicitly define "groups", which might be given the more long-winded label "models of group theory". (But that's probably not a useful description, since theorems talking about relationships between groups are of greater interest than theorems talking about relationships between models of ZFC, until you're doing mathematics that goes far beyond working in ZFC itself.) What we don't have, however, is an explicit definition of the sets in ZFC. We only have an implicit definition, in the form of axioms making claims about what they're like.

So, are axioms claims of the form "if X then Y" rather than just Y? I suppose you could formulate each ZFC axiom as "this holds if we're taking about sets". But even if we do, I don't think that addresses the point Russell was trying to make. His point was that if mathematics has to assume something to prove Y, but X suffices, "if X then Y" needs nothing to be proven (again, this either folds rules of inference into the claim or doesn't consider them to be a "something" we're assuming), but Y does. This usually only matters if you're either getting your head around your first X-violator, or trying to understand why any philosopher would say your favourite theorem, as you would word it without so many assumptions spelled out, isn't a tautology.

So as you see, foundations are a complex interplay of rules of inference, axioms and definitions, and the last two overlap.

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    $\begingroup$ The distinction you seem to be wanting to make is that a group by definition is a model of the theory of a group, while a set isn't a model of set theory by definition. To say it the other way, the individuals of the theory of a group are not groups, while the individuals of set theory are sets. What ZFC describes is a directed graph whose nodes we call "sets". What those nodes are is no more determined than the elements of a group. $\endgroup$ – Derek Elkins left SE May 18 '19 at 18:41
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    $\begingroup$ Separately, you can't condition the axioms of ZFC with "this holds if we're talking about sets" in any formal sense because there's no formal definition of set, unless your intent was that you'd say "this axiom holds of all the things that satisfy this list of axioms (which includes this axiom)". $\endgroup$ – Derek Elkins left SE May 18 '19 at 18:41
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    $\begingroup$ @DerekElkins Good points. Just to be clear, "this holds of sets" was intended as a claim that's true in ZFC, not as a claim that's provable from some more fundamental set definition, because we don't have one. In particular, it wasn't intended as a useful reformulation or justification of the axioms, just as a way that they can be rephrased if you're insistent on fitting Russell's mould. $\endgroup$ – J.G. May 18 '19 at 19:26

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