I might be misunderstanding your intent here, but here’s one “inconsistent” system you can construct trivially. (It is, of course, only trivial in the sense that after you know a powerful theorem it is obvious, with the theorem here being incompleteness.)

One can construct a provability predicate for ZFC. Call it F(x).
Then both ZFC + F(0=1) and ZFC + ~F(0=1) are consistent if ZFC is.

The second system trivially tacks on another statement, which happens to be true, but the first one adds a false axiom. This axiom cannot “interact” with the arithmetic component of ZFC, so no formal inconsistency arises.

I’ll just leave this thought here for posterity’s sake. After all, I stumbled onto this thread.

The “requires induction” idea is surely not rigorous, whatever that means, but on the other hand, I am convinced that it could be made so, to such a degree that rigor actually has a value.

But somehow I don’t think “ZFC” is going to be the slightest bit of help. For one thing, ZFC concerns the whole of mathematics, whereas what I am thinking about concerns only (finite) combinatorics [it probably extends to combinatorial set theory in some limited form, or whatever].

As such I have no idea how one might, even if you wanted to, come up with a new, general pan-mathematical meaning to the word “equivalent” that would be more wise.

But I think it is important that implications can “just seem wrong”. Earlier (while googling what lead me to this thread) I read that Hall’s theorem implies Dilworth’s theorem and the max-flow-min-cut theorem. These implications, though logically valid, just seem wrong somehow, and seem even more wrong when viewed through the spectacles of induction.

This is why you could say something like, Dilworth’s theorem is a special case of the perfect graph theorem in graph theory, and in turn Hall’s theorem is a special case of Dilworth’s theorem. You don’t need to induct to prove these implications, they sort of “spill out”, in not too many lines. I honestly think that if you tried to reverse these implications, you would end up using induction, and that is what makes these theorems stronger. I am particularly sure about the impossibility of getting from Dilworth’s theorem to the perfect graph theorem, in such a way.

I think this fits the sense I get that, for example, proving Dilworth’s theorem by induction is doing the donkey work, and deriving Hall’s theorem from Dilworth’s theorem is just fiddling. Fiddling fiddling fiddling.

1For those who want to keep track, A is known as strong superadditivity of entanglement of formation, B is additivity of entanglement of formation, C is additivity of classical capacity of a quantum channel, and D is additivity of minimum entropy output of a quantum channel. These names are irrelevant to the point of the story.

Consider a finite POSET (partially ordered set) X.

Dilworth’s theorem asserts that X can be covered by chains whose number is the size of the largest antichainin X.

A much easier theorem asserts that X can be covered by antichains whose number is the size of the largest chain in X.

(A chain is a set of elements so that every two are comparable. An antichain is a set of elements so that every two (distinct) are incomparable.)

Now, these two theorems are equivalent because of a Theorem by Lovasz which asserts that the complement of a perfect graph is also perfect.

A (finite) graph G is perfect if for every induced subgraph H of G, H can be covered by cliques (=complete subgraphs) where is the independence number of H.

Starting from a post X we can desribe two graphs: the comperability graph where x and y are adjecent if x>y or y>x, and (its complement) the uncomparability graph where x and y are adjacent if neither x>y nor y>x. Lovasz’s theorem gives an equivalent between two theorems one which is easy and another which is harder.

In this example we are talking about equivalence of absolute statements (and not equivalence of properties of graphs that sometimes hold and sometimes not like the perfect matching example), so it is more similar to the equivalence between theorems like at the end of the post. But here the equivalence is “genuine” in some sense.

There are various ways to start from a statement A and get another statement B. One can, for example, deduct B from A, negate A to get B, add quantifiers to A to get B. Here, we are studying the interaction of statements with respect to deduction. However, it may be interesting to consider another form of ‘interaction’ between statements. Specifically I was thinking about dualisation. Indeed, many statements have natural dual statements. I often found that to better understand a statement, it is revealing to consider its dual statement. Also, at least empirically, dual statements are often true.

So suppose one has two equivalent statements A and B (with respect to deduction), with A having a natural dual statement, but not B. Then one could say A is stronger than B with respect to dualisation. Indeed, A is more prone than B to “producing revealing and often true statements”. This is very vague and based on an intuition. I’ll try to find an example illustrating this.

Let be a semigroup. Consider the following three conditions that might satisfy:
(a) is a group;
(b) for all in there is a unique in such that ;
(c) for all in there is a unique in such that and .

The main exercise is to show that (a) is equivalent to (b). The implication (a) implies (b) is essentially trivial, but (b) implies (a) is much more fun.

A superficial comparison of (b) and (c) suggests that (c) might be stronger than (b), but in fact it is strictly weaker: (c) defines an inverse semigroup.

Joel

One suggestion that came out of the FOM discussion was that work in “dynamic epistemic logic” is relevant. That is, “RH might be true” is interpreted as “We don’t know that RH is false,” and so we formalize “RH might be true” by formalizing the concept of knowledge. Note that knowledge can change over time even if mathematical facts can’t; hence the adjective “dynamic.”

I haven’t invested the effort to understand the details of dynamic epistemic logic so I’m not sure how much the theory has to offer, but it seems that it might be relevant to the current discussion too. As has already been pointed out, the term “easy to prove” seems to have strong psychological overtones, and so maybe “easy proof” can be plausibly formalized along the same lines that “knowledge” can be formalized.

Some observations

* Being a skew field can be explicitly made part of the definition of being a field – this is not normally pointed out (or if it is, it is quickly forgotten) as one tends to learn about fields first and skew fields later. One could do something similar with Hall’s theorem defining matched graphs and paired graphs (or whatever, with fairly obvious definitions) and saying ‘every finite matched graph is a paired graph’. How natural would it be to say that a paired graph is a matched graph + another condition? – I don’t think it quite works the same.

* With Wedderburn there are some fairly obvious finite and infinite examples to hand: with Hall, the language of marriage and the way I learned graph theory both give me a psychological bias to thinking of finite cases first.

* I quite like what Benjamin posted above, which raises the question what properties beyond those explicit in the definition of Q are needed to prove R?

* Are there any natural/obvious examples of this phenomenon which don’t depend on a finiteness condition?

with respect to your example -

I’d claim that the definition of ‘finite’ is not totally trivial. Of course it is necessary: if one takes an infinite set A, a set B of the same cardinality, a set C with one element, and D the empty set, then there exist a bijection f from A to B (by definition) and g from A union C to B union D (via well-ordering) but of course not h from C to D.

So, then, here is an easy proof, using Peano and ZF (but of course not C).

Let r be an enumeration of B union D, i.e. a bijection from B union D to [k] for some integer k (I see no other definition of ‘finite’ that doesn’t involve a non-trivial proof to make sense of the definition).

Let s be an enumeration of C.

Consider the set r(h(A union C)-f(A)). This is a set of integers; let q(j) be the j-th integer of this set in increasing order (which can be defined by functions via summation of an indicator).

Let g(c)=r^{-1}(q(s(c))) for c in C.

If g fails to be well-defined, then this can only be because q fails to be well-defined, which it does not (Peano arithmetic).

It is clear that g is injective and maps onto D; to see that it is surjective is again Peano arithmetic.

I think the fact that g(b) as given here is actually a bijection requires about as much thought as the fact that the iterative procedure you give actually terminates – that is, both fail if there are inappropriate infinite quantities floating around, and for both proving that they succeed requires some idea of ‘finite’, which in turn requires Peano arithmetic to make sense. The difference is I think the proof I give is ‘obvious’ in the sense that it’s the first thing you’d try if you didn’t care for elegance and were told that the word ‘finite’ in the statement is necessary: which for me means the proof is trivial.

Returning to your comment, I am of course much happier with the idea of nonstandard models than I am with inconsistent models, and very interested and impressed that there is such a beautifully precise answer to the question I asked about Hall’s theorem. At first, the idea of an inconsistent model seems itself to be inconsistent (after all, the existence of a model surely guarantees consistency). But I suppose there might conceivably be a way of carrying out Carrie’s suggestion. First, one would need to define some notion like an “imperfect model” for some axioms. The axioms wouldn’t be true in the model, but neither would they be blatantly false. If that could be done, it’s conceivable that it might be easier to construct such a model than to show syntactically that an implication was not trivial.

Suppose we can specify in other terms (i.e. without making reference to what we happen to find obvious) a set of necessary and sufficient conditions for a deduction to count as ‘trivial’. I agree it’s not straightforward but let’s suppose. We then use it to specify the relevant class of worlds.

The point is that even if the new specification of the relevant worlds ends up delivering exactly the same worlds (same extension) as if you had just said ‘holding fixed the trivial deductions but varying other things’, provided you have specified it in other terms (different hyperintension) you have given an account of the sense in which one statement is stronger than the other which does not make reference to what we happen to find plausible/obvious, and could therefore reasonably be counted as objective.

Here’s an analogy. Suppose I like all and only green things. And someone is wondering if there is an objective difference between these things and all the others. Obviously a non-objective difference is that I only like some of them. But then we notice that everything I like is green, and everything else isn’t. Then we can say that there is an objective colour difference between the two groups. The factor we have identified marks an objective difference although it coincides with something non-objective (my preferences).

I have this question that is beyond me but might just be appropriate now. The FLT’s equation:

a^n + b^n = c^n and gcd(c^n – 1, a^n + b^n -1) = 1

is similar. I mean the second equation implies the first one. Can you show or give counterexample of positive ODD factor on the second equation? What’s interesting about the second equation is that it is based on Fermat’s Little theorem. The question is if the second equation is true then the first one is true. But we all knew that the first one is true. It doesn’t automatically implies the second one to be true.

Thanks

In the case of Hall’s theorem, Cook and Nguyen’s work means that we’re in good shape. Take the very weak logical theory VTC_0. In this theory we can *state* Hall’s theorem, and in the standard model of VTC_0, Hall’s theorem means what we think it means. However, only one direction of Hall’s theorem is known to be *provable* in VTC_0. The provability of the other direction in VTC_0 is an open problem, but let’s suppose that it’s not provable. This means that there is some nonstandard model of VTC_0 in which one direction of Hall’s theorem holds but the other doesn’t. In this nonstandard model, “Hall’s theorem” doesn’t quite mean what we normally think of it as meaning, because we’re interpreting the words slightly differently from usual. Nevertheless, to the extent that VTC_0 captures a natural subset of correct mathematical reasoning about finite combinatorics, this gives a precise and natural meaning to the statement that one side of Hall’s theorem is stronger than the other.

It’s worth noting that Cook and Nguyen’s logical systems are very closely related to standard computational complexity classes. Thus, “stronger” in this context means something like “harder to compute.” Of course, the informal notion of “easy” doesn’t *exactly* coincide with “easy to compute,” but low computational complexity is at least one reasonably plausible way to interpret “easy.”

My first problem is that I can devise such a world in an uninteresting way, simply by adding to ZF the statement “there exists a graph that satisfies Hall’s condition but that does not contain a perfect matching”. Such a world is plausible in the sense that the contradiction doesn’t jump out, but in order to specify it I haven’t done anything interesting. So one needs to decide what counts as an interesting construction of an impossible mathematical world.

My second problem arose from an attempt to deal with the first. Here is a possible reason for regarding the easy direction of Hall’s theorem as trivial: it’s that we can give a proof in a high-level language using little more than easy syllogisms. Such a proof would go something like this. (I think I’ve made it needlessly complicated actually, but that doesn’t affect the point I want to make.)

Bijections preserve cardinality.

The restriction of a bijection to a subset is a bijection.

Therefore, the restriction of a bijection to a subset preserves the cardinality of that subset.

A perfect matching is a bijection such that the image of any element is a neighbour of that element.

Given any subset, its neighbourhood contains the images of all its elements.

Therefore, given any subset, its neighbourhood contains the image of that subset.

By the earlier remark, the image of that subset has the same size as the subset.

Therefore, given any subset, its neighbourhood contains a set of the same size as that subset.

If a set contains another set, then the first set is at least as big as the second.

Therefore, given any subset, its neighbourhood is at least as big as the subset.

Anyhow, suppose I could carry out some analysis like that, and argue on the basis of it that the easy direction of Hall’s theorem was trivial because it all took place at a superficial level. I could then perhaps define a “plausible mathematical world” as being one where all superficial deductions were valid, or something like that. It might be necessary to restrict the statements that one was even allowed to consider, to avoid the objection that any proof can be broken down into steps that are superficial. Or one might wish to argue that “trivially implies” is not a transitive relation between statements. Again, these are problems, but they are not the main difficulty, it seems to me. The main problem is that if I could give an analysis that did justice to our idea of what counts as a trivial deduction, then it’s not clear what would be added by talk of plausible mathematical worlds: I could just define P to be stronger than Q if P trivially implies Q and Q does not trivially imply P.

Both my main difficulties can be summed up as follows: how would one get the notion of a plausible mathematical world to do any work? I’m not saying that I think it couldn’t, but just that I don’t at the moment see how it could.

The purely psychological differences are important; I think that they are really all that’s responsible for the persistant interest in what’s now known as the Church-Fitch paradox (aka the paradox of knowability or Fitch’s paradox).

But I wanted to suggest a way of thinking about the (first) kind of more objective difference you’re looking for. The thought I have is to take seriously the temptation you feel to use modal language (“Is there some precise sense in which an efficient check that a graph satisfies Hall’s condition “has to” find a perfect matching?”) and apply an idea from the philosophy of modality.

This is the idea of varying worlds whilst ‘holding fixed’ certain facts. Normally this is done with possible worlds but to apply the idea to mathematics we’d need to use impossible worlds too. The idea would be that as we hold fixed facts which are particularly mathematically important and *vary* other mathematical facts, we get a range of non-actual worlds in which one of the two equivalent statements always implies the other but not vice versa. This could give you a sense in which the ‘stronger’ one has more ‘strength’ than the ‘weaker’ one: the ‘stronger’ one keeps its power in other salient (impossible) worlds where the mathematical facts are different. Whereas the ‘weaker’ one loses its power in some of those worlds.

1) P=>(R=>Q)

Now also suppose that the “ideas” in R are contained within the ideas in Q s.t.:

2) Q=>R – A somewhat trivial example might be that Q is simply a composite statement of (R & S).

So given system P, Q and R are equivalent. But it happens that I can derive statements with Q that I cannot derive with R. (Using my trivial case above, I can derive S from Q but I cannot derive S from R.) Hence, Q is “stronger” than R.

The possibility that Q is equivalent to R, yet stronger than R, is contingent on the fact that Q and R are embedded within a system P, at least as far as this example is concerned. Separated from that system, Q and R are simply not equivalent.

As a first approximation, one could take “difficulty” to be “length,” but it has been remarked above that there are many examples in which a lengthy proof can be simple and a short proof can be hard.

Suppose, however, that B is only known to have a routine but lengthy proof, and that someone discovers a clever and non-trivial trick giving a short proof that A=>B; it may be incorrect to say that A=>B has a “less difficult” proof than B, but certainly the short proof of A=>B shows that there is some relation between A and B. (So even the naive choice of formalizing difficulty as length does not lead to completely meaningless conclusions.)

By the way, even results in reverse mathematics can be framed this way, if one defines “difficulty” not as a quantitative measure but as a the strength of the adopted proof system.