Implications of unprovability of $P\neq NP$

Question

I was reading "Is P Versus NP Formally Independent?" but I got puzzled.

It is widely believed in complexity theory that $\mathsf{P} \neq \mathsf{NP}$. My question is about what if this is not provable (say in $ZFC$). (Let's assume that we only find out that $\mathsf{P} \neq \mathsf{NP}$ is independent from $ZFC$ but no further information about how this is proven.)

What will be the implications of this statement? More specifically,

hardness

Assuming that $\mathsf{P}$ capture the efficient algorithms (Cobham–Edmonds thesis) and $\mathsf{P} \neq \mathsf{NP}$, we prove $\mathsf{NP\text{-}hardness }$ results to imply that they are beyond the present reach of our efficient algorithms. If we prove the separation, $\mathsf{NP\text{-}hardness}$ means that there is no polynomial time algorithm. But what does an $\mathsf{NP\text{-}hardness }$ result mean if the separation is not provable? What will happen to these results?

efficient algorithms

Does unprovability of the separation mean that we need to change our definition of efficient algorithms?

Answer 1

Your question might better be phrased, "How would complexity theory be affected by the discovery of a proof that P = NP is formally independent of some strong axiomatic system?"

It's a little hard to answer this question in the abstract, i.e., in the absence of seeing the details of the proof. As Aaronson mentions in his paper, proving the independence of P = NP would require radically new ideas, not just about complexity theory, but about how to prove independence statements. How can we predict the consequences of a radical breakthrough whose shape we currently can't even guess at?

Still, there are a couple of observations we can make. In the wake of the proof of the independence of the continuum hypothesis from ZFC (and later from ZFC + large cardinals), a sizable number of people have come around to the point of view that the continuum hypothesis is neither true nor false. We could ask whether people will similarly come to the conclusion that P = NP is "neither true nor false" in the wake of an independence proof (for the sake of argument, let's suppose that P = NP is proved independent of ZFC + any large cardinal axiom). My guess is not. Aaronson basically says that he wouldn't. Goedel's 2nd incompleteness theorem hasn't led anyone that I know of to argue that "ZFC is consistent" is neither true nor false. P = NP is essentially an arithmetical statement, and most people have strong intuitions that arithmetical statements—or at least arithmetical statements as simple as "P = NP" is—must be either true or false. An independence proof would just be interpreted as saying that we have no way of determining which of P = NP and P $\ne$ NP is the case.

One can also ask whether people would interpret this state of affairs as telling us that there is something "wrong" with our definitions of P and NP. Perhaps we should then redo the foundations of complexity theory with new definitions that are more tractable to work with? At this point I think we are in the realm of wild and unfruitful speculation, where we're trying to cross bridges that we haven't gotten to and trying to fix things that ain't broke yet. Furthermore, it's not even clear that anything would be "broken" in this scenario. Set theorists are perfectly happy assuming any large cardinal axioms that they find convenient. Similarly, complexity theorists might also, in this hypothetical future world, be perfectly happy assuming any separation axioms that they believe are true, even though they're provably unprovable.

In short, nothing much follows logically from an independence proof of P = NP. The face of complexity theory might change radically in the light of such a fantastic breakthrough, but we'll just have to wait and see what the breakthrough looks like.

Answer 2

This is a valid question, even though perhaps a little unfortunately phrased. The best answer I can give is this reference:

Scott Aaronson: Is P versus NP formally independent. Bulletin of the European Association for Theoretical Computer Science, 2003, vol. 81, pages 109-136.

Abstract: This is a survey about the title question, written for people who (like the author) see logic as forbidding, esoteric, and remote from their usual concerns. Beginning with a crash course on Zermelo Fraenkel set theory, it discusses oracle independence; natural proofs; independence results of Razborov, Raz, DeMillo-Lipton, Sazanov, and others; and obstacles to proving P vs. NP independent of strong logical theories. It ends with some philosophical musings on when one should expect a mathematical question to have a definite answer.

Answer 3

As proved in this paper:

http://www.cs.technion.ac.il/users/wwwb/cgi-bin/tr-get.cgi/1991/CS/CS0699.revised.pdf

If $P \neq NP$ can be shown to be independent of Peano Arithmetic, then NP has extremely-close-to-polynomial deterministic time upper bounds. In particular, in such a case, there is a $DTIME(n^{log^*(n)})$ algorithm that computes SAT correctly on infinitely many huge intervals of input lengths.

Answer 4

As Timothy Chow explains, just knowing that a theorem is independent from a theory doesn't say much about the truth/falsity of that statement. Most non-experts confuse formal unprovability in a fixed theory (like $[ZFC][1]$) with impossibility of knowing that answer to the truth/falsity of the statement (or sometimes meaninglessness of the statement). Independence and formal unprovability always means independence/unprovability in a theory. It simply means that the theory can prove neither the statement nor its negation. It doesn't mean that the statement does not have a truth value, it doesn't mean that we cannot know the truth value of the statement, we might be able to add new reasonable axioms that will make the theory strong enough to be able to prove the statements or its negation. At the end, provability in a theory is a formal abstract concept. It is related to our real world experience only as a model.

Same applies to the thesis that efficient computation is captured by complexity class $\mathsf{P}$. See this post.

Now you can ask if it is possible for a formal statement to not have a truth value. Generally in practice in principle, we can affirm $\Sigma_1$ (a.k.a. r.e.) properties and refute $\Pi_1$ (a.k.a. co-r.e.) properties by observations. Any statement more complex than this is not directly observable, i.e. no (finite) observation will allow you to affirm or refute the statement. However we can look at the observable logical consequences of these statements and try to use them to decide whether a statement is true or false. (For more on finitely observable properties see Samson Abramsky's Ph.D. thesis "Domain Theory and the Logic of Observable Properties", 1987 and Steven Vickers' "Topology via Logic", 1996.)

For most mathematicians statements of higher logical complexity are also meaningful and have a truth value, but this goes into the philosophical issues in mathematics. Almost all mathematicians believe that statements in the arithmetical hierarchy are meaningful and have definite truth values, and in some sense they view the truth value of these statements to be more definite than statements of higher logical complexity (like CH). The statement $\mathsf{P} \neq \mathsf{NP}$ can be stated as a $\Sigma_2$ statement and therefore is an arithmetical statement. As such, almost all mathematicians would believe that it is meaningful and has a definite truth value. You may want have a look at this MO question, and search the posts on FOM mailing list.

Answer 5

Just some rambling thoughts about this. Feel free to criticize.

Let Q = [cannot prove (P = NP) and cannot prove (P /= NP)]. Suppose Q for a contradiction. I will also assume that all known discoveries about P vs NP are still viable. In particular, all NP problems are equivalent in the sense that if you can solve one of them in polynomial time, you can solve all others in polynomial time. So let W be an NP complete problem; W equally represents all problems in NP. Because of Q, one cannot obtain an algotithm A to solve W in polynomial time. Otherwise we have proof that P = NP, which contradicts Q (1)(*). Note that all algorithms are computable by definition. So saying that A cannot exist implies that there is no way to compute W in polynomial time. But this contradicts Q (2). We are left with rejecting either (1) xor rejecting (2). Either case leads to a condradiction. Thus Q is a contradiction, which means that the proof of whether or not (N = NP) must exist.

(*) You might say, "Aha! A might exist, but we just cannot find it". Well, if A existed, we can enumerate through all programs to find A by enumerating from smaller programs to larger programs, starting with the empty program. A must be finite because it is an algorithm, so if A exists, then the enumeration program to find it must terminate.