What is the strongest portion of arithmetic $F$ such that we know that it is consistent with it that P=NP, provided this fragment $F$ is consistent ? Here, P=NP is understood as a second order formula in the language of arithmetic $$\{0,1,+,\cdot,S,<\}.$$
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${0,1,+}$ gets you all of integer arithmetic. – Him Apr 27 '20 at 21:00
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@Scott Even without the axiom of induction ? – user122424 Apr 27 '20 at 21:03
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I think that Peano arithmetic without the axiom of induction is a "larger" system in some sense. I.e. the theorems of Peano arithmetic + axiom induction are consistent with Peano arithmetic - axiom induction. – Him Apr 27 '20 at 21:08
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You're going to need a finite system, I'm pretty sure. – Him Apr 27 '20 at 21:10
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@Scott: taking away axioms gives you more models and fewer theorems. – Rob Arthan Apr 27 '20 at 21:10
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@RobArthan exactly. – Him Apr 27 '20 at 21:10
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@Scott So you claim that ${0,1,+}$ is probably inconsistent with P=NP , right ? Even in the mere theory of the successor $S$ like $S(0)=1$ and $S$ is injective ? – user122424 Apr 27 '20 at 21:12
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I'm not making that claim. Whether $P=NP$ is consistent with our usual logical systems is currently not known. – Him Apr 27 '20 at 21:16
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@Scott Right, therefore I've asked of some very weak theory for which we could say that it is consistent P=NP with it. Like the theory of ${0,1,S,+}.$ – user122424 Apr 27 '20 at 21:18
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Like I said, you'll need a finite system, I'm pretty sure. But that's kind of cheating, because with bounded input, all algorithms are constant time. – Him Apr 27 '20 at 21:20
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Once you get infinity, the insanity ensues. :) – Him Apr 27 '20 at 21:21
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@Scott ${0,1,S,+}.$ is infinite even without induction, right ? – user122424 Apr 27 '20 at 21:23
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It would be really helpful to be clear about what we mean by "finite system" and "consistent with". The languages and theories we are talking about are all infinite. Some of these theories have finite models (e.g., the theory of $(0, 1, S, +)$ with a suitable subset of the Peano axioms has $\Bbb{Z}/2\Bbb{Z}$ as a model). In a weak theory, a standard encoding of a statement like $P = NP$ doesn't necessarily mean what we mean by $P = NP$ and so the question of whether the theory is "consistent with" $P = NP$ becomes a question about encodings. – Rob Arthan Apr 27 '20 at 21:36
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3... and (by the way) there are very tractable, e.g., decidable, theories, like the theory of an unbounded discrete linear order that only have infinite models. So we can have infinity without "insanity" $\ddot{\smile}$. – Rob Arthan Apr 27 '20 at 21:40
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@Scott Addition is only sufficient if we use second-order logic. In first-order logic, addition is very weak indeed. – Noah Schweber Apr 27 '20 at 23:22
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You might want to have a look at An axiomatic approach to algebrization. – ShyPerson May 06 '20 at 23:16
1 Answers
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Razborov showed that - under a mild assumption - a particular theory of bounded arithmetic is not capable of proving (a suitable formulation of) $P\not=NP$. Bounded arithmetics are extremely weak indeed - basically, they have extremely limited induction abilities (well below being able to prove that exponentiation is well-defined, for example). These were first introduced by Sam Buss; stronger fragments of arithmetic (e.g. $I\Sigma_n$, $I\Delta_0$, etc.) were previously studied.
Of course since Razborov's result was conditional, this doesn't quite constitute an example of the desired phenomenon. However, I don't believe there really are any of those currently known. Part of the issue is methodological:
For us to even ask "Is $P=NP$ consistent with $T$?" the language of $T$ needs to be rich enough to formulating the $P=NP$ question appropriately in the first place. E.g. it doesn't make sense to ask whether arithmetic with only addition proves $P\not=NP$, because that system - Presburger arithmetic - is too weak to even make sense of the question in the first place (e.g. we can't even define a pairing function in Presburger arithmetic - see here). In fact, mere richness of language isn't enough on its own: $T$ needs to be able to prove basic facts about the symbols in the language, enough for the proposition we're expressing in that language to "mean what it should." Robinson arithmetic, for example, can't even prove that addition is commutative; given that we'll be formulating complexity theory in terms of addition and multiplication, the inability to prove such basic facts indicates that "Robinson arithmetic is consistent with $P=NP$" isn't a very meaningful proposition.
So how weak can we go? Well, the bounded arithmetics Razborov considers are to my knowledge among the weakest theories we currently think can implement complexity theory in any meaningful way. So I would tentatively say that since we don't yet have an unconditional consistency proof even over those, we're left with no meaningful instances of known consistency of $P=NP$.
(Admittedly Razborov's result is quite old now - from $1995$ - so it may have been superceded by an unconditional result. However, to the best of my knowledge it hasn't been yet.)
Noah Schweber
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