A consistency result in the computational complexity theory

Question

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 $$\{0,1,+,\cdot,S,<\}.$$

Answer 1

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(Admittedly Razborov's result is quite old now - from $1995$ - so it may have been superceded by an unconditional result.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.)