aa r X i v : . [ m a t h . L O ] A p r A limitation on the KPT interpolation
Jan Kraj´ıˇcekFaculty of Mathematics and PhysicsCharles University ∗ Abstract
We prove a limitation on a variant of the KPT theorem proposedfor propositional proof systems by Pich and Santhanam [7], for all proofsystems that prove the disjointness of two NP sets that are hard to dis-tinguish.
For a coNP property ψ ( x ), given n ≥
1, we can construct a size n O (1) propositional formula || ψ || n ( x, y ) with n atoms x = ( x , . . . , x n ) and n O (1) atoms y such that for any a ∈ { , } n , ψ ( a ) is true iff || ψ || n ( a, y ) ∈ TAUT. Thisis just a restatement of the NP-completeness of SAT. In addition, if ψ ( x ) isdefined in a suitable language of arithmetic and has a suitable logical form, thetranslation can be defined purely syntactically without a reference to machinesor computations. This then allows to transform also a possible first-order proofof ∀ xψ ( x ) into a sequence of short propositional proofs of tautologies || ψ || n , n = 1 , , . . . ; if the original proof uses axioms of theory T (essentially anysound r.e. theory) then the propositional proofs will be in a proof system P T associated to T . Many standard proof systems are of the form P T for some T , and this is often the most efficient way how to construct short P T -proofs ofuniform sequences of tautologies. Although the unprovability of ∀ xψ ( x ) in T does not imply lower bounds for P T -proofs of the tautologies, a method used inestablishing the unprovability sometimes yields an insight how the lower boundcould be proved. All this is a well-established part of proof complexity and thereader can find it in [4, Chpt.12] (or in references given there).The translation is, however, not entirely faithful for formulas of a certainlogical form, and this is a block for transforming the conditional unprovabilityresult for strong universal theories in [3] into conditional lower bounds for strongproof systems. To explain the problem in some detail assume ψ ( x ) has the form ∃ i < | x |∀ y ( | y | = | x | ) ϕ ( x, i, y ) (1)where ϕ is a p-time property. The provability of ∀ xψ ( x ) in a universal T canbe analyzed using the KPT theorem (cf.[6] or [4, Sec.12.2]). The same methoddoes not, however, work in the propositional setting. To illustrate this assume ∗ Sokolovsk´a 83, Prague, 186 75, The Czech Republic, [email protected] || ψ || n has a proof in proof system P T attached to T and from that we candeduce in T that _ i Let P be a propositional proof system. The system has KPT interpolation if there are a constant k ≥ and k p-time functions f ( x, z ) , f ( x, z, w ) , . . . , f k ( x, z, w , . . . , w k − ) such that whenever π is a P-proof of a disjunction of the form A ( x, y ) ∨ . . . ∨ A m − ( x, y m ) where x is a n -tuple of atoms and y , . . . , y m are disjoint tuples of atoms, thenfor all a ∈ { , } n the following is valid for all b , . . . , b m of the appropriatelengths: • either A i ( a, y i ) ∈ TAUT for i = f ( a, π ) or, if A i ( a, b i ) is false, • A i ( a, y i ) ∈ TAUT for i = f ( a, π, b i ) or, if A i ( a, b i ) is false, • . . . , or • A i k ( a, y i k ) ∈ TAUT for i k = f k ( a, π, b i , . . . , b i k − ) . An illuminating interpretation of the definition can be made using the interactivecommunication model of [5] involving Student and Teacher. Student is a p-timemachine while Teacher has unlimited powers. At the beginning Student gets a ∈ { , } n and the proof π and computes from it his first candidate solution:index i such that A i ( a, y i ) is - he thinks - a tautology. Teacher either approvesor she provides Student with a counter-example: an assignment b i for y i which2alsifies the formula. In the next round Student can use this counter-exampleto propose his next candidate solution, etc. Functions f , . . . f k in the definitionform a strategy for Student so that he solves the task for all a and π in k stepsin the worst case.Unfortunately, we show in this note that this property fails for strong proofsystems (above a low depth Frege system) for essentially same reasons whyordinary feasible interpolation fails for them (cf.[4, Sec.18.7])). For a set U ⊆{ , } ∗ and n ≥ U n := U ∩ { , } n . LK / is the Σ-depth 1 subsystem ofsequent calculus (cf.[4, Sec.3.4]). Theorem 2 Let P be a proof system containing LK / . Assume that U, V are disjointNP sets such that:1. Propositional statements U n ∩ V n = ∅ have p-size P-proofs.2. For any constant c ≥ , for all large enough n there is a distribution D n on { , } n supported on U n ∪ V n such that there is no size n c circuit C n for whichP rob x [( x ∈ U n ∧ C n ( x ) = 1) ∨ ( x ∈ V n ∧ C n ( x ) = 0)] ≥ / n − c where samples x in the probability are chosen according to D n .Then P does not admit the KPT interpolation. Remarks: 1. An example of a pair of two NP sets U, V that are conjectured to satisfythe second condition can be defined using a one-way permutation (moregenerally an injective one-way function with output length determined byinput length) and its hard bit: U (resp. V ) are the strings in the rangeof the permutation whose hard bit is (resp. ). Distribution D n is inthis case generated by the permutation from the uniform distribution onthe seed strings, i.e. it is uniform itself.2. It is known that the hypothesis of the theorem can be fulfilled for sys-tems such as EF, F, TC -F and, under stronger hypothesis about non-separability of U and V , also for AC -F above certain small depth; see thecomprehensive discussion in [4, Sec.18.7]. Proof of the theorem occupies the rest of this note.Write U ( x, y ) for a p-time relation that y witnesses x ∈ U and similarly V ( x, z ) for V , with the length of both y and z p-bounded in the length of x . Let n, m ≥ m strings x , . . . x m of length n each consider the following 2 m propositional formulas translating the predicates U ( x, y ) and V ( x, z ) (which weshall denote also U and V in order to ease on notation):3 U ( x i , y i ): x i is an n -tuple of atoms for bits of x i and y i is an n O (1) -tupleof atoms for bits of a witness associated with x i together with bits neededto encode U as propositional formula suitable for P (e.g. as 3CNF), • V ( x i , z i ): analogously for V , • where all x i , y i , z i are disjoint.Consider the induction statement: x ∈ U ∧ ( ∀ i < m, x i ∈ U → x i +1 ∈ U ) → x m ∈ U (3)and write it as a disjunction with m + 1 disjuncts: x / ∈ U ∨ _ i ( x i ∈ U ∧ x i +1 / ∈ U ) ∨ x m ∈ U . (4)Now replace x i ∈ U by x i / ∈ V and x m ∈ U by x m / ∈ V and write it proposi-tionally: ¬ U ( x , y ) ∨ _ i [ ¬ V ( x i , z i ) ∧ ¬ U ( x i +1 , y i +1 )] ∨ ¬ V ( x m , z m ) . (5)Note that except the x -variables the m + 1 disjuncts are disjoint. Claim 1: (5) has a p-size proof in P. To see this note that induction (3) can be proved by simulating modus ponens(here we use that P contains LK / ). Disjunction (4) follows from it becausewe assume that the disjointness of U n , V n has short P-proofs, i.e. U ( x, y ) →¬ V ( x, z ) has a short proof.Now apply the supposed KPT interpolation to (5). W.l.o.g. we shall assume(and arrange that in the construction below) that x ∈ U and x m ∈ V (withwitnesses y and z m , respectively). Hence Student in the KPT computation issupposed to find i < m for which the i -th disjunct A i := [ ¬ V ( x i , z i ) ∧ ¬ U ( x i +1 , y i +1 )] , i = 1 , . . . , m − i to i + 1 fails). We shall showthat the existence of such a KPT p-time Student allows to separate U n from V n with a non-negligible advantage violating the hypotheses of the theorem.Take any m such that 32 k − ≤ m ≤ n O (1) (the upper bound implies thatthe proof in Claim 1 is of size n O (1) ). For 1 ≤ i < m define: W i [ m ] := U i × V m − i and W [ m ] := [ i W i [ m ] . Note that any string w = ( w , . . . , w m ) ∈ W [ m ] satisfies w ∈ U and w m ∈ V .Let k ≥ f , . . . , f k be the constant and the p-time functions providedthe assumed KPT interpolation for P. Assume that 1 ≤ i < m is the most4requent value f computes on inputs from W [ m ] (thinking of a P-proof π asfixed). This maximal frequency γ is at least 1 /m . (Here the frequency meanswith respect to the product of distributions D n on { , } n for which it is assumedthat U n , V n are hard to separate.) Claim 2: The frequency on W i [ m ] is at least γ − n ω (1) , i.e. it is at least /m modulo a negligible error. Note that for any i < j the frequency for W i [ m ] , W j [ m ] can differ only negligiblybecause otherwise we could use the usual triangle inequality argument to find anon-negligible discrepancy between frequencies on W t [ m ] and W t +1 [ m ] for some i ≤ t < j , and use it to separate U n from V n (on position t + 1, after fixing therest of coordinates by averaging). Because all W i [ m ] are disjoint, the frequencymust be γ up to a negligible difference.Now we describe a process that transforms the assumed successful strategyfor Student into a p-time algorithm with p-size advice, separating U n , V n witha non-negligible advantage.Assume first i < m/ 2. By averaging there are u , . . . u m/ ∈ U n s.t. f ( w ) = i with frequency at least 1 / (2 m ) (the factor 2 in the denominator allows us toforget about the ”up to the negligible error” phrase) for all w of the form: { u } × . . . × { u m/ } × W [ m/ . Fix such u , . . . , u m/ and also witnesses a , . . . , a m/ for their membership in U . These will be used as advice for the eventual algorithm.If i ≥ m/ m/ V n and include the relevant witnesses in the advice. W.l.o.g. we assume that thefirst case i < m/ k − W [ m/ w = ( w , . . . , w m/ ) ∈ W [ m/ 2] define: ˜ w := ( u , . . . , u m/ , w , . . . , w m/ ) ∈ W [ m ] (6)and run f on ˜ w . If f ( ˜ w ) = i , declare failure. Otherwise use the advicewitnesses to produce a falsifying assignment for A i : U ( u i +1 , a i +1 ) holds.After this first step use functions f , f , . . . and as long as they give values j < m/ a j . EventuallyStudent proposes value j ≥ m/ 2: choose the most frequent such value i ≥ m/ i , further restricting domain (6) as in binary search.Repeating this at most ( k − W [ m/ (2 k − )] which is at least W [3].In fact, we shall arrange in the last step that exactly W [3] remains (byfilling in more positions by elements of U n or V n , respectively, if needed)and hence the inputs before applying the last KPT function f k are of theform ( w , w , w ) with w ∈ U and w ∈ V .Note that Student gets to use f k because if he succeeded earlier it wouldviolate Claim 2. 5. The last function f k has to find a gap in the induction, and this itself willviolate Claim 2. In particular, the gap is either between w and w andthen w ∈ V , or between w and w and then w ∈ U .3. This process has the probability ≥ / (2 m ), i.e. non-negligible, of notfailing in any of the k − w in U or V with a non-negligibleprobability. In all cases when the process fails output random bit 0 or 1with equal probability.This proves the theorem.We conclude by pointing out that the KPT theorem enters propositional proofcomplexity also via notions of pseudo-surjective and iterable maps in the theoryof proof complexity generators, cf.[2] or [4, Sec.19.4]. Acknowledgements: I thank J. Pich (Oxford) for comments on an earliernote. References [1] S. A. Cook and N. Thapen, The strength of replacement in weak arithmetic, ACM Transactions on Computational Logic , , (2006).[2] J. Kraj´ıˇcek, Dual weak pigeonhole principle, pseudo-surjective functions,and provability of circuit lower bounds, J. of Symbolic Logic , , (2004),pp.265-286.[3] J. Kraj´ıˇcek, On the proof complexity of the Nisan-Wigderson generatorbased on a hard NP ∩ coNP function, J. of Mathematical Logic , ,(2011), pp.11-27.[4] J. Kraj´ıˇcek, Proof complexity , Encyclopedia of Mathematics and Its Appli-cations, Vol. , Cambridge University Press, 2019.[5] J. Kraj´ıˇcek, P. Pudl´ak, and J. Sgall, Interactive Computations of OptimalSolutions, in: B. Rovan (ed.): Mathematical Foundations of Computer Sci-ence (B. Bystrica, August ’90), Lecture Notes in Computer Science ,Springer-Verlag, (1990), pp. 48-60.[6] J. Kraj´ıˇcek, P. Pudl´ak and G. Takeuti, Bounded arithmetic and the polyno-mial hierarchy, Annals of Pure and Applied Logic ,52