Polynomial-Time Satisfiability Tests for ‘Small’
Membership Theories
∗†Domenico Cantone[0000−0002−1306−1166] and Pietro Maugeri Dept. of Mathematics and Computer Science, University of Catania, Italy
{domenico.cantone,pietro.maugeri}@unict.it
Abstract. We continue here our investigation aimed at the identifica- tion of ‘small’ fragments of set theory that are potentially useful in au- tomated verification with proof-checkers based on the set-theoretic for- malism, such asÆtnaNova. More specifically, we provide a complete tax- onomy of the polynomial and theNP-complete fragments comprising all conjunctions that may involve, besides variables intended to range over the von Neumann set-universe, the Boolean set operators∪,∩,\and the membership relators∈ and ∈. This is in preparation of combining the/ aforementioned taxonomy with one recently developed for similar frag- ments, but involving, in place of the membership relators∈and∈, the/ Boolean relators ⊆,6⊆,=,6=, and the predicates ‘· = ∅’ and ‘Disj(·,·)’
(respectively meaning ‘the argument set is empty’ and ‘the arguments are disjoint sets’), along with their opposites ‘·6=∅’ and ‘¬Disj(·,·)’.
Keywords: Satisfiability problem, Computable set theory,NP-completeness, Proof verification
1 Introduction
Since the late seventies, the satisfiability problem for fragments of set theory, namely the problem of algorithmically determining for any formula in a given fragment whether or not there exists an assignment of sets in the von Neumann universe to its free variables that makes the formula true, has been thoroughly studied within the research field namedComputable Set Theory(see [4,6,15,14,9]
for a fairly comprehensive account). The initial goal envisaged an automated proof verifier, based on the set-theoretic formalism, capable to carry out the formalization of extensive parts of classical mathematics (e.g., the Cauchy inte- gral theorem of complex analysis). To dispense users with the very tiny details, typical of low-level logic-oriented proofs, such a proof verifier should have been able to automatically guess the ‘obvious’ deduction steps left as tacit, through an inferential core embodying an extensive library of decision procedures.
∗We gratefully acknowledge support from “Universit`a degli Studi di Catania, Piano della Ricerca 2016/2018 Linea di intervento 2”.
†Copyright c2019 for this paper by its authors. Use permitted under Creative Com- mons License Attribution 4.0 International (CC BY 4.0)
However, a foundational quest to locate the precise frontier between the de- cidable and the undecidable in set theory (and in other fundamental mathemat- ical theories as well) began long before the proof verifier came into being, and inspired much of the subsequent work. The progenitor fragments in this quest, whose decision problems were addressed in the seminal paper [11], are the so- called Multi-Level Syllogistic (MLS, for short) and its extensionMLSSwith the singleton operator {·}. We recall that MLS is the quantifier-free fragment con- sisting of the propositional combination of literals of the following types:
x=y∪z, x=y∩z, x=y\z, x∈y.
As shown in [5], the satisfiability problem for MLS is NP-complete, even when restricted to conjunctions of flat literals of the above types, plus negative literals of typex /∈y. Then,a fortiori, all decidable extensions ofMLShave anNP-hard satisfiability problem (even hyperexponential in some cases; see [2,7,8]).
Nevertheless, despite itsNP-hardness, the decision procedure for a slightly extended variant of MLSS, implemented in the form of a decidable tableaux calculus (see [10]), has become the core of the most central inference primitive of theÆtnaNova/Refproof-checker [15], viz. theELEMrule. Optimization of the MLSS-decision test, at least in favourable cases, is therefore of utmost importance to limit occasional poor performances of ÆtnaNova originating from the full- strength decision test.
For such reason, we recently undertook an investigation aimed at identifying
‘small’, yet useful, decidable fragments of set theory (which so far are all subfrag- ments of MLS) endowed with efficientpolynomial-time decision tests. In [3], we have recently reported about the complexity taxonomy of all subfragments of the set theory fragment denotedBST and consisting of the conjunctions of literals involving, besides set variables, the Boolean set operators ∪,∩,\, the Boolean relators⊆,*,=,6=, and the predicates (both affirmed and negated) ‘·=∅’ and
‘Disj(·,·)’, expressing respectively that a specified set is empty and that two specified sets are disjoint.
Here we examine the sublanguages ofMST(∪,∩,\,∈, /∈),1the fragment of set theory consisting of the conjunctions of literals involving, besides set variables, the Boolean set operators∪,∩,\and the relators∈and∈./
Of a fragment ofMST(∪,∩,\,∈, /∈), we say that it is NP-complete if it has anNP-complete satisfiability problem (see [12]). Likewise, we say thatit is poly- nomial if its satisfiability problem has polynomial complexity. As in [3], as a first approximation, at least as far as polynomial fragments are concerned, it is enough to discover theminimal NP-complete fragments (namely theNP-complete frag- ments of MST(∪,∩,\,∈, /∈) that do not strictly contain any NP-complete frag- ment) and the maximal polynomial fragments (namely the polynomial fragments of MST(∪,∩,\,∈, /∈) that are not strictly contained in any polynomial frag- ment of MST(∪,∩,\,∈, /∈)). Indeed, any MST(∪,∩,\,∈, /∈)-fragment either is contained in some maximal polynomial fragment or contains some minimalNP- complete fragment.
1 The acronym MST stands for ‘membership set theory’, whereas BST stands for
‘Boolean set theory’.
The paper is organized as follows. Preliminarily, in Section 2, we introduce the syntax and semantics of the fragmentMST(∪,∩,\,∈, /∈) of our interest. Then, in Section 3, we provide polynomial decision procedures for its maximal polynomial subfragmentsMST(∪,∈, /∈),MST(∪,∈, /∈), andMST(∪,∩,\, /∈), whereas in Sec- tion 4 we prove theNP-completeness of its minimalNP-complete subfragments MST(∪,∩,∈) andMST(\,∈). Section 5 contains some closing considerations and hints for future investigations.
2 Syntax and Semantics
The fragments of set theory of which in this paper we are investigating the satisfiability problem are parts, syntactically delimited, of the quantifier-free language MST(∪,∩,\,∈, /∈).This is the collection of all conjunctions of literals of the two typess∈t ands /∈t, where sandt stand for terms built up from a denumerably infinite supply of set variablesx1,x2,x3, . . .by means of the Boolean set operators of union∪, intersection∩, and set difference\.
More generally, we shall denote byMST(op1, . . . ,pred1, . . .) the subtheory of MST(∪,∩,\,∈, /∈) involving exactly the set operators op1, . . . (drawn from the set{∪,∩,\}) and the predicate symbolspred1, . . .(drawn from the set{∈, /∈}).
For any MST(∪,∩,\,∈, /∈)-conjunction ϕ, we shall denote by Vars(ϕ) the collection of set variables occurring in ϕ; Vars(τ) is defined likewise, for any termτ.
Aset assignment M is any function sending a collection of set variables V (called thedomainofM and denoteddom(M)) into the von Neumann universeV of well-founded sets. We recall that thevon Neumann universe(see [13, pp. 95–
102]) is built up in stages as the union V := S
α∈OnVα of the levels Vα :=
S
β<αP(Vβ), with αranging over the class On of all ordinal numbers, where P(·) is the powerset operator. For any set S ∈ V, the least ordinal α such S⊆ Vα is therank ofS, denotedrk(S).
To shorten proofs, we shall sometimes make use ofurelements in the defi- nition of set assignments. These are objects that do not own any element and yet are distinct from the empty set (and, therefore, from any set) and distinct among them. However, in all cases under consideration, it will always be possible, without disrupting the correctness of any of the proofs, to replace all urelements by ‘proper’ sets, all of which either sharing the same sufficiently high rank, or having a conveniently large cardinality.
Natural designation rules attach recursively a value to every term τ of MST(∪,∩,\,∈, /∈) such that Vars(τ) ⊆ dom(M), for any set assignment M; here is how:
M(s∪t) :=M s∪M t, M(s∩t) :=M s∩M t, and M(s\t) :=M s\M t.
We also put: M(s ∈ t) = true ↔ M s ∈ M tand M(s /∈ t) := ¬M(s ∈ t), and then, recursively,M(ϕ∧ψ) :=M ϕ∧M ψ,whenϕ, ψ areMST(∪,∩,\,∈, /∈)- conjunctions such thatVars(ϕ∧ψ)⊆dom(M).
For convenience, we shall represent terms of the form x1 ∪. . .∪xh and y0∩. . .∩yk as S{x1, . . . , xh} and T{y0, . . . , yk}, respectively. Thus, for a set
assignmentM and a finite nonempty collection of set variablesL⊆dom(M), we shall have M(SL) =S
x∈LM x andM(TL) = T
x∈LM x. We also putM L = {M x|x∈L}, so thatM(SL) =SM Land M(TL) =TM Lhold.
Given a conjunctionϕofMST(∪,∩,\,∈, /∈) and a set assignmentMsuch that Vars(ϕ)⊆dom(M), we say thatM satisfiesϕ, and writeM |=ϕ, ifM ϕ=true.
WhenM satisfiesϕ, we say thatM is amodel ofϕ.
A conjunctionϕissatisfiableif it has some model, otherwise it isunsatisfiable.
Any two conjunctionsϕandψareequisatisfiableif they are either both satisfiable or both unsatisfiable.
Since MST(∪,∩,\,∈, /∈) is a subtheory of MLS, it plainly has a solvable satisfiability problem, namely there is an algorithm (called adecision procedure or a satisfiability test) that, for any given conjunctionϕ of MST(∪,∩,\,∈, /∈), establishes in an effective manner whetherϕis satisfiable or not.
In the following sections, we shall find out the maximal polynomial and the minimalNP-complete fragments ofMST(∪,∩,\,∈, /∈).
Remark 1. Our complexity results will implicitly refer tolinearly bounded sub- classes ofMST(∪,∩,\,∈, /∈), whose conjunctionsϕmeet the condition
max{j|xj ∈Vars(ϕ)} −min{j |xj∈Vars(ϕ)}=O(|ϕ|). (1) For instance, it is immediate to see that the class of all MST(∪,∩,\,∈, /∈)- conjunctionsϕsuch thatVars(ϕ) ={x1, . . . , x|Vars(ϕ)|}is linearly bounded. For lists Lϕ of sets of variables occurring in any conjunction ϕ belonging to some linearly bounded subclass ofMST(∪,∩,\,∈, /∈), it turns out that the duplicates ofLϕ can be detected in linear timeO(|ϕ|).
3 The maximal polynomial fragments of MST(∪, ∩, \, ∈, / ∈)
The maximal polynomial fragments of MST(∪,∩,\,∈, /∈) are MST(∪,∈, /∈), MST(∩,∈, /∈), andMST(∪,∩,\, /∈), whose satisfiability problems can be solved in linear, quadratic, and constant time, respectively.
3.1 The fragment MST(∪,∈, /∈)
We prove that the satisfiability problem forMST(∪,∈, /∈)-conjunctions admits a linear-time solution by first showing thatMST(∪,∈) is linear and then proving that the satisfiability problem forMST(∪,∈, /∈) can be reduced in linear time to that forMST(∪,∈).
A linear-time satisfiability test for MST(∪,∈) To start with, we provide a decision procedure forMST(∪,∈) and then prove that it runs in linear time in the size of the input formula.
Theorem 1. Let ϕ be a MST(∪,∈)-conjunction of the form Vp i=1
SLi ∈SRi, where theLi’s and theRi’s are finite nonempty collections of set variables. Then:
(a) if ≺is a linear ordering of Vars(ϕ)satisfying the condition
max(Li,≺) ≺ max(Ri,≺), fori= 1, . . . , p, (2) thenϕhas a modelM such that
(a1) M x ={ux} ∪ {SM Li | x= max(Ri,≺), fori = 1, . . . , p}, where the ux’s are pairwise distinct urelements; and
(a2) S
M L /∈ M x, for every L ⊆ Vars(ϕ) and x ∈ Vars(ϕ) fulfilling the
condition p
^
i=1
(L=Li −→x /∈Ri); (3)
(b) if ϕ is satisfiable, then there is a linear ordering ≺ of Vars(ϕ) such that condition (2)holds.
Hence, the satisfiability problem for MST(∪,∈)-conjunctions is solvable.
Proof. Let us first assume that there is a linear ordering≺ofVars(ϕ) such that (2) holds. Following the ordering≺, forx∈Vars(ϕ) we put
M x:={ux} ∪S
M Li |x= max(Ri,≺), i= 1, . . . , p , (4) where theux are pairwise distinct urelements. For any literalSLi∈SRi in ϕ, withi∈ {1, . . . , p}, setting xi := max(Ri,≺), we haveS
M Li∈M xi ⊆S M Ri, so that M |= S
Li ∈ S
Ri. Thus, M models correctly all conjuncts of ϕ, and it plainly satisfies condition (a1). In fact, also condition (a2) is true for M. Preliminarily, we observe that we clearly have
SM L=S
M L0 if and only ifL=L0, (5) forL, L0 ⊆Vars(ϕ), as the urelements inS
M Lare the same as those inS M L0 if and only if L = L0. Next, let L ⊆ Vars(ϕ) and x ∈ Vars(ϕ) be such that Vp
i=1(L=Li −→x /∈Ri) holds, but assume, by way of contradiction,S M L∈ M x. Then, by (5) and (4), L = Li0 for some i0 ∈ {1, . . . , p} such that x = max(Ri0,≺)∈Ri0, contradicting (3). ThereforeM satisfies condition (a2) too.
Concerning condition (b), let us now assume that our conjunctionϕis satis- fiable, and letM be a model ofϕ. Also, let≺be any linear ordering ofVars(ϕ) such that
rk M x
<rk M y
−→x≺y, forx, y∈Vars(ϕ), (6) Let us check that (2) holds for the ordering≺. Leti∈ {1, . . . , p}. Therefore SM Li ∈ SM Ri, so that Ri 6= ∅. Hence if Li = ∅, max(Li,≺) ≺ max(Ri,≺) holds trivially. On the other hand, ifLi 6=∅, we have:
max{rk M x
|x∈Li}=rk SM Li
<rk SM Ri
= max{rk M x
|x∈Ri}, so that, by (6), max(Li,≺)≺max(Ri,≺) holds also in this case. Therefore, (2) holds, completing the proof of (b).
Conditions (a) and (b) yield that the conjunctionϕis satisfiable if and only if there is a linear ordering≺ofVars(ϕ) satisfying (2), from which the decidability of the satisfiability problem forMST(∪,∈) readily follows.
Towards a linear satisfiability test forMST(∪,∈), we derive next two condi- tions, which can be tested in linear time and whose conjunction is equivalent to (2). To begin with, in connection with any formula ϕof MST(∪,∈), we define the left-variables (resp., right-variables) of ϕ as those variables inϕ occurring in the left-hand sideL(resp., right-hand sideR) of some literalSL∈SRinϕ.
Right-variables ofϕthat are not left-variables are calledproper right-variables.
Let nowϕbe any satisfiableMST(∪,∈)-conjunction and let≺be any linear ordering ofVars(ϕ) fulfilling condition (2). Plainly, no left-variable inϕcan be
≺-maximal. Hence
(A) ϕhas some proper right-variable.
In addition, by letting ϕ− be the result of dropping from ϕall conjuncts that involve some proper right-variable, we clearly have
(B) ϕ− is satisfiable.
It turns out that conditions (A) and (B) are also sufficient for the satisfiability of ϕ. Indeed, assume that conditions (A) and (B) hold for a givenMST(∪,∈)- conjunctionϕ, and let≺−be a linear ordering ofVars(ϕ−) fulfilling condition (2) of Lemma 1 as applied toϕ−. Then, any extension≺of≺−to a linear ordering of Vars(ϕ−) such thatx≺y, for every left-variablexand proper right-variableyof ϕ, fulfils condition (2) of Lemma 1 as applied toϕ, proving that the conjunction ϕis satisfiable.
The above considerations readily yield the following satisfiability test for MST(∪,∈):
Algorithm 1Satisfiability tester forMST(∪,∈) while ϕcontains some proper right-variable do
drop fromϕall the conjuncts involving some proper right-variable;
if ϕis the empty conjunction then return“satisfiable”;
else
return“unsatisfiable”;
A straightforward implementation of Algorithm 1 is quadratic. An alterna- tive implementation consists in counting the number of left-occurrences of each variablex∈Vars(ϕ), while also maintaining, for each right-variabley inϕ, the list of the conjuncts containing y. Variables with a zero counter are exactly the proper right-variables. While the conjunctionϕ contains right-variables with a zero counter, drop from it all conjuncts containing some right-variable with a zero counter and, accordingly, decrement the counters of the left-occurrences in the dropped conjuncts. If eventually one ends up with the empty conjunction, then the initial conjunction ϕ is declared satisfiable. Otherwise, ϕ is declared unsatisfiable.
It is not hard to see that the above implementation has a linear-time com- plexity. Hence, we have:
Theorem 2. The satisfiability problem for MST(∪,∈) can be solved in linear time.
A linear-time reduction of MST(∪,∈, /∈) to MST(∪,∈) We shall now prove that the satisfiability problem for MST(∪,∈, /∈) can be reduced in linear time to that ofMST(∪,∈), thereby yielding a linear-time decision procedure for the satisfiability problem of the extended fragmentMST(∪,∈, /∈).
Lemma 1. The satisfiability problem forMST(∪,∈, /∈)can be reduced in linear time to that forMST(∪,∈).
Proof. Letϕbe a conjunction ofMST(∪,∈, /∈) of the form Vp
i=1
SLi∈SRi ∧ Vn j=p+1
SLj∈/SRj,
where theLi’s andRi’s are finite collection of set variables. For eachi= 1, . . . , p, let
R0i:=Ri\[
{Rj|Lj=Li, forj=p+ 1, . . . , n}, (7)
and put ϕ0 :=
(Vp i=1
SLi∈SR0i if no R0i is empty x0∈x0 otherwise,
for a fixed, but otherwise arbitrary, set variablex0. It is an easy matter to verify that ifϕis satisfiable, then (ϕ0≡Vp
i=1
SLi∈SR0i and) any model forϕis also a model forϕ0, i.e.,|=ϕ−→ϕ0.
Conversely, ifϕ0 is satisfiable (and therefore ϕ0 ≡ Vn i=1
SLi ∈S
R0i), then, by Theorem 1(b), there is a linear ordering≺ofVars(ϕ0) fulfilling condition (2) and in addition, by Theorem 1(a), the conjunctionϕ0 has a modelM satisfying conditions (a1) and (a2) of the same theorem. We extend M to the variables x ∈ Vars(ϕ)\Vars(ϕ0), if any, by setting M x := {ux}, where the ux’s are pairwise distinct urelements new to the assignmentM. Since S
M Ri0 ⊆S M Ri, fori= 1, . . . , p, we plainly haveM |=Vp
i=1
SLi ∈S
Ri,namelyM satisfies all the positive conjuncts ofϕ.
Let us now show thatM satisfies the negative partVn j=p+1
SLj ∈/ SRj ofϕ as well. Indeed, if this were not the case, there would exist aj ∈ {p+ 1, . . . , n}
and a variablex∈Rj such thatSM Lj ∈M x. SinceSM Lj is not a urelement, M x would contain some set and therefore x ∈ Vars(ϕ0). Thus, by condition (a1) of Theorem 1, M x={ux} ∪S
M Li|x= max(R0i,≺), fori= 1, . . . , p , and so we would have SM Lj = SM Li, for some i ∈ {1, . . . , p} such that x = max(R0i,≺), so that x ∈ R0i. The pairwise distinctness of the urelements would yieldLj=Li. Hence, by (7),x /∈R0i, a contradiction.
In conclusion,M must also satisfy the negative partVn j=p+1
SLj ∈/ S Rj of ϕ, and thus, in view ofM |=Vp
i=1
SLi∈S
Ri seen above,M satisfiesϕ.
Summing up, we have proved that ifϕ0 is satisfiable, so isϕ, which, in view of|=ϕ−→ϕ0 observed above, yields the equisatisfiability ofϕandϕ0.
To complete the proof, it is enough to observe that the conjunctionϕ0 can be constructed inO(|ϕ|) time, where|ϕ|denotes the size ofϕ, using a suitable linear time algorithm to detect the duplicates in the list of setsL1, . . . , Ln.
In view of Theorem 2, Lemma 1 yields the following complexity result:
Theorem 3. The satisfiability problem forMST(∪,∈, /∈)can be solved in linear time.
3.2 The fragment MST(∩,∈, /∈)
Letϕbe aMST(∩,∈, /∈)-conjunction of the form
p
^
i=1
TLi∈TRi ∧
`
^
j=p+1
TLj∈/TRj, (8) where theLu’s and theRu’s arenonempty finite collections of set variables, for u= 1, . . . , `. We shall denote byϕ+ and ϕ− thepositive part Vp
i=1
TLi ∈T Ri
and the negative part V` j=p+1
TLj ∈/ TRj of ϕ, respectively. For decidability purposes and without loss of generality, we may assume that the following con- ditions hold forϕ:
(C0) the positive partϕ+ ofϕis nonempty, namelyp≥1, (C1) Lh6=Li, for any two distincth, i∈ {1, . . . , p}.
Indeed, as for (C0), if the positive part ofϕwere empty, thenϕwould be always satisfiable (e.g., ϕwould be satisfied by the null assignment M∅ over Vars(ϕ), which maps every set variable in ϕ to the empty set ∅). In addition, without disrupting satisfiability, condition (C1) can be enforced by replacing, for each set of variablesL∈ {L1, . . . , Lp}, the collection of conjunctsTLi ∈TRi inϕ+ such that Li =L by the single conjunct TL ∈ TR, where R := S{Ri |Li = L, fori= 1, . . . , p}, since|=TTRi|Li=L, fori= 1, . . . , p =TR.
Notice that the duplicates in the list of setsL1, . . . , Lp can be suitably de- tected in linear time, and therefore condition (C1) can be enforced in timeO(|ϕ|).
Theorem 4. Letϕbe aMST(∩,∈, /∈)-conjunction of the form (8)and fulfilling conditions (C0) and (C1). Then ϕis satisfiable if and only if the following two conditions hold:
(a) Li=Lj −→ Rj *Ri,
fori= 1, . . . , pandj=p+ 1, . . . , `;
(b) there is an indexing of the conjuncts ofϕ+ such that, forh, i= 1, . . . , p,
Li⊆Rh −→ h < i. (9)
Proof. (Necessity).Let us first assume thatϕis satisfiable, and letM |=ϕ.
Concerning condition (a), let Li = Lj, for some i ∈ {1, . . . , p} and j ∈ {p+ 1, . . . , `}. Hence we haveT
M Li∈T
M Ri\T
M Rj, soRj *Ri must hold.
As for condition (b), any indexing of the conjuncts of the positive partϕ+ ofϕsuch that
rk(TM Lh)<rk(TM Li) −→ h < i, (10) forh, i= 1, . . . , p, satisfies (9), since, forh, i= 1, . . . , p,
Li⊆Rh −→ \
M Lh∈\
M Rh⊆\ M Li
−→ rk\ M Lh
<rk\ M Li
−→ h < i.
(Sufficiency).Conversely, let us assume that conditions (a) and (b) of the theo- rem hold forϕ.
A pair of (distinct) indicesu, v ∈ {1, . . . , `}such that Lu6=Lv is said to be R+-indistinguishable (w.r.t. ϕ)ifLu ⊆Ri ←→Lv⊆Ri, for everyi= 1, . . . , p.
Otherwise, we say that the pairu, v isR+-distinguishable.
For each R+-indistinguishable pairu, v, we set
kuv := min {u|Lu*Lv} ∪ {v|Lv *Lu} , and let kuv be the index such that {kuv, kuv} ={u, v}, so that Lk
uv * Lkuv. Also, we let Kbe the collection of all thekuv’s so defined.
Next, with eachk∈K we associate a distinct urelementuk, and put recur- sively, forv= 1, . . . , `:
Ii:={Iu|Lv⊆Ri, fori= 1, . . . , p} ∪ {uk |Lv⊆Lk, fork∈K}. (11) We also put, forx∈Vars(ϕ):
M x:={Ii|x∈Ri, fori= 1, . . . , p} ∪ {uk |x∈Lk, fork∈K}. (12) The rest of the proof is devoted to showing that the assignment M just defined satisfiesϕ. We shall use the following claims, whose proofs are omitted for lack of space.
Claim 1. Lu=Lv ←→ Iu=Iv, for u, v= 1, . . . , `.
Claim 2. If Iu ∈TM S, where u∈ {1, . . . , `} andS ⊆Vars(ϕ), then S ⊆Ri, for somei∈ {1, . . . , p} such thatLi=Lu.
Claim 3. TM Lv=Iv, for v= 1, . . . , `.
We start with the positive conjuncts. Thus, letTLi ∈TRi be any positive conjunct in ϕ, with i ∈ {1, . . . , p}. By Claim 3, TM Li = Ii. In addition, by (12), Ii ∈ M xfor every x∈ Ri, and thereforeTM Li =Ii ∈TM Ri, proving that M models correctly the conjunct TLi∈TRi, and in turn all the positive conjuncts ofϕ.
Next, we show that also the negative conjuncts ofϕare satisfied byM. To this purpose, let TLj ∈/ TRj be any negative conjunct in ϕ, with j ∈ {p+ 1, . . . , `}. Again by Claim 3, TM Lj = Ij. By way of contradiction, assume that TM Lj ∈TM Rj, namely Ij ∈ TM Rj. Then, by Claim 2, Rj ⊆Ri and Lj = Li for some i ∈ {1, . . . , p}, contradicting condition (a) of the theorem.
ThusTM Lj ∈/ TM Rj, proving thatM satisfies also the negative conjuncts of ϕ. Hence,M satisfies the conjunctionϕ.
Complexity issues The complexity of the satisfiability problem forMST(∩,∈ , /∈) can be estimated as follows. Given aMST(∩,∈, /∈)-conjunction of the form (8), condition (a) of Theorem 4 can be tested in O(ϕ) time by detecting the duplicates in the list of setsL1, . . . , Lp. Next, we observe that condition (b) of Theorem 4 is equivalent to the aciclicity of the oriented graphGϕ+= (Vϕ+, Eϕ+), where Vϕ+ := {1, . . . , p} and Eϕ+ := {(h, i) | Li ⊆ Rh, fori, h ∈ Vϕ+}. The graphGϕ+ can be constructed in timeO(p· |ϕ+|) and its aciclicity tested in time O(p2). Hence, condition (b) can be tested in timeO(p· |ϕ+|), yielding an overall time complexityO(p· |ϕ+|+|ϕ|) for the satisfiability problem ofMST(∩,∈, /∈).
Summing up:
Lemma 2. The satisfiability problem forMST(∩,∈, /∈)-conjunctions can be solved in quadratic time.
3.3 The trivial fragment MST(∪,∩,\, /∈)
Since any MST(∪,∩,\, /∈)-conjunctionϕis trivially satisfied by the null assign- mentM∅ over Vars(ϕ), defined byM∅x=∅ for everyx∈Vars(ϕ), we readily have:
Theorem 5. The satisfiability problem for the fragmentMST(∪,∩,\, /∈)can be solved in constant time.
4 The minimal NP-complete fragments
4.1 The fragment MST(∪,∩,∈)
We shall prove that the satisfiability problem for MST(∪,∩,∈)-conjunctions is NP-complete by reducing the problem 3-SAT to it.
LetF :=Vm
i=1(Li1 ∨Li2 ∨ Li3) be an instance of 3-SAT, where theLij’s are propositional literals, and let P1, . . . , Pn be the distinct propositional variables occurring inF. Also, letx, X1, X1, . . . , Xn, Xn be 2n+ 1 distinct set variables.
Fori= 1, . . . , m,j= 1,2,3, andk∈ {1, . . . , n}such thatLij ∈ {Pk,6=Pk}, put Tij :=ifLij =Pk thenXk else Xk endif .
Finally, let ΦF :=Vm
i=1(x∈Ti1∪Ti2∪Ti3) ∧Vn
k=1(x∈Xk∪Xk ∧ Xk∩Xk∈x).
Theorem 6. A 3-SAT instanceF is propositionally satisfiable if and only if the correspondingMST(∪,∩,∈)-conjunctionΦF is satisfied by a set assignment.
Proof. (Necessity). To begin with, let us assume that F is propositionally sat- isfiable, and let v be a Boolean valuation that satisfies it. Let Mv be the set assignment induced overVars(ΦF) byvand defined as follows, fork= 1, . . . , n:
MvXk := if v(Pk) = truethen {{∅}} else ∅ endif and MvXk := {{∅}} \ MvXk, and such thatMvx:={∅}. Hence,Mv(Xk∪Xk) ={{∅}}andMv(Xk∩ Xk) =∅for each k, so thatMv |=Vn
k=1(x∈Xk∪Xk ∧ Xk∩Xk ∈x).
Let i ∈ {1, . . . , m}. Since, by hypothesis, v satisfies F, then v(Li1∨Li2∨ Li3) =t, so thatv(Liji) =tfor someji ∈ {1,2,3}. Let k∈ {1, . . . , n} be such that Liji ∈ {Pk,¬Pk}. If Liji = Pk, then v(Pk) = t and Tiji = Xk, so that MvTiji ={{∅}}. On the other hand, ifLiji =¬Pk, thenv(Pk) =f and Tiji = Xk; hence, again, MvTiji = {{∅}}. Thus, Mvx = {∅} ∈ {{∅}} = Mv(Tiji) ⊆ Mv(Ti1∪Ti2∪Ti3),proving thatMv satisfies the literalx∈Ti1∪Ti2∪Ti3.
The arbitrariness of i ∈ {1, . . . , m}, together with Mv |= Vn
k=1(x ∈ Xk ∪ Xk ∧ Xk∩Xk ∈x) proved before, yields that the set assignmentMv satisfies the conjunctionΦF corresponding toF.
(Sufficiency).Let us now assume that theMST(∪,∩,∈)-conjunctionΦF is satisfi- able, and letM be a set assignment that satisfies it. Then, for eachk= 1, . . . , n, we have M x ∈ M Xk∪M Xk ∧ M Xk ∩M Xk ∈ M x, and therefore M x /∈ M Xk∩M Xk. Hence, either M x ∈ M Xk or M x ∈ M Xk, but not both. We now define a Boolean valuationvM induced onP1, . . . , Pn by the set assignment M, by putting,vM(Pk) :=M(x∈Xk), fork = 1, . . . , n. SinceM |=ΦF, then
M |=Vm
i=1x∈Ti1∪Ti2∪Ti3. Let i∈ {1, . . . , m}. ThenM x∈M Tiji for some ji∈ {1,2,3}. Letk∈ {1, . . . , n}be such thatTiji∈ {Xk, Xk}. IfTiji=Xk, then Liji =Pk and vM(Pk) =t. On the other hand, if Tiji =Xk, thenLiji =¬Pk andvM(Pk) =f. In any case,vM(Liji) =t, and sovM(Li1∨Li2∨Li3) =t. The arbitrariness of i∈ {1, . . . , m} yields that the Boolean valuationv satisfies the 3-SAT instanceF.
From the previous theorem, and since the conjunctionΦF corresponding to F can be constructed in timeO(|F|), we can conclude that
Lemma 3. The satisfiability problem forMST(∪,∩,∈)-conjunctions is NP-complete.
4.2 The fragment MST(\,∈)
We shall prove that the satisfiability problem forMST(\,∈)-conjunctions isNP- complete by reducing the 3-SAT problem to it.
Thus, let F := Vm
i=1(Li1 ∨ Li2 ∨ Li3) be an instance of 3-SAT, where theLij’s propositional literals, and let P1, . . . , Pn be the distinct propositional variables occurring inF. Also, letx, X, X1, . . . , Xnben+2 distinct set variables.
Fori = 1, . . . , m, j = 1,2,3, andk∈ {1, . . . , n} such thatLij ∈ {Pk,¬Pk}, let Tij :=if Lij =Pk thenXk else X\Xk endif. Finally, put
ΦF :=Vm
i=1(X\Ti1\Ti2\Ti3)∈x ∧x∈X.
Theorem 7. A 3-SAT instanceF is propositionally satisfiable if and only if the MST(\,∈)-formulaΦF is satisfied by a set assignment.
Proof. (Necessity).First, let us assume thatF is propositionally satisfiable, and letvbe a Boolean valuation overP1, . . . , Pn that satisfiesF. We define the set assignmentMv, induced overVars(ΦF) byv, by setting:
MvX :={{∅}}, Mvx:={∅}, MvXk :=
({{∅}} ifv(Pk) =t
∅ otherwise, for k = 1, . . . , n. Plainly Mvx∈ MvX. Let i ∈ {1, . . . , m}. As, by hypothesis, the Boolean valuationvsatisfiesF, thenv(Li1 ∨ Li2 ∨ Li3) =t, so that there exists a ji ∈ {1,2,3} such that v(Liji) = t. Let k ∈ {1, . . . , n} be such that Liji ∈ {Pk,¬Pk}. If Liji =Pk, then v(Pk) =tand Tij =Xk, so that MvTiji = {{∅}}. On the other hand, ifLiji=¬Pk, thenv(Pk) =fandTiji=X\Xk and MvXk =∅, and againMvTiji ={{∅}}. Hence,
MvX\MvTi1\MvTi2\MvTi3=MvX\(MvTi1∪MvTi2∪MvTi3) =∅ ∈ {∅}=Mvx.
Thus, the arbitrariness ofi∈ {1, . . . , m}together withMvx∈MvX yields that the induced set assignmentMv satisfiesΦF.
(Sufficiency). Next, let us assume that the MST(\,∈)-conjunction ΦF corre- sponding to F is satisfiable, and let M be a set assignment over its variables
that satisfies it, so thatM x∈M Xholds. We define the Boolean valuationvM, induced onP1, . . . , Pn byM, by settingvM(Pk) :=M(x∈Xk) and prove that it satisfies the 3-SAT instance F. Thus, leti ∈ {1, . . . , m}. Since M |=ΦF, it holds that M |=X\Ti1\Ti2\Ti3 and therefore
M X\(M Ti1∪M Ti2∪M Ti3) =M X\M Ti1\M Ti2\M Ti3∈M x∈M X.
Hence, the well-foundedness of∈yields thatM x /∈M X\(M Ti1∪M Ti2∪M Ti3) must hold, so that M x ∈ M Ti1∪M Ti2∪M Ti3 must hold as well. From the latter, it follows thatvM(Li1 ∨ Li2 ∨ Li3) =t. Finally, by the arbitrariness of i∈ {1, . . . , m}, we have thatvM(F) =t, proving thatF is satisfiable.
Since the formulaΦF can be constructed in timeO(|F|), for any given 3-SAT instanceF, from the previous theorem we immediately conclude that
Lemma 4. The satisfiability problem forMST(\,∈), is NP-complete.
\ ∪ ∩ ∈ ∈/ Complexity
? ?
NP-complete? ? ?
NP-complete? ? ?
O(n)? ? ?
O(n2)? ? ? ?
constantTable 1.Maximal polynomial and minimalNP-complete fragments ofMST(\,∪,∩,∈, /∈)
5 Conclusions
With in mind applications in automated proof checking with verifiers based on the set-theoretic formalism, in this paper we identified the maximal polyno- mial and the minimal NP-complete sublanguages of the fragment of set theory MST(∪,∩,\,∈, /∈), as reported in Table 1. These allow one to easily pinpoint the 14 polynomial fragments (of which 2 are quadratic, 4 are linear, and 8 are constant) and the 10 NP-complete fragments among the 24 sublanguages of MST(∪,∩,\,∈, /∈).
We plan to extend our analysis when also the singleton operator {·}, the Boolean relators ⊆,6⊆,=,6=, and the predicates ‘· = ∅’ and ‘Disj(·,·)’ (both affirmed and negated) are allowed. A library of the related polynomial decision tests will then be implemented and integrated within the inferential core of the ÆtnaNovaproof-checker.
Acknowledgements
We are particularly grateful to E.G. Omodeo for very fruitful discussions and suggestions. We also wish to thank the anonymous reviewers for their helpful comments.
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Foreword by M. Davis.
