First-order axiomatisations of representable relation algebras need formulas of unbounded quantifier depth
aa r X i v : . [ m a t h . L O ] A ug FIRST-ORDER AXIOMATISATIONS OF REPRESENTABLERELATION ALGEBRAS NEED FORMULAS OF UNBOUNDEDQUANTIFIER DEPTH
ROB EGROT AND ROBIN HIRSCH
Abstract.
We prove that RRA, the class of all representable relation algebras ,cannot be axiomatised by any first-order theory of bounded quantifier depth.The proof uses of significant modification of the standard rainbow construction.We also discuss and correct a strategy proposed elsewhere for proving thatRRA cannot be axiomatised by any first-order theory using only finitely manyvariables. Introduction
It is known that RRA cannot be axiomatised by any finite theory [Mon64] norby any equational theory using only finitely many variables [J´on91, theorem 3.5.6].Moreover, any axiomatisation of RRA must involve infinitely many non-canonicalequations [HV05]. To prove that RRA cannot be axiomatised by any c -variabletheory would yield the first two of these results as corollaries and would significantlystrengthen what is known. Although the problem remains open, some progress ismade here. Our main new result is that there can be no axiomatisation of RRA ofbounded quantifier depth.This paper arises, partly, from a difficulty with [HH02, problem 1, page 625].The problem is to prove that RRA cannot be axiomatised by any c -variable first-order theory for finite c . As mentioned above, the problem remains open. Theproblem statement in the book includes a proposed solution. The proposal, which ismentioned in [SA05, p491], is to find graphs G and H with no homomorphism from G to H , but indistinguishable in a certain c -colour graph game. It is claimed thatsuch graphs could be used to prove that there is is no c -variable axiomatisation ofRRA. Readers of the book, who failed to solve the problem, cannot be rebuked sincethere was a flaw in the proposed solution. Fortunately, by noticing the problem andattempting to correct it, we are led to a modification of the rainbow construction,and this modification leads not only to a correction for [HH02, problem 1, page625], but also to the proof of the main result here.Those who are not familiar with the open problem, the rainbow construction andthe flaw in the suggestion from [HH02] may prefer to jump ahead to our main result,theorem 1.1, whose proof occupies most of the paper. The paper is structures asfollows. In section 2 we introduce the modified rainbow construction over pairs ofbinary relational structures, and prove a theorem relating a certain property of thesestructures to complete representability of the construction. As a corollary to this Mathematics Subject Classification.
Primary 03G15; Secondary 05C90, 05C60.
Key words and phrases.
RRA, Representable relation algebras, Finite variable axiomatisation,Bounded quantifier depth axiomatisation. e obtain the result that classes of atom structures intermediate between those ofcompletely representable and representable relation algebras cannot be axiomatizedusing only finitely many variables (see corollary 2.2). In section 3 we define thevertex colouring games, building up to a proof of theorem 1.1. Section 4 correctsthe proposal from [HH02]. In the special case of graphs, these games are connectedto the famous reconstruction conjecture (see e.g. [Sto88, Bon91, LS16] for surveys),and are discussed in more detail in [EH]. For the rest of the introduction we brieflydiscuss the rainbow construction, how it was hoped it could help solve the openproblem of the existence of finite variable axiomatisations for RRA, the flaw in theargument, and the modifications to the construction needed for our new resultshere.Suppose we can find graphs G, H , with no homomorphism from G to H , thatare indistinguishable in a certain colouring game using c colours (call them c -indistinguishable for short). The problem statement uses a ‘rainbow relation al-gebra’ A G,H built from the two graphs. It can be shown that A H,H is representablebut A G,H is not (see [HH02, theorem 16.5]). It is claimed in the problem statementthat, since
G, H are c -indistinguishable, it follows that A H,H and A G,H could not bedistinguished by any c -variable formula (i.e. that A G,F ≡ c A H,F ). Unfortunately,that implication is false. The problem is that there are white atoms w S ∈ A G,H forevery set of G -nodes S of size at most two, which would be represented by binarypredicates over graph nodes, but only monadic predicates are used in the graphcolouring game. Thus the proof cannot be completed.To fix that, the idea is to let B G,H be obtained from A G,H by deleting all whiteatoms w S (and deleting any forbidden triple involving these deleted atoms). Thissolves one problem, because now it is true that if G and H are c -indistinguishablethen it can be shown B G,F ≡ c B H,F . But it creates another, since ∃ really neededatoms w S in her winning strategy for the representation game over A H,H , so wecan no longer be sure that B H,H is representable. However, if it happens that everypartial homomorphism from H to H of size two extends to a homomorphism, thenit can be shown that B H,H is representable. This property is satisfied by completegraphs, for example, and we use this in the proof of theorem 1.1.We now give a formal statement of our main result.
THEOREM 1.1. If Σ is a set of first-order formulas defining RRA, then Σ in-cludes formulas of arbitrary quantifier depth. The proof is developed in the next two sections.2.
The modified rainbow construction
Let
G, H be structures in a signature consisting of only binary predicates, whichwe refer to as binary structures . For most purposes we can assume G and H tobe directed graphs, but it will be convenient at one point later to be able to forcegraph homomorphisms to preserve non-edges, and so we phrase our results here interms of binary structures so we can formally handle this without issue.Given two binary structures G and H , we define an atomic relation algebra B G,H by defining its atom structure. A partial homomorphism is a partial map h from G to H such that if i = i ′ ∈ G and ( i, i ′ ) belongs to a binary predicate interpreted in G then ( h ( i ) , h ( i ′ )) also belongs to that predicate interpreted in H . The atoms are { ′ , b , w , y } ∪ { g i , : i ∈ G } ∪ { r j,j ′ : j, j ′ ∈ H } he non-identity atoms are considered to be black, white, yellow, green or red. Allatoms are self-converse, except r ⌣j,j ′ = r j ′ ,j . Forbidden triples of atoms are Peirceantransforms of(I) (1 ′ , a, b ) where a = b (II) ( g i , g i ′ , g i ′′ ) , ( g i , g i ′ , w ), any i, i ′ , i ′′ ∈ G (III) ( y , y , y ) , ( y , y , b )(IV) ( r j ,j , r j ′ ,j ′ , r j ∗ ,j ∗ ), unless j = j ∗ , j = j ′ , j ′ = j ∗ .(V) ( g i , g i ′ , r j,j ′ ) unless { ( i, i ′ ) , ( j, j ′ ) } is a partial homomorphism.(VI) ( g i , g i , r j,j ′ ) any j, j ∈ H ′ .Observe for later that the only combinations of three colours where some but notall triples of atoms with those colours are forbidden are green-green-red and red-red-red, see (IV), (V) and (VI).The relation algebra B G,H is the complex algebra of this atom structure. Notethat here a triple ( a, b, c ) being forbidden corresponds to setting ( a ; b ) · c = 0 in thecomplex algebra. This is slightly different from the approach taken in [HH02], butthe differences are entirely superficial.To build complete representations for these algebras, we will employ a gameplayed by two players, ∀ and ∃ , over a kind of labeled complete digraph. Wedescribe this game now, basing our exposition on the material in [HH02, section 11].An atomic network N for an atomic relation algebra A consists of a set of nodes(denoted nodes ( N )) and a map (also denoted N ) from pairs of nodes to atoms of A , such that N ( x, x ) ≤ ′ , N ( y, x ) = N ( x, y ) ⌣ and ( N ( x, y ) , N ( y, z ) , N ( x, z )) isnot forbidden, for all x, y, z ∈ nodes ( N ). This property of not containing edgelabels forming a forbidden triple ( N ( x, y ) , N ( y, z ) , N ( x, z )) is often referred to asthe consistency of N . For atomic networks M, N we write M ⊆ N if nodes ( M ) ⊆ nodes ( N ) and for all x, y ∈ nodes ( M ) we have N ( x, y ) = M ( x, y ). If N i : i < λ is asequence of atomic networks where i < j → N i ⊆ N j then the limit N = S i<λ N i is the atomic network with nodes S i<λ nodes ( N i ) and where N ( x, y ) = N i ( x, y )where x, y ∈ nodes ( N i ) (and this label does not depend on i , since the sequence isnested).Given an atomic relation algebra A , the complete representation game for A hasmin( ω, |A| ) rounds. ∃ is trying to build an atomic network for A , and ∀ is tryingto force a situation where this is impossible. In a play of this game, let the currentatomic network be N . Then ∀ picks nodes x, y ∈ nodes ( N ) and atoms α, β suchthat ( α, β, N ( x, y )) is not forbidden. In response, ∃ is required to extend N to N ′ such that there is a node z ∈ nodes ( N ) ′ where N ( x, z ) = α, N ( z, y ) = β . Thedifficulty is that she must label all new edges induced by adding z without causingthe resulting network to be inconsistent. In other words, without creating anytriangles corresponding to forbidden triples (we also call these forbidden triangles ).We can assume that no suitable witness z is already in N , else the move is trivialas ∃ does not need to add any extra nodes to the network. If ∀ has no non-trivialmove to make then ∃ wins, as the network now reveals the required completerepresentation. This is how N ′ is obtained from N at rounds indexed by successorordinals. At a round indexed by a limit ordinal the network is simply the limitof all the previous networks, which is well defined since the networks are nested.The game starts with ∀ playing a non-identity atom α , and ∃ creating a two nodenetwork { x , y } such that the edge ( x , y ) is labeled by α . We say ∀ wins if insome round before min( ω, |A| ) he makes a move such that ∃ cannot extend the etwork consistently, and we say ∃ wins if she survives min( ω, |A| ) rounds, or if ∀ cannot make a non-trivial move at some point. We say ∃ has a winning strategy ifshe can play so that her victory is guaranteed. The key result, as proved in [HH02,theorem 11.7] in the countable case, and discussed in [HH02, exercise 11.4.3] forhigher cardinalities, is that A is completely representable if and only if ∃ has awinning strategy in the complete representation game over A .We return now to the relation algebra B G,H defined above. Given an atomicnetwork N for B G,H and nodes x, y ∈ nodes ( N ) let R N ( x, y ) = { z ∈ nodes ( N ) : N ( x, z ) is green and N ( y, z ) = y } . Observe that R N ( x, y ) depends only on the green and yellow edge labels of N . Aset of nodes of a network where every edge between distinct nodes has a red labelis called a red clique. For any x, y , by forbidden triples (II) and (III), R N ( x, y ) is ared clique. In a red clique C of size at least two, by forbidden triple (IV), each node z ∈ C has a well-defined index ρ C ( z ) ∈ H such that N ( z , z ) = r ρ C ( z ) ,ρ C ( z ) , for z = z ∈ C . So ρ C is defined on z ∈ R N ( x, y ) by taking the first subscript in thelabel of ( z, z ′ ) where z ′ ∈ R N ( x, y ) \ { z } is arbitrary. By rule (IV), this subscriptdoes not depend on choice of z ′ . By consistency of N and (VI), for each i ∈ G there can be at most one node z such that z ∈ R N ( x, y ) and N ( x, z ) = g i .Similarly, if θ is a complete representation of B G,H over base X , then for x, y ∈ X ,let R θ ( x, y ) = { z ∈ X : ( x, z ) ∈ [ i ∈ G g θi ∧ ( z, y ) ∈ y θ } . As with networks, if | R θ ( x, y ) | >
1, then each point z ∈ R θ ( x, y ) has an index ρ ( θ,x,y ) ( z ) ∈ H . This is defined by noticing that if z = z ∈ R θ ( x, y ), then( z , z ) ∈ g θi ; g θi ′ ∩ y θ ; y θ for some i, i ′ ∈ G . As θ is a complete representation,there is an atom α with ( z , z ) ∈ α θ (see [HH02, theorem 2.21]) and from theforbidden triple rules we see that α must be r j,j ′ for some j, j ′ ∈ H . We define ρ ( θ,x,y ) ( z ) to be j , which does not depend on the choice of z ∈ R θ ( x, y ) \ { z } . THEOREM 2.1.
Let
G, H be binary structures. The following are equivalent.(1) For all i = i ′ ∈ G there are j, j ′ ∈ H such that { ( i, i ′ ) , ( j, j ′ ) } is a partialhomomorphism, and every partial homomorphism { ( i, j ) , ( i ′ , j ′ ) } where i = i ′ from asubstructure of G into H extends to a homomorphism G → H .(2) B G,H is completely representable.Proof.
Suppose B G,H is completely representable, say θ is a complete representa-tion. Since θ is complete, for every pair of points ( x, y ) ∈ θ in the base of therepresentation there is a unique atom α such that ( x, y ) ∈ α θ (by [HH02, theorem2.21]).Let i = i ∈ G . Find points x, y in the base of the representation such that( x, y ) ∈ w θ , see the first part of figure 1. Since ( g i t , y , w ) is not forbidden (for t = 1 , z , z such that ( x, z t ) ∈ g θi t and ( z t , y ) ∈ y θ , for t = 1 , z , z ) cannot be the identity by forbidden triple(I), nor green, white, yellow or black, by forbidden triples (II), (III), hence it mustbe red, say r j ,j . We also have ( z , z ) ∈ g θi ; g θi , and so, by forbidden triple (V),the map { ( i , j ) , ( i , j ) } is a partial homomorphism of size two.To show that partial homomorphisms of size two extend, let { ( i , j ) , ( i , j ) } bea partial homomorphism from G to H , where i = i . Let z i , z i be distinct points y ❄❄❄❄ ❄❄❄❄❄❄❄❄ r j ,j / / ❴❴❴❴ z y x wg i g i ⑧⑧⑧⑧ ⑧⑧⑧⑧⑧⑧⑧ y z i y ❄❄❄ ❄❄❄❄❄❄❄❄ r j ,j / / r j ,j " " t ♠ ❢ ❴ ❳ ◗ ❏ z i y r j ,j / / ❴❴❴❴ z i x α ❴❴❴❴❴ g i g i ⑧⑧⑧⑧ ⑧⑧⑧⑧⑧⑧⑧ g i ♦♦♦♦♦ ♦♦♦♦♦♦♦♦♦♦♦♦♦♦♦♦ y y ⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧ Figure 1.
From the representation, a partial homomorphism { ( i , j ) , ( i , j ) } exists and extends to i .in the base of the representation such that ( z i , z i ) ∈ r θj ,j (see the second partof figure 1). Since { ( i , j ) , ( i , j ) } is a partial homomorphism it follows from rule(V) that ( g i , g i , r j ,j ) is not forbidden, so there is a point x where ( x, z i ) ∈ g θi and ( x, z i ) ∈ g θi . Also, ( y , y , r j ,j ) is not forbidden, so there is a point y where( y, z i ) , ( y, z i ) ∈ y θ , and clearly x = y . Since the representation is complete, theremust be an atom α such that ( x, y ) ∈ α θ (by [HH02, theorem 2.21]), and as x = y this atom cannot be 1 ′ (by rule (I)).We have shown that z i = z i ∈ R θ ( x, y ). Write ρ for ρ ( θ,x,y ) , so for w ∈ R θ ( x, y ), ρ ( w ) denotes the index of w in H , and for w, w ′ ∈ R θ ( x, y ) we have( w, w ′ ) ∈ r θρ ( w ) ,ρ ( w ′ ) . Since ( z i , z i ) ∈ r θj ,j , we have ρ ( z i ) = j , ρ ( z i ) = j .Regardless of which non-identity atom α is, for each node i of G \ { i , i } , thetriple ( g i , y , α ) is not forbidden, from which it follows that ( x, y ) ∈ g i ; y , and so theremust be a point z ∈ R θ ( x, y ) where ( x, z ) ∈ g θi and ( z, y ) ∈ y θ . This point is unique,as if z ′ is a point with the same properties, then we have ( z, z ′ ) ∈ ( g θi ; g θi ) ∩ ( y θ ; y θ ).As θ is complete, ( z, z ′ ) is contained in the interpretation of some atom, and theforbidden triple rules imply that this atom must be the identity. Thus z = z ′ , bydefinition of the identity in proper relation algebras. The map from G to H thatsends i ∈ G to ρ ( z i ) is therefore well defined, is a homomorphism, by (V), andextends { ( i , j ) , ( i , j ) } as required.Now we check the converse. Assume the first condition in the theorem. Asdiscussed above, it is sufficient to show she has a winning strategy in the completerepresentation game for B G,H . The basic idea behind ∃ ’s strategy is that she will, asfar as possible, use labels which obviously do not interfere with the integrity of thenetwork, and never any labels that are not either white, black or red. The difficultcases turn out to be when she is forced to use a red atom. This occurs only when z is in R ′ N ( x, y ) (or R N ′ ( y, x )), and | R N ′ ( x, y ) | > | R N ′ ( y, x ) | > N to N ′ so that a new red cliqueof form R N ′ is created with | R N ′ ( x, y ) | >
1. The first is if | R N ( x, y ) | = 1 and ∀ plays ( x, y, g i ′ , y ) for some appropriate i ′ ∈ G . In this case, R N ′ ( x, y ) = { w, z } for some w ∈ N , with N ( x, w ) = g i for some i = i ′ ∈ G , and by the first partof our assumption there are j, j ′ ∈ H such that h = { ( i, j ) , ( i ′ , j ′ ) } is a partialhomomorphism. Here ∃ sets N ′ ( w, z ) = r j,j ′ . The second way such a red cliqueof size greater than one can be created is where w ∈ N , z is the new node, and ∀ ’s move is ( x, y, y , y ), with N ( w, x ) = g i and N ( w, y ) = g i ′ for some i = i ′ ∈ G . ince ∃ does not use green or yellow labels, R N ′ ( w, z ) = { x, y } . Here the map h from green subscripts to red indices is a partial homomorphism of size two, byconsistency of the previous network and of ∀ ’s move. The third way is similar tothe second, except ∀ plays ( x, y, g i , g i ′ ), and N ( w, x ) = N ( w, y ) = y . Here a partialhomomorphism h is defined as in the second case.In all three cases, the second part of our assumption tells us that h extends toa homomorphism h + : G → H . In later rounds, whenever a new node is addedto R N ( x, y ), ∃ will use h + to get the index of the new node and hence to labelall edges in R N ( x, y ) incident with the new node z . The fact that red labels aredefined by node indices will ensure that (IV) is not violated and the fact that h + isa homomorphism will ensure that (V) is not violated. Although in general it turnsout that two distinct red cliques can intersect in up to two points, crucially the newpoint z belongs to at most a single clique in the round when it is added, so, ∃ isnever conflicted about which homomorphism to use, as we shall see.Now we have sketched out ∃ ’s plan, we must check that it works. Since the initialround and rounds indexed by limit ordinals present no difficulties, we considerrounds indexed by successor ordinals. Suppose the current network is N with | N | ≥
2, and ∀ ’s move is ( x, y, α, β ), with ∃ responding by adding a new node z tocreate N ′ = N ∪ { z } . We have two induction hypotheses:(H1) For all x ′ , y ′ ∈ N where | R N ( x ′ , y ′ ) | ≥ h : G → H such that for all z = z ′ ∈ R N ( x ′ , y ′ ) we have( N ( x ′ , z ) = g i ∧ N ( x ′ , z ) = g i ′ ) → N ( z, z ′ ) = r h ( i ) ,h ( i ′ ) . (H2) For any x = y ∈ N , if α and β are green or yellow then there is at most asingle node z ∈ N such that N ( x, z ) = α, N ( z, y ) = β .Given N , for each x ′ , y ′ ∈ N where | R N ( x ′ , y ′ ) | ≥ h x ′ y ′ be a homomorphismsatisfying ((H1)). For each w ∈ N \ { x, y } she must assign N ′ ( w, z ) in such a waythat N ′ is a network, and the induction hypothesis is maintained. She proceeds asfollows:(a) If N ( w, x ) and α are not both green, and N ( w, y ) , β are not both green, shelets N ′ ( w, z ) = w .(b) If N ( w, x ) , α are both green but N ( w, y ) , β are not both yellow, or if N ( w, y ) , β are both green but N ( w, x ) , α are not both yellow, she lets N ′ ( w, z ) = b .(c) The remaining case is where N ( x, w ) = g i , α = g i ′ , N ( w, y ) = β = y (or similarwith x, y swapped). Note that i = i ′ , by the ‘no trivial moves’ assumption.Here z, w ∈ R N ( x, y ), and she is forced to choose N ′ ( w, z ) = r j,j ′ for some j, j ′ ∈ H . If h xy is already defined for N she lets N ′ ( w, z ) = r h xy ( i ) ,h xy ( i ′ ) ,thereby maintaining (H1) for ( x, y ). Otherwise, R N ( x, y ) = { w } and she maypick any j, j ′ ∈ H such that { ( i, j ) , ( i ′ , j ′ ) } is a partial homomorphism andextend it to a homomorphism h xy : G → H (using both parts of (1)) and againlet R N ′ ( w, z ) = r h xy ( i ) ,h xy ( i ′ ) , establishing (H1) for ( x, y ) in N ′ . Note that if z ∈ R N ′ ( x ′ , y ′ ), then, as ∃ never uses green or yellow labels, it’s easy to showthat x = x ′ and y = y ′ , so the above strategy is well defined.First we show that N ′ is a consistent network by checking that the labeling ofeach triangle ( w, w ∗ , z ) for w = w ∗ ∈ N is not forbidden. If { w, w ∗ } = { x, y } thenthe triangle is consistent (else ∀ ’s move would be illegal), so without loss of gener-ality we assume that w / ∈ { x, y } . Observe that N ′ ( w, z ) must be either white, blackor red, as, if N ′ ( w, z ) is green or yellow it follows that w ∈ { x, y } , contradicting our ssumption. If N ′ ( w, z ) = w , then the only possibility that the triangle ( w, w ∗ , z )could be forbidden comes from (II), but this requires that N ( w ∗ , z ) and N ( w, w ∗ )be green, and thus that w ∗ ∈ { x, y } . But then the conditions of (a) would not havebeen met, so N ′ ( w, z ) could not be w after all. Similarly, if N ′ ( w, z ) = b then thepossibility of violating (III) is ruled out by case (b) conditions.In the remaining case, N ′ ( w, z ) is red, and the only forbidden triples involvingred atoms are (IV), (V) and (VI). We assume N ( x, w ) = g i , α = g i ′ , and thecase where x, y are swapped follows by symmetry. A triangle ( w, w ∗ , z ) could onlyviolate forbidden triple (IV) if all three edges were red, which only happens when w, w ∗ , z ∈ R N ′ ( x, y ). In this case, by (H1) we have N ′ ( w, w ∗ ) = r h xy ( i ) ,h xy ( i ∗ ) forsome i ∗ ∈ G and some h xy , and according to ∃ ’s strategy the other edge labels are N ′ ( w, z ) = r h xy ( i ) ,h xy ( i ′ ) and N ′ ( z, w ∗ ) = r h xy ( i ′ ) ,h xy ( i ∗ ) , so (IV) is not violated. Forforbidden triple (V), the only possible green-green-red triangle incident with z and w is ( x, w, z ) and the edge labels ( g i , g i ′ , r h xy ( i ) ,h xy ( i ′ ) ) do not violate (V), since h xy is a homomorphism. The only triangle containing { w, z } which could violate (VI)is ( w, x, z ) (only this can be green-green-red), but in this case i = i ′ (else w is awitness to the current move, contrary to the ‘no trivial moves’ assumption) so (VI)is not violated. Hence N ′ is a consistent network.It remains to check the induction hypotheses. (H2) is clear, since ∃ never addsa new node to the network if a suitable witness is already in N . We check (H1).Suppose ∃ is playing according to the strategy we have described, and she adds z to N to obtain N ′ in response to a move ( x, y, α, β ) by ∀ . We say that a pair( x ′ , y ′ ) is safe if either x ′ , y ′ ∈ N and R N ′ ( x ′ , y ′ ) = R N ( x ′ , y ′ ) , or | R N ′ ( x ′ , y ′ ) | ≤ x ′ , y ′ ) is safe then (H1) is true for ( x ′ , y ′ ) in N ′ either trivially because the sizeof R N ′ ( x ′ , y ′ ) is less than two, or inductively, since (H1) is assumed true for N .We check induction hypothesis (H1) according to whether α and β are green,yellow or neither.(i) If α = g i , β = y then R N ′ ( x, y ) = R N ( x, y ) ∪ { z } and all other red cliques aresafe.(ii) If α = y , β = g i then R N ′ ( y, x ) = R N ( y, x ) ∪ { z } and all other cliques aresafe.(iii) If α = g i , β = g i ′ and N ( x, w ) = N ( w, y ) = y for some w ∈ N , then R N ′ ( z, w ) = { x, y } , and, by (H2), all other red cliques are safe.(iv) If α = β = y , N ( x, w ) = g i , N ( y, w ) = g i ′ for some w ∈ N , then R N ′ ( w, z ) = { x, y } , and, (H2), all other red cliques are safe.(v) Else { α, β } 6⊆ { g i , g i ′ , y } (for any i, i ′ ∈ G ) and all red cliques are safe.For case (i), if R N ′ ( x, y ) = { w, z } (some w ∈ N ) then this red clique has size twoin this round, for the first time. Say N ( x, w ) = g i ′ , where i ′ = i else w is alreadya witness. By assumption (1) there is a partial homomorphism defined on { i, i ′ } which extends to a homomorphism, and her strategy chooses such a homomorph-ism h xy and lets N ′ ( w, z ) = r h xy ( i ′ ) ,h xy ( i ) , as required for the induction hypothesis.If | R N ′ ( x, y ) | ≥ | R N ( x, y ) | ≥
2, so inductively there is already a homo-morphism h xy : G → H determining red labels in R N ( x, y ). In this case, for each w ∈ R N ′ ( x, y ), her strategy defines N ′ ( w, z ) = r h xy ( i ′ ) ,h xy ( i ) , where N ( x, w ) = g i ′ ,thereby maintaining (H1) for ( x, y ) in N ′ . Case (ii) is similar.In case (iii) when R N ′ ( z, w ) = { x, y } , since it follows from the stated conditionsthat ( g i ; g i ′ ) · ( y ; y ) ≥ N ( x, y ), we know that N ( x, y ) = r j,j ′ for some j, j ′ ∈ H , and,by (VI), we must have i = i ′ . By (V) and the legality of ∀ ’s move, { ( i, j ) , ( i ′ , j ′ ) } s a partial homomorphism. By assumption (1) this extends to a homomorphism h zw : G → H , as required by the induction hypothesis. Case (iv) is similar, andcase (v) is trivial. So the strategy described above is indeed a winning one for ∃ . (cid:3) The class of atom structures of representable relation algebras is known to beelementary [Ven97]. We can apply the result above to say something about classesof atom structures intermediate between this class and that of the atom structuresof completely representable relation algebras.
COROLLARY 2.2. If K is a class of relation algebra atom structures includingall atom structures of completely representable relation algebras, and contained inthe class of atom structures of representable atomic relation algebras, then K cannotbe defined by any theory in the language of RA atom structures using only finitelymany atom valued variables.Proof. A digraph may be considered as a binary structure with a single predicatedenoting edges. Given m ≥ K m be the complete irreflexive digraph with m vertices, and consider the algebras B K m ,K m and B K m +1 ,K m . By theorem 2.1,the former is completely representable while the latter is not, and, since they arefinite, representability is the same as complete representability. Thus B K m ,K m ∈ K and B K m +1 ,K m / ∈ K . However, ∃ has a winning strategy in the m -pebble, ω -roundEhrenfeucht Fra¨ıss´e-game over the atom structures of B K m ,K m and B K m +1 ,K m whichwe describe now (see e.g. [Imm99, chapter 6] for a full description of the game).Whenever ∀ picks a non-green atom, ∃ picks the corresponding non-green atom inthe other atom structure. If ∀ places a pebble where another pebble is alreadyplaced, then ∃ covers the corresponding pebble in the other algebra. If ∀ picks agreen atom, not already in play, then ∃ picks any green atom in the other atomstructure not currently selected. There are always enough green atoms for this.This strategy is a winning one because the underlying graphs are complete, so atriple ( g i , g i ′ , r j,j ′ ) will be forbidden iff i = i ′ or j = j ′ , in either atom structure.It follows that the two atom structures agree on all m -variable formulas (see e.g.[Imm99, theorem 6.10]). (cid:3) The corollary shows that the atom structures of B K m ,K m and B K m +1 ,K m cannotbe distinguished in the language of atom structures restricted to m atomic variables.If we use formulas with variables that range over arbitrary elements of a relationalgebra, much more can be expressed. Consider, for example, the formula φ k ( x )with variables x, y of which only x appears free, which we will define shortly. It isintended to express that x is above at least k atoms, in an atomic relation algebra.So φ ( x ) is ¬ ( x = 0). Recursively, suppose φ k ( x ) and φ k ( y ) have been defined (thevariables x and y are swapped throughout in the latter formula), and suppose theformula holds exactly when the free variable denotes an element above at least k atoms. Let φ k +1 ( x ) be the formula ∃ y ( y < x ∧ φ k ( y )). If k is the number of atomsin B K m +1 ,K m then ∃ xφ k ( x ) is true in B K m +1 ,K m but not in B K m ,K m . As well asdefining the finite cardinality of an algebra, two variable formulas can express manyother properties. Indeed, it is conceivable that any pair of non-isomorphic finiterelation algebras can be distinguished by a two variable formula; this remains anopen problem. The proof of corollary 2.2 is a kind of warm up for the proof of heorem 1.1, which involves formulas with variables ranging over arbitrary relationalgebra elements, and occupies most of the next section.3. A vertex colouring game
We now define a vertex colouring game played by ∀ and ∃ over a pair of binarystructures using a finite number of colours, which are used to colour sets of vertices,rather than the individual vertices used in the pebble game used in the proof ofcorollary 2.2. This game is studied in more detail for digraphs in [EH], where it iscalled a Seurat game in reference to the pointillist style of painting. Let c < ω, n ≤ ω . Given two binary structures G, H we define the c -colour, n -round colouring game G cn ( G, H ) to test equivalence of the binary structures using c monadic predicates, n times. A G -interpretation is a map { , . . . , c − } → ℘ ( G ) to subsets of the verticesof G , and an H -interpretation is a mapping { , . . . , c − } → ℘ ( H ). Intuitively,these maps associate vertices in G and H with the different colours. A position in the game consists of a G -interpretation and an H -interpretation. A play of thegame is a sequence of 1 + n positions ( g , h ) , . . . , ( g i , h i ) , . . . ( i ≤ n ). Round 0begins with the starting position ( g , h ), where g ( t ) = h ( t ) = ∅ , for all t < c (i.e.initially all vertices are uncoloured). If n = 0 then neither player does anything,and the result of the game is determined completely by the starting position (whichfor us is always the same, though this need not be required). For n >
0, at the startof round r < n , if the current position is ( g r , h r ), then ∀ chooses t < c and a subsetof the vertices of G or a subset of the vertices of H , ∃ responds with a subset ofthe nodes of the other binary structure. The intuition here is that ∀ is colouringsome set of vertices in one of the structures, and ∃ is responding by colouring a setof vertices of the other structure with the same colour. If ∀ reuses a colour that hasalready been used, then its previous use is first erased from both binary structures.To reflect the new situation, the position is updated to ( g r +1 , h r +1 ) from ( g r , h r )by changing g r +1 ( t ) ⊆ G and h r +1 ( t ) ⊆ H according to these choices. If n is finite,the final position is ( g n , h n ).A palette π is a subset of { , . . . , c − } . Given a G -interpretation g , we mayinterpret π by π g = { x ∈ G : ∀ t ∈ π ( x ∈ g ( t )) and ∀ t ∈ { , . . . , c − } \ π ( x / ∈ g ( t )) } . Intuitively, π g tells us which vertices of G are coloured according to g with exactlythe combination of colours defined by π . Observe that the set of vertices of G isthe disjoint union of the sets π g , as π ranges over palettes. A position ( g, h ) is awin for ∀ if either(C1) there is a palette π where π g is empty but π h is not or the other way round,or(C2) there are palettes π, π ′ and a binary predicate b such ( π g × ( π ′ ) g ) ∩ b G isempty but ( π h × ( π ′ ) h ) ∩ b H is not, or the other way round.We say that ∀ wins in round k if ( g k , h k ) is the first winning position for him.For n < ω , if ∀ does not win in any round i ≤ n , then ∃ is the winner. If n = ω ,then ∃ wins if ( g k , h k ) is not a win for ∀ for all k < ω .In addition to the game described above, we will use the following minor vari-ation of the classic Ehrenfeucht-Fra¨ıss´e game used in the proof of corollary 2.2.Given two relation algebras A , B we define the c -pebble, n -round equivalence gameΓ cn ( A , α , B , β ). The pair ( α , β ) defines the starting position of the game, and onsists of two partial maps α : { , . . . , c − } → A and β : { , . . . , c − } → B .We require that α and β have the same domains (i.e. that they are defined forthe same elements). If the initial position is defined by maps with empty domains,then we may refer to the game just as Γ cn ( A , B ) for brevity.If n = 0, then the game is entirely determined by the starting position, andneither player does anything. For n ≥
1, in each round k < n , if the positionis ( α k , β k ), ∀ picks t < c and an element of A or of B , and then ∃ picks anelement of the other algebra. At the end of the round, the position is updated bychanging α k ( t ) ∈ A , β k ( t ) ∈ B according to these choices, but leaving other valuesunchanged. This defines α k +1 and β k +1 ready for the start of the next round.Let A α and B β denote the subalgebras of A and B generated by the images of α and β , respectively. At the start of round k , consider the binary relation α ⌣k ◦ β k = { ( α k ( t ) , β k ( t )) : t ∈ dom ( α k ) } (here ◦ and ⌣ denote, respectively, ordinarycomposition and conversion of relations). If α ⌣k ◦ β k is a partial function then itinduces a homomorphism h α ⌣k ◦ β k i from A α k to B β k , else if α ⌣k ◦ β k is not a functionthen let h α ⌣k ◦ β k i be an arbitrary non-isomorphism. We say ∀ wins the game inround k if h α ⌣k ◦ β k i is not an isomorphism, and the maps have been isomorphismsin all previous rounds. On the other hand, ∃ wins if the maps we have describedare isomorphisms for all k ≤ n . The value of these modified Ehrenfeucht-Fra¨ıss´egames is given by the following definition and lemma. DEFINITION 3.1.
Let A and B be relation algebras, and let α and β be partialmaps from { , . . . , c − } to A and B , respectively, and suppose also that α and β have the same domains. We say ( A , α ) ≡ cn ( B , β ) if whenever φ is a first-order formula in the language of relation algebras with theadditional restrictions that the quantifier depth in φ be at most n , that φ involvesonly variables from the set { x , . . . , x n − } , and that the free variables of φ are allindexed by values from the domain of α and β , we have A , α | = φ ⇐⇒ B , β | = φ. Here, for example, A , α | = φ means that A | = φ if all variables x i occurring free in φ are assigned to α ( i ) in A . LEMMA 3.2.
For ≤ n ≤ ω , if ∃ has a winning strategy in Γ cn ( A , α , B , β ) thenwe must have ( A , α ) ≡ cn ( B , β ) .Proof. This is half the well known result for relational signatures (see e.g. [Imm99,theorem 6.10]). Having functions in the signature blocks the proof of the converse.We induct on n . For the base case, ∃ has a winning strategy in Γ c ( A , α , B , β ) if andonly if the induced map h α ⌣ ◦ β i is an isomorphism, if and only if ( A , α ) , ( B , β )agree on all equations using appropriate variables, if and only if ( A , α ) ≡ c ( B , β ).For the inductive step, suppose ∃ has a winning strategy in Γ cn +1 ( A , α , B , β ).Let φ = ∃ x i ψ be a formula of quantifier depth at most n + 1, where the variablesoccurring free in ψ are either indexed by values for which α and β are defined, orare x i . If A , α | = ∃ x i ψ then there is an x i -variant α of α such that A , α | = ψ .If ∀ plays α ( x i ) in the game, then since ∃ has a winning strategy there is an x i -variant β of β where ∃ has a winning strategy in Γ cn ( A , α , B , β ). Inductively, B , β | = ψ , hence B , β | = ∃ x i ψ . Since the argument is symmetric, it follows that( A , α ) agrees with ( B , β ) on all c -variable formulas ∃ x i ψ where ψ has quantifier epth at most n , hence they agree on all c -variable formulas of quantifier depth atmost n + 1. By induction, the lemma holds for all finite n . For the case n = ω thena winning strategy for ∃ in Γ cω ( A , α , B , β ) entails a winning strategy in all finitelength games, so ( A , α ) ≡ cn ( B , β ) for all finite n . Hence ( A , α ) ≡ cω ( B , β ), asrequired. (cid:3) Now we return to the Seurat game G . This has a connection to the game Γwhich we will exploit in the proof of lemma 3.4 to come. First, given m >
0, let K m be the complete irreflexive directed graph with m vertices. LEMMA 3.3.
Let ≤ c, n < ω , and let m, m ′ ≥ n . Then ∃ has a winningstrategy in G cn − ( K m , K m ′ ) .Proof. Note that, since the graphs are complete, ∀ wins in a round if and only if atthe start of that round there is a palette π ⊆ { , . . . , c − } where | π g | 6 = | π h | and either | π g | < | π h | < . ( ∗ )Suppose ∃ plays according to the following principle. She ensures that:For all palettes π , if either | π g r | < n − r or | π h r | < n − r then | π g r | = | π h r | . ( † r )If ∃ can maintain ( † r ) while r ≤ n − n −
1, when ( g n − , h n − ) is checked and the games ends. Note that if thegame were to continue for another round then she might lose, because with r = n she is not insured against violating ( ∗ ) above.We now prove by induction that ∃ can indeed always play so as to ensure theabove condition holds up to and including r = n −
1. The condition is obviouslysatisfied at the beginning of the game, because all vertices of both binary structuresare uncoloured, so the empty palette is interpreted as the whole structure, and allother palettes are interpreted as the empty set. Suppose now that the condition issatisfied going into round r , for some r < n −
1. That is, if ( g r , h r ) was the stateof play at the start of round r , we can assume that ∃ has played so that ( † r ) issatisfied.Suppose that ∀ picks t < c and, without loss of generality, a subset X of G (the case where he chooses a subset of H is similar). Let ( g ′ r , h ′ r ) be the positionidentical to ( g r , h r ) except g ′ r ( t ) = h ′ r ( t ) = ∅ (so we essentially ‘delete’ colour t from both graphs). Observe that ( † r ) remains true for ( g ′ r , h ′ r ). We also have X = S π ⊆{ ,...,c − } ( X ∩ π g ′ r ).For ∃ ’s response Y ⊆ H to ∀ ’s move ( t, X ) she will pick disjoint subsets Y π ⊆ π h ′ r for each palette π , and then she will define Y = S π ⊆{ ,...,c − } Y π . How she chooses Y π is explained below. Let π be any palette not containing t , as if t ∈ π we have π h ′ r = ∅ so her choice is trivial. She defines Y π as follows: • If | π g ′ r ∩ X | < n − ( r +1) and | π g ′ r \ X | < n − ( r +1) , then | π g ′ r | < n − r , soinductively | π g ′ r | = | π h ′ r | . Here she lets Y π be any subset of π h ′ r of size | π g ′ r ∩ X | , and it follows immediately that π h ′ r \ Y π has the same size as π g ′ r \ X . • If | π g ′ r ∩ X | < n − ( r +1) but | π g ′ r \ X | ≥ n − ( r +1) then she lets Y π be anysubset of π h ′ r of the same size as π g ′ r ∩ X . It follows that π h ′ r \ Y π will be ofsize at least 2 n − ( r +1) . The case where | π g ′ r \ X | < n − ( r +1) but | π g ′ r ∩ X | ≥ n − ( r +1) is similar. Here she chooses Y π so that | π h ′ r \ Y π | = | π g ′ r \ X | . Finally, if both π g ′ r ∩ X and π g ′ r \ X have size at least 2 n − ( r +1) , then | π g ′ r | ≥ n − r , so inductively | π h ′ r | ≥ n − r . She lets Y π ⊆ π h ′ r be any subsetof size 2 n − ( r +1) , so π h ′ r \ Y π must have size at least 2 n − ( r +1) .Then ∃ lets h r +1 be the same as h r except h r ( t ) = Y = S π ⊆{ ,...,c − } Y π . Now,given a palette π we have π g r +1 = ( π g ′ r \ X if t / ∈ ππ g ′ r ∩ X if t ∈ ππ h r +1 = ( π h ′ r \ Y = π h ′ r \ Y π if t / ∈ ππ h ′ r ∩ Y = Y π if t ∈ π By the definition by cases given above, the cardinalities of these sets agree whennecessary. Thus ( † r +1 ) is established for the new position ( g r +1 , h r +1 ). (cid:3) Let
F, G, H be finite digraphs. We are interested in the games Γ cn ( B G,F , B H,F ).Note that B G,F and B H,F are both finite (and thus atomic), and differ only withrespect to their sets of green atoms (we identify the non-green atoms between eachalgebra in the obvious way). Given an element x ∈ B G,F , we define the set ofvertices of G indexing green atoms under x to be G x . Similarly, given y ∈ B H,F ,we define the set of vertices of H indexing green atoms under y to be H y . LEMMA 3.4.
Let ≤ c, n < ω and let F be any digraph. If m, m ′ ≥ n then B K m ,F ≡ cn − B K m ′ ,F .Proof. By the previous lemma, we know that ∃ has a winning strategy in the game G cn − ( K m , K m ′ ). To prove the equivalence it suffices, by lemma 3.2, to prove that ∃ has a winning strategy in Γ cn − ( B K m ,F , B K m ′ ,F ). Her strategy is to simulatea corresponding play of G cn − ( K m , K m ′ ) in which she uses her winning strategy,and to maintain a correspondence between the plays of the games. So, if ∀ selects x ∈ B K m ,F , then she selects y ∈ B K m ′ ,F whose non-green part is identical to thatof x and whose green part is defined by her response to the ∀ -move G x in the playof G cn − ( K m , K m ′ ), and similar when he picks y ∈ B K m ′ ,F .Let ( α, β ) be a position in a play of Γ cn − ( B K m ,F , B K m ′ ,F ) in which ∃ uses thisstrategy, and let ( g, h ) be the corresponding position in a play of G cn − ( K m , K m ′ ).Using our assumption that ( g, h ) is not a win for ∀ we have( † ) either | π g | = | π h | or | π g | , | π h | ≥
2, for all palettes π, and we have to prove that ∃ ’s move does not result in a losing position in the gameΓ cn − ( B K m ,F , B K m ′ ,F ), by showing that there is a relation algebra isomorphismfrom ( B K m ,F ) α to ( B K m ′ ,F ) β (recall that these are the subalgebras generated bythe ranges of α and β respectively).Consider the boolean subalgebra B of B K m ,F generated (using boolean operators)by the range of α and all non-green atoms. For any palette π , let π B = P t ∈ π g g t ∈B , the corresponding green element of B . If π g = ∅ then the sum is empty and π B = 0, else the sum is non-empty and π B is a an atom of B . A little thoughtreveals that all green atoms of B arise in this way. Similarly, let B ′ be the booleansub-algebra of B K m ′ ,F generated by the non-green atoms and rng ( β ), and let π B ′ = P t ∈ π h g t . Since π g = ∅ ⇐⇒ π h = ∅ , for all palettes (by ( † )), the map φ = { ( π B , π B ′ ) : π ⊆ { , . . . , c − }} \ { (0 , } s a bijection from the green atoms of B to those of B ′ which extends to a uniqueboolean isomorphism ˆ φ : B → B ′ fixing non-green atoms. B contains the identity and is closed under conversion, since all green elementsare self-converse. In order to show that it is also closed under composition we haveto check that whenever x, y, z are boolean atoms in B , we have either ( x ; y ) · z = 0or x ; y ≥ z . We indicate the former case by writing × and the latter case by writing X .The cases where 1 ′ ∈ { x, y, z } are easy, so suppose x, y, z are non-identity atomsof B (so each has a colour). Recall that the only sets of three colours where some butnot all triples of atoms of those colours are forbidden, are green-green-red and red-red-red. In all other cases either all triples of atoms of those colours are forbiddenand we have × or none is forbidden and we have X . If z is red we get X or × since z is an atom of B K m ,F . So, without loss of generality suppose that x = π B and z = π B are green and y = r j ′ ,j is red (for some j, j ′ ∈ F ). So we are interested in( π B ; r j ′ ,j ) · π B . We want to show that either for every green atom g i (of B K m ,F )below π B there is a green atom g i ′ below π B such that ( g i ′ , r j ′ ,j , g i ) is not forbidden(for X ), or that for every green atom g i (of B K m ,F ) below π B and for every greenatom g i ′ below π B , the triple ( g i ′ , r j ′ ,j , g i ) is forbidden (for × ).Applying the Peircean equivalences, the triple under consideration here is equi-valent to ( g i , g i ′ , r j,j ′ ). In this case, if either ( j, j ′ ) or ( j ′ , j ) is not an edge of F , thenwe have × since K m is complete, which means the triple will be forbidden when i = i ′ , and the triple is always forbidden when i = i ′ anyway. Suppose then thatboth ( j, j ′ ) and ( j ′ , j ) are edges. Here we also have × if x = z and | π g | = | π g | = 1,since ( g i , g i , r j,j ′ ) is always forbidden. However, if x = z and | π g | ≥
2, or if x = z ,we get X , since K m is complete.It follows that B is closed under all relation algebra operations, and is a sub-relation algebra of B K m ,F . Since rng ( α ) ⊆ B we have the inclusion of relationalgebras, ( B K m ,F ) α ⊆ B ⊆ B K m ,F . Similarly, the boolean subalgebra B ′ of B K m ′ ,F generated by non-green atoms and { β ( t ) : t ∈ dom ( β ) } is a sub-relation algebra of B K m ′ ,F extending ( B K m ′ ,F ) β .To show that the map ˆ φ : B → B ′ is a relation algebra isomorphism, it is sufficientto show that for all atoms x, y, z of B , we have( x ; y ) · z = 0 ⇐⇒ ( φ ( x ) ; φ ( y )) · φ ( z ) = 0 . In other words, that ( x, y, z ) corresponding to X or × means ( φ ( x ) , φ ( y ) , φ ( z ))corresponds to X or × appropriately. By what we have just proved, the divisionof triples of diversity atoms ( x, y, z ) in B into X and × depends on the coloursof x, y, z , whether x = z in the case where both x and z are green, and, if x = z = π B is green, on whether | π g | is at least two or not. In all cases, this will bematched by ( φ ( x ) , φ ( y ) , φ ( z )), as ∃ is using a winning strategy in the parallel game G cn − ( K m , K m ′ ), and thus ˆ φ is an isomorphism as required.Moreover, for all t in the domain of α we have ˆ φ ( α ( t )) = β ( t ), as ∃ ’s strategyin Γ cn − ( B K m ,F , B K m ′ ,F ) ensures this is true. It follows that the restriction of ˆ φ to ( B K m ,F ) α , which, as we have just proved, is an isomorphism onto ( B K m ′ ,F ) β , is h α ⌣ ◦ β i . Thus ∃ survives the game Γ cn − ( B K m ,F , B K m ′ ,F ) for another round, asrequired. (cid:3) ow let A = B K n +1 ,K n +1 and B = B K n +1 ,K n +1 . Then A ∈
RRA but
B 6∈
RRA , by theorem 2.1, and
A ≡ cn B by lemma 3.4. It follows that RRA cannotbe axiomatised by any theory consisting of c -variable formulas of quantifier depthat most n . This proves theorem 1.1, and so RRA cannot be defined by formulas ofbounded quantifier depth as claimed. REMARK 3.5.
This suggests that a similar construction could be used to provethat the class of representable cylindric algebras of dimension n cannot be defined bya theory of bounded quantifier depth, however we have not succeeded in demonstrat-ing this. There is a way of constructing a rainbow cylindric algebra of dimension n ≥ from two graphs G, H given in [HH97, § . The atoms of this cylindricalgebra are labelled hypergraphs on n nodes. The two-dimensional edges of thesehypergraphs have green, red and other labels generalising the green, red, yellow andblack atoms of the rainbow relation algebra A G,H , but the white atoms w S are re-placed by ( n − -ary hyperlabels in the cylindric version. Our modified rainbowconstruction, in which the atom w S are all deleted, can be used to make a slightlyeasier cylindric algebra C nG,H from two graphs whose atoms are labelled graphs on n -nodes, and no hyperlabels are needed. It follows that C nG,H is generated by its relation algebra reduct . By considering the graphs K m +1 and K m we can showthat C nK m +1 ,K m is not in RCA n but C nK m ,K m is in RCA n . The problem is that al-though the atom structures of these two cylindric algebra agree on all m -variableatom structure formulas, we cannot prove that the two cylindric algebras are equi-valent with respect to unrestricted formulas of quantifier depth at most log m . Thusour attempt to extend to various algebras of higher order relations using the knownconnections between relation algebras and cylindric algebras was not successful. A corrected strategy for proving no finite variableaxiomatisation exists for RRA
It turns out that if ∃ has a winning strategy in the infinite game G c +3 ω ( G, H ),then she also has a winning strategy in Γ cn ( B G,F , B H,F ) for all n < ω , and this canbe used to correct the claims of [HH02, problem 1, page 625]. We will prove thissoon, but first we will need the following lemmas.
LEMMA 4.1.
Let G and H be binary structures and let c ≥ . Then, if ∃ isplaying G cω ( G, H ) according to a winning strategy, whenever ∀ colours a set in oneof the binary structures, ∃ must respond by colouring a set of nodes of the otherbinary structure with the same cardinalityProof. If the set of nodes with a certain colour is bigger in one structure than theother, then ∀ may use a second colour to colour all but one node in the largerset (and ∃ must colour a proper subset of the smaller set to avoid losing straightaway), and he may repeat by re-using his first colour to colour all but one nodeof the larger set, and so on, until he colours a non-empty set of nodes in the firstgraph but ∃ has only the empty set to choose in the other graph, so ∀ wins. See[EH, Proposition 2.3] for the details. (cid:3) Now consider the relation algebra equivalence game Γ cω ( B G,F , B H,F ). As in theproof of lemma 3.4, we suppose that ∃ maintains a private corresponding play ofa colouring game over ( G, H ), but now we suppose she is playing according to awinning strategy in the infinite game G c +3 ω ( G, H ), which has three extra colours. o, to recap, if, for example, ∀ picks an element x ∈ B G,F , she picks the element y ∈ B H,F with the identical non-green part and green part determined by H y , where H y is the response to G x in the parallel play of G c +3 ω ( G, H ). These moves in theplay of G c +3 ω ( G, H ) are determined by the play of Γ cω ( B G,F , B H,F ) and only involvethe first c colours. Our assumption is that at each position ( g, h ) occurring in theplay of G c +3 ω ( G, H ), ∃ has a winning strategy in the game proceeding from ( g, h ),even if ∀ decides to use the three additional colours. In other words, ∃ cannot makeany move in the parallel game resulting in a position from which ∀ could force awin.Suppose ( α, β ) is a position in Γ cω ( B G,F , B H,F ), played as described above, andlet ( g, h ) be the corresponding position in G c +3 ω ( G, H ). We will need to interpretterms in the language of relation algebras with finite variable set { x , . . . , x c − } inthe algebras B G,F and B H,F . For any variable x i where i ∈ dom ( α ), we interpret x i in B G,F by defining x αi = α ( i ) ∈ B G,F . If b is a relation algebra constant, wedefine b α to be the interpretation of b in B G,F . Thus, any relation algebra term t involving only variables with indices in dom ( α ) and relation algebra constants hasan obvious interpretation t α ∈ B G,H , and similarly t β ∈ B H,F . For any such term t we define γ ( t ) = { x ∈ G : g x ≤ t α } and η ( t ) = { y ∈ H : g y ≤ t β } . LEMMA 4.2.
Let ( α, β ) be a position arrived at in a game Γ cn ( B G,F , B H,F ) duringwhich ∃ plays using a winning strategy in a parallel game G c +3 ω ( G, H ) , as describedabove. Let t be a term involving only variables indexed by values from dom ( α ) = dom ( β ) . Then:(1) The sets of non-green atoms below t α and t β are identical.(2) In the parallel game G c +3 ω ( G, H ) , if ∀ were to use a colour not previouslyused to colour γ ( t ) , then ∃ would have to respond by colouring η ( t ) , other-wise ∀ could force a win, and similar with γ ( t ) and η ( t ) switched.(3) | γ ( t ) | = | η ( t ) | .Proof. For convenience we match the colours { , . . . , c − } in Γ cn ( B G,F , B H,F ) and G c +3 ω ( G, H ) in the obvious way, and we refer to the colours { c, c + 1 , c + 2 } used in G c +3 ω ( G, H ) as additional colours , or words to that effect.We will use induction on t to prove (1) and (2), and we note that (3) followsfrom (2), because if | γ ( t ) | 6 = | η ( t ) | , then by colouring γ ( t ) with one of the extracolours, ∀ could force ∃ to colour a set with a different size, and thus force a win in G c +3 ω ( G, H ) (see Lemma 4.1). In the base case, (1) is automatic. For (2), suppose t = x i , and t α = α ( i ) for some i ∈ dom ( α ). So γ ( t ) = γ ( x i ) = { x ∈ G : g x ≤ α ( i ) } is already coloured by the i th colour. Moreover, η ( t ) = { y ∈ H : g y ≤ β ( i ) } mustalso be coloured by this colour, as according to ∃ ’s strategy β ( i ) is defined to makethis true. If ∀ uses one of the additional colours to colour γ ( t ), then ∃ must colourall of η ( t ) in response, otherwise there will be a palette mismatch between G and H . The cases where t is one of the relation algebra constants are also easy. For theinductive step, we proceed as follows (assuming the result for terms s, s , s ): t = − s : As the non-green parts of the two relation algebras are identical, (1)holds for t . For (2), if ∀ uses an additional colour to colour γ ( − s ) ⊆ G then ∃ mustcolour a set Y ⊆ H with the same colour. If he goes on to colour γ ( s ) ⊆ G with asecond additional colour, then by inductive assumption ∃ will colour η ( s ) ⊆ H withthat colour. We know that γ ( s ) , γ ( − s ) are disjoint and cover H . Since the positionis a winning position it must be that Y, η ( s ) are disjoint and cover G , hence Y isthe complement in H of η ( s ), so Y = η ( − s ), as required. = s · s : For (1), if a is a non-green atom, then, appealing to the inductivehypothesis, we have a ≤ ( s · s ) α ⇐⇒ a ≤ s α ∧ a ≤ s α ⇐⇒ a ≤ s β ∧ a ≤ s β ⇐⇒ a ≤ ( s · s ) β . For (2), if ∀ uses an additional colour to colour γ ( s · s ) ⊆ G then ∃ must coloursome set Y ⊆ H with the same colour. If he went on in the next two rounds to usethe other two additional colours to colour γ ( s ) and then γ ( s ), then inductivelywe know that ∃ colours η ( s ) , η ( s ) ⊆ H . Since the position at the end of this isnot a win for ∀ we must have Y = η ( s ) ∩ η ( s ) = η ( s · s ), as required. t = s ⌣ : For any non-green atom a we know a ⌣ is also a non-green atom and so a ≤ ( s ⌣ ) α ⇐⇒ a ⌣ ≤ ( s α ) ⌣ ⇐⇒ a ⌣ ≤ s α ⇐⇒ a ⌣ ≤ s β ⇐⇒ a ≤ ( s ⌣ ) β , proving (1). Part (2) is easy since all green elements are self-converse, so γ ( s ⌣ ) = γ ( s ) and η ( s ⌣ ) = η ( s ). t = s ; s : We show first that (1) holds. If a ≤ ( s ; s ) α is a non-green atomthen there are atoms b ≤ s α , b ≤ s α where a ≤ b ; b . Not all triples of atomsof the colours of b , b , a are forbidden, as we are assuming a ≤ b ; b . If no tripleof atoms of the colours of b , b , a is forbidden, then a ≤ ( s ; s ) β , because, by theinductive hypothesis with (1) and (3), the non-green atoms below s α and s β arethe same, and likewise for s , and | γ ( s ) | = | η ( s ) | , and likewise for s .This leaves the case where some but not all triples of atoms of the colours of b , b , a are forbidden. Recall that the only such colour combinations are red-red-red and green-green-red. If b , b , a are all red then, by the inductive hypothesison (1), we have b ≤ s β , b ≤ s β , so a ≤ b ; b ≤ ( s ; s ) β . This leaves the casewhere a = r j ,j is a red atom, and b = g x , b = g x are both green. Since( g x , g x , r j ,j ) is not forbidden, we know that x = x and { ( x , j ) , ( x , j ) } isa partial homomorphism. Since ( g, h ) is a position from which ∃ has a winningstrategy in G c +3 ω ( G, H ), after a short but technical argument to come, we shall seethat there must be y ∈ η ( s ) , y ∈ η ( s ) such that { ( y , j ) , ( y , j ) } is a partialhomomorphism.To this end, suppose for contradiction that there are no such y , y . In theparallel game G c +3 ω ( G, H ), if ∀ were to colour γ ( s ) with one of the extra colours,and then to use another to colour γ ( s ), then, by the inductive hypothesis on (2), ∃ would have to respond by colouring η ( s ) and η ( s ) respectively. Suppose that ∀ then uses the final additional colour to colour { x , x } . Then ∃ must colour some { y , y } ⊆ H . Since x ∈ γ ( s ) and x ∈ γ ( s ), we can suppose without loss ofgenerality y ∈ η ( s ) and y ∈ η ( s ). Suppose that, in a further move, ∀ usesone of the original colours (i.e. { , . . . , c − } ) to colour { x } , and without loss ofgenerality we can assume that ∃ responds by colouring { y } (if ∃ can colour { y } then we must have x , x ∈ γ ( s ) ∩ γ ( s ) and y , y ∈ η ( s ) ∩ η ( s ), in which case wecan just switch the labels of y and y ). But then { ( y , j ) , ( y , j ) } is not a partialhomomorphism but { ( x , j ) , ( x , j ) } is a partial homomorphism, and it followsthat { ( x , y ) , ( x , y ) } is not a partial isomorphism, indicating that (C2) holds.Thus ∀ wins, contrary to our assumption that ∃ is following a winning strategy.The implication a ≤ ( s ; s ) β ⇒ a ≤ ( s ; s ) α , for non-green atoms a , is provedsimilarly. This proves (1).For (2), the cases where either s α or s α is zero or the identity are trivial, so as-sume not. First we suppose that s α is either a single non-green atom, or pure green(i.e. above only green atoms), and the same for s α . By our induction hypothesis, β is the same non-green atom in the former case and green in the latter case, andsimilar for s β . Suppose the colour of s α is c , and that the colour of s α is c . Ifall triples of atoms of colours c - c -green are forbidden (i.e. c is green, c is greenor white, or the other way round) then γ ( s ; s ) = η ( s ; s ) = ∅ . If no triple ofatoms of colours c - c -green are forbidden then γ ( s ; s ) = G and η ( s ; s ) = H . Inboth these cases, if ∀ colours γ ( s ; s ), then ∃ must obviously respond by colouring η ( s ; s ). The only colour combinations where some but not all triples of atoms ofthose colours are forbidden are red-red-red and green-green-red. So the remainingcases are where c is green, c is red, or the other way round. Without loss ofgenerality, suppose s α is green and s α = r j,j ′ is a red atom. Observe that γ ( s ; s ) = { x ′ ∈ G : ∃ x ∈ γ ( s ) s.t. { ( x, j ) , ( x ′ , j ′ ) } is a partial homomorphism } , and η ( s ; s ) = { y ′ ∈ H : ∃ y ∈ η ( s ) s.t. { ( y, j ) , ( y ′ , j ′ ) } is a partial homomorphism } , using a Peircean transformation with (V). If either of these is empty then the resultis trivial, so we assume not.If ∀ makes an additional move by colouring γ ( s ; s ) ⊆ G with k where c ≤ k Suppose G, H are finite binary structures such that1. ∃ has a winning strategy in G c +3 ω ( G, H ) ,2. every partial homomorphism of H of size two extends to a full homomorphismfrom H into itself, and3. eithera. there is no homomorphism G → H , orb. there are i = i ′ ∈ G and j, j ′ ∈ H such that { ( i, j ) , ( i ′ , j ′ ) } is a partialhomomorphism that does not extend to a homomorphism G → H .Then RRA cannot be defined by any c -variable first-order theory.Proof. First note that by the partial homomorphism extension property assumedof H , there being no homomorphism G → H is equivalent to there being i = i ′ ∈ G such that for all j, j ′ ∈ H the partial map { ( i, j ) , ( i ′ , j ′ ) } is not a homomorphism.So it follows from lemma 3.2 and corollary 4.3 that B G,H and B H,H would agreeabout all first-order formulas with at most c variables, but by theorem 2.1 thelatter relation algebra would be completely representable while the former wouldnot. Since both algebras are finite, all representations are complete, and it wouldfollow immediately that B H,H ∈ RRA, B G,H RRA . (cid:3) In the result above, conditions a) and b) seem to be incomparable. This issuggested by the following example. EXAMPLE 4.5. Let G be the cyclic graph with three vertices C , and let H bethe discrete graph with a single vertex D . Then H trivially satisfies condition 2,and ( G, H ) satisfies condition 3.a. However, ( G, H ) does not satisfy condition 3.bas there are no partial homomorphisms of size two from G into H .Alternatively, let G be the three element walk (AKA chain) graph W (this hasvertices { v , v , v } and edges ( v , v ) and ( v , v ) ), and let H be the disjoint unionof two copies of C . Then H satisfies condition 2, and ( G, H ) does not satisfy 3.a,as there’s an obvious homomorphism W → C , but does satisfy 3.b, as the partialhomomorphism taking v to a vertex in one copy of C and v to a vertex of theother copy of C does not extend. The above example is not conclusive as in neither case does ( G, H ) satisfy con-dition 1. However, it’s not clear if non-isomorphic binary structures satisfying thiscondition even exist. In the special case of graphs, the existence of such a non-isomorphic pair would disprove the reconstruction conjecture, as we discuss brieflyafter corollary 4.6 below. On the subject of graphs, we can apply theorem 4.4 toobtain two ways to correct the graph based formulation of [HH02, problem 1]. Thefirst way is given by simply replacing ‘binary structure’ with ‘graph’ or ‘digraph’in the statement of that theorem. The second way is given by the result below. COROLLARY 4.6. Let G, H be finite digraphs such that1. G, H cannot be distinguished in a modified infinitely long ( c + 3) -colour gamewhere ∀ can also win at position ( g, h ) if there are two palettes π, π ′ and everypair from π g × ( π ′ ) g is an edge but not every pair from π h × ( π ′ ) h is an edge, orthe other way round, . every partial embedding of H into itself of size two extends to an automorphismof H , and3. there is no embedding of G into H .Then RRA cannot be defined by any c -variable first-order theory.Proof. Two digraphs G, H may be considered as binary structures with three pre-dicates, one for edges, another for non-edges and a third for ‘non-equality’ given by { ( u, v ) : u = v } . The modified game for digraphs understood as binary structureswith a single edge relation (the standard setting) is equivalent to the original gameplayed over the same digraphs understood as binary structures with ‘edge’, ‘non-edge’ and ‘non-equality’ relations. To see this, note that for palettes π and π thenon-edge relation holding between π g and π g and not between π h and π h resultsin a win for ∀ according to the modified rules, even if we only consider the ‘edge’relation. Moreover, when π = π the ‘non-equality’ relation holds between π g and π g whenever these are both non-empty, as interpretations of distinct palettes aredisjoint. Similar holds for π h and π h , and so accommodating ‘non-equality’ in thestandard digraph setting does not require any modification to the game rules.Now, homomorphisms in the ‘three relation’ setting clearly correspond to embed-dings in the standard digraph setting, and so condition 2 of theorem 4.4 translatesinto condition 2 here. Moreover, if there exists an embedding G → H , then, byassumption of condition 2, every partial embedding G → H of size two must extendto a full embedding of G into H , so condition 3 here covers both conditions 3.a and3.b from theorem 4.4. (cid:3) The result of corollary 4.6 and the graph result obtained from theorem 4.4 bywriting ‘digraph’ for ‘binary structure’ seem to be incomparable in strength, thoughagain the uncertainty around condition 1 prevents us from being sure. We get someindication of this by examining the conditions on the graph H in the two results.Respectively, these are:(1) Every partial homomorphism of size two of H into itself extends to a full ho-momorphism.(2) Every partial embedding of H into itself of size two extends to an automorph-ism.Observe that the cyclic graph C satisfies (2) but not (1), and we can construct agraph satisfying (1) but not (2) as follows (we work with undirected graphs herefor simplicity). Let W and W be walk (AKA chain) graphs, and let R be thegraph with a single reflexive vertex v . Define H by taking the disjoint union of W , W and R , and adding an edge ( w, v ) for each w ∈ W ∪ W . Then H satisfies(1), as every partial homomorphism can be extended by sending every other vertexto v . On the other hand, H does not satisfy (2) as, if we suppose the verticesof W and W are { u , u } and { w , w , w } respectively, the partial embedding { ( w , u ) , ( w , u ) } cannot be extended to an automorphism.Looking at the conditions on H may well be beside the point however, as, asmentioned previously, we do not know whether non-isomorphic graphs G, H existsuch that ∃ can win even G ω ( G, H ). The power of the game G to distinguishbetween graphs is investigated in more detail in [EH]. In particular, it is shownthere in section 6 that finding non-isomorphic graphs indistinguishable in the 3-colour game would disprove the reconstruction conjecture for graphs (or one of its ariations, in the case of directed graphs). This indicates that, at the very least,such graphs will likely be difficult to find. References [Bon91] J. A. Bondy. A graph reconstructor’s manual. In Surveys in combinatorics, 1991 (Guild-ford, 1991) , volume 166 of London Math. Soc. 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Topics in graph automorphisms and reconstruction , volume432 of London Mathematical Society Lecture Note Series . Cambridge University Press,Cambridge, second edition, 2016.[Mon64] J. Monk. On representable relation algebras. Michigan Mathematics Journal , 11:207–210, 1964.[SA05] T. Sayed Ahmed. Algebraic logic, where does it stand today? Bull. Symbolic Logic ,11(4):465–516, 2005.[Sto88] P.K. Stockmeyer. Tilting at windmills, or My quest for nonreconstructible graphs.volume 63, pages 188–200. 1988. 250th Anniversary Conference on Graph Theory (FortWayne, IN, 1986).[Ven97] Y. Venema. Atom structures. In M. Kracht, M. De Rijke, H. Wansing, and M. Za-kharyaschev, editors, Advances in Modal Logic ’96 , pages 291–305. CSLI Publications,Stanford, 1997., pages 291–305. CSLI Publications,Stanford, 1997.