Alan H. Mekler, Evelyn Nelson and Saharon Shelah- A Variety with Solvable, but not Uniformly Solvable, Word Problem

8/3/2019 Alan H. Mekler, Evelyn Nelson and Saharon Shelah- A Variety with Solvable, but not Uniformly Solvable, Word Prob

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A Variety with Solvable, but not Uniformly

Solvable, Word Problem

Alan H. MeklerDepartment of Mathematics and Statistics

Simon Fraser UniversityBurnaby, B.C. V5A 1S6 CANADA

Evelyn Nelson

Department of Mathematics and StatisticsMcMaster University

Hamilton, Ontario L8S 4K1, CANADA

Saharon ShelahInstitute of MathematicsThe Hebrew University

Jerusalem, Israel


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Abstract

In the literature two notions of the word problem for a variety occur. Avariety has a decidable word problem if every nitely presented algebra inthe variety has a decidable word problem. It has a uniformly decidable word problem if there is an algorithm which given a nite presentation produces analgorithm for solving the word problem of the algebra so presented. A varietyis given with nitely many axioms having a decidable, but not uniformlydecidable, word problem. Other related examples are given as well.


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0. INTRODUCTIONThe following two options occur in the literature for what is meant bythe solvability of the word problem for a variety V :

(1) There is an algorithm which, given a nite presentation P in nitelymany generators and relations, solves the word problem for P relative to thevariety V .

(2) For each nite presentation P in nitely many generators and rela-tions, there is an algorithm which solves the word problem for P relative tothe variety V .

We say that V has uniformly solvable word problem if (1) holds. Itis the rst notion that is studied in Evans [2, 3], where it is called justthe word problem for V , and the second coincides with the terminology inBurris and Sankapannavar [1]. Benjamin Wells has informed us that Tarskiwas interested in the existence of varieties with solvable but not uniformlysolvable word problem.

Varieties with uniformly solvable word problem include commutative semi-groups and abelian groups (each of these are equivalent to the existence of an algorithm for solving systems of linear equations over the integers whichis due to Aryabhata, see chapter 5 of [10]), any nitely based locally nite orresidually nite variety, and the variety of all algebras of a given nite type(see [4]).

The examples which appear in the literature, of varieties with unsolvableword problem, all provide a nite presentation P for which the word problemfor P relative to that variety is unsolvable. These include semigroups [9],groups [8], and modular lattices [5].

Here, we present a nitely based variety V of nite type which does nothave a uniformly solvable word problem, but which nevertheless has solvableword problem. We also present a recursively based variety of nite type,which is dened by laws involving only constants (i.e., no variables), withsolvable but not uniformly solvable word problem. This second result is thebest possible one can provide for varieties dened by laws involving no vari-

ables: every nitely based such variety has uniformly solvable word problem.If one would be satised with varieties with innitely many operations, thenit is relatively easy to produce an example of a recursively based varietywith solvable but not uniformly solvable word problem; we present such anexample, essentially due to B. Wells [11], at the end of the paper.

Our proof uses the unsolvability of the halting problem for the universal

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Turing machine, and the laws dening the variety precisely allow us to modelthe action of the universal Turing machine in the variety. In the usual proofsthat the variety of semigroups has an undecidable word problem, a nitelypresented congruence is given so that for any initial Turing machine cong-uration, the instantaneous descriptions of the Turing machine calculationsall lie in the same congruence class. Then laws are added which make thehalting state a right and left zero. So, the undecidability of the word problemin this algebra comes from not being able to decide whether a given word(initial conguration) is congruent to the halting state. This algebra containsall possible Turing machine calculations. In our variety each calculation willbe modeled by a single algebra.

Our approach is based on a different picture of a Turing machine calcu-lation than the sequence of instantaneous descriptions used in semigroups.We view a Turing machine calculation as taking place on a Z grid, wherethe copy of Z with second coordinate n represents the Turing machine tapeat time n. To understand the calculation, we must know the alphabet con-tent of each square, which square the head is reading for each time n, andthe state the machine is in at time n. There are various possible ways toformalize this insight, so that each Turing machine calculation correspondsto a nitely presented algebra.

To ensure that the word problem doesnt have a uniform solution, weintroduce a function which has value 1 when applied to any state the machinereaches and value 0 on the halting state. Then a decision procedure whichgiven a nite presentation, determines whether 1 is congruent to 0 wouldsolve the halting problem. There are considerable technical difficulties inimplementing this idea in such a way that we can prove that each nitelypresented algebra in the variety has a decidable word problem.

It seems to us that there are two interesting directions that research canfollow in light of the results in this paper. There remains the question of whether a nitely based variety of unary algebras with solvable word problemhas uniformly solvable word problem. Another direction research could take

is to consider subvarieties of interesting natural varieties. This could eitherbe understood as varieties of X where X is a favourite class of algebras or,say, congruence modular varieties. This second problem was suggested to usby the persistent question of everyone to whom we told the result, namelyIs there a natural example?

The research for this paper was begun out while the latter two authors

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were visitors at the Department of Mathematics and Statistics at SimonFraser University. We gratefully acknowledge nancial support from theNatural Science and Engineering Research Council of Canada. This is paper#291 on Shelahs publication list. We also wish to thank the referee for athorough job of reading the paper.

1. DEFINITIONS

We assume is some nitary type of algebras (with possibly innitelymany operations). A presentation is a pair P = ( X, R ) consisting of a setX (of generators) and a set R F X F X (of relations), where F X is the

(absolutely) free -algebra over X . A nite presentation is a presentationP = ( X, R ) where both X and R are nite.Given a variety V and a presentation P = ( X, R ), there is an algebra

A V and a homomorphism h : F X A with R Ker (h), such that anyhomomorphism g : F X B with B V and R Ker (g) factors uniquelythrough h. The algebra A is unique up to isomorphism, and is called thealgebra given by the presentation P relative to the variety V .

The word problem for P relative to the variety V is to determine, givens, t F X , whether ( s, t ) Ker (h). Note that Ker (h) is the congruenceon F X generated by R V , where V consists of all equations in variablesfrom X satised by the variety V ; equivalently, V is the kernel of the uniquehomomorphism from F X to the V -free algebra on X mapping the elementsof X identically.

Next, we introduce the notion of a partial subalgebra, and state one resultwhich will be proved and used in 6. The proof bears a familial resemblanceto the more complicated proof in 5.

Denition. A partial subalgebra is a pair (A, A ) where A is a subset of F X which is closed under formation of subterms, and A is an equivalencerelation on A which is a partial congruence , i.e., has the property that foreach operation of arity n, if a i A bi for 1 i n and (a1, . . . , a n ),

(b1, . . . , bn ) A then(a1, . . . , a n ) A (b1, . . . , bn ).

Proposition 1.1 If for a partial subalgebra (A, A ),(1) membership in A is decidable (for elements of F X ),

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(2) membership in A is decidable (for pairs of elements of F X ),(3) there is an algorithm which, given an operation of arity n and a1, . . . ,an A, determines whether there exist b1, . . . , bn A for which a i A bi for 1 i n and (b1, . . . , bn ) A, then , the congruence on F X generated by A , is decidable. Further this decision procedure is uniform in the algorithms for deciding (1), (2), (3) .

Remark: The conclusion of this result says that there is a solution to the wordproblem for the presentation ( X, A ) relative to the variety of all algebras of the given type.

Denition A partial subalgebra satifying the hypotheses of Proposition 1.1is called decidable.

We delay the proof of Proposition 1.1 until the end of 6.Corollary 1.2. (Evans) Let V be the variety of all algebras in some nitelanguage. Then V has uniformly decidable word problem.

Proof. Suppose we are given some nite presentation. Let A be thenite set consisting of the terms appearing in the presentation and theirsubterms. By brute search through the nitely many possibilities, we cannd A , the smallest partial congruence on A containing all the relations inthe presentation. Now we can apply Proposition 1.1.

Corollary 1.2 implies that any variety in a nite language which is denedby nitely many laws involving only constants has a uniformly decidable wordproblem. Any presentation can be viewed as a new presentation in the varietywithout the laws, by viewing each law as a relation in the new presentation.

2. THE FINITELY BASED VARIETY.

2.1 MODIFICATION OF THE UNIVERSAL TURING MACHINE

Suppose that we are given a universal Turing machine with a unique halt-ing state h which, at each move, prints some letter on the scanned square,

moves one square either left or right (denoted respectively by 1 or 1) andenters a new (or the same) state. We are rst going to adjust the machine byadding right and left end markers eR and eL , (as new members of the alpha-bet), and adding, for each state, two new states qL and qR , and appropriateinstructions so that the adjusted machine does the following: if it is scanningeR in state q, it prints B (blank), moves right (into state qR ), prints eR and

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then moves left and returns to state q; if it is scanning eL in state q it printsB, moves left, prints eL , and moves right and returns to state q. That is, tothe Turing ow chart we add the following

qLeL :B : 1

eL :B :1

qeR :B :1

eR :B : 1qR

The resulting machine, if started on a nite tape inscription with the leftand right endmarkers at the appropriate ends, and the rest of the tape blank,does what the original machine would have done if placed on that inscriptionwith the rest of the tape blank, except that whenever the adjusted machinehits an endmarker it rst moves it out one square, leaving behind a blanksquare.

Suppose that the resulting machine has state set Q and alphabet , andthat its action is given by the functions , , and operating on Q ,which specify the next state, the motion (either left or right) and the printinstruction, so that

: Q Q : Q { 1, 1} : Q .For simplicity we will assume , and are total functions and so dened

even if we reach the halting state.2.2 DEFINITION OF THE VARIETY

Our variety V has the following operations:CONSTANTS: c, all elements of Q , 0, 1, 0F , 1F UNARY: T , S , S 1, H , P , C , C Q , U , E BINARY: F , R, K , K , C ,TERNARY: N H , N Q , N The elements of the range of P will be the space-time elements (in the

intended interpretation they represent the tape squares); the action of S

and S 1

represents stepping right and left respectively though space, of T represents moving ahead one time period, and of H represents moving to thehead position. C gives the letter in the square and C Q gives the state themachine is in while scanning the square. The action of the ternary operationN H gives head position at the next time instant, while N gives the letter in

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the square scanned by the head at the next time position and C Q gives thestate at the next time.The intended interpretations of F (x, y) and R(x, y) are y follows x and

y is to the right of x. K and K are comparison functions, and U is amodied addition by 1. The intended interpretation is explained more fullyin paranthetical comments below and in the proof of Theorem 3.1.

The laws dening our variety are the following:I P P (x) P (x)

P T (x) T P (x) T (x)P S (x) SP (x) S (x)

P S 1

(x) S 1

P (x) S 1

(x)P H (x) HP (x) H (x)P N H (x,y,z ) N H (x, y , P (z)) N H (x,y,z )P K (x, y) K (x, P (y)) K (x, y)

HS (x) HS 1(x) HH (x) H (x)HT H (x) HT (x)T S (x) ST (x)T S 1(x) S 1T (x)SS 1(x) S 1S (x) P (x)

N H (x, y , H (z)) N H (x,y,z )HN H (x,y,z ) HT (z).

II N Q (q,a,H (x)) (q, a) for all qQ, a N (q,a,H (x)) (q, a) for all qQ, a N H (q,a,H (x)) S (q,a )T H (x) for all qQ QLRN Q (qL , C H (x), H (x)) q N Q (qR , C H (x), H (x))

N H (qL , C H (x), H (x)) S 1T H (x)N H (qR , C H (x), H (x)) ST H (x)

N (qL , C H (x), H (x)) eL

N (qR , C H (x), H (x)) eRIII C T H (x) N (C Q H (x), C H (x), H (x))

C Q T H (x) N Q (C Q H (x), C H (x), H (x))HT (x) N H (C Q H (x), C H (x), H (x))

IV C Q P (x) C Q H (x) C Q (x)

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C (x) C P (x)C T P (x) C (P (x), R(P (x), H (x)))C T P (x) C (P (x), R(H (x), P (x)))C P (x) C (P (x), 1)(C ensures the symbol in a square, which is either to the right or to

the left of the square being scanned, remains unchanged at the next time.)

V R(x, y) R(P (x), P (y))F (x, y) F (Px ,Py )R(P (x), P (x)) 0

R(P (x), SP (y)) UR(P (x), P (y))F (P (x), P (y)) F (H (x), H (y))F (P (x), P (x)) 0F F (P (x), T P (y)) UF (P (x), P (y))

U (0) U (1) 1U (0F ) U (1F ) 1F

VI For all operations f except T , S , S 1, H , P , K , and N H , P f is constantwith value P (c).

VII (i) K (0, P (x)) P (x)K (1, P (x)) P (c)K (0F , P (x)) P (x)K (1F , P (x)) P (c)(K ensures that if 0 = 1 or 0 F = 1 F in an algebra then all the space time

elements are identical or in other words that space-time is degenerate.)

(ii) K (d, d) 0 for all constants dK (d, e) 1 for all constants d, e with d = e.K (P (x), d) 1 for all constants d = cK (C P (x), d) 1 for all constants d /K (C Q P (x), d) 1 for all constants d /QK (R(P (x), P (y)) , d) 1 for all constants d = 0 , 1K (F (P (x), P (y)) , d) 1 for all constants d = 0 F , 1F K (t, t ) 0 and K (s, t ) 1 for all s = t where both s, t belong to{P (x1), C P (x2), C Q P (x3), R(P (x4), P (y4)), F (P (x5), P (y5))}

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(K

ensures that space-time is degenerate if there is any undesired equal-ities between constants or if there is a nonempty intersection between theranges of certain operations.)

VIII EC Q P (x) 1E (h) 0(E ensures that space-time is degenerate if the halting state is reached.)

2.3 NORMAL FORM FOR SPACE-TIME ELEMENTSThe terms which are in the image of the operations P , S , S 1, T , H , N H

and K are called space-time terms. For each such term t, P (t) is equivalent,modulo the laws of our variety, to t. For all terms t in the images of theother operations, P (t) is equivalent, modulo the laws of our variety, to P (c).Thus a term t is a space-time term if and only if t and P (t) are equivalent,modulo the laws of our variety.

We are going to develop a normal-form representation for space-timeterms.

First, for a space-time term t dene

t = {S n T m (t)|m, n Z , m 0} {S n T m HT k (t)|n,m,k Z ,m,k 0} {S n T m N H (s,u,T k(t)) |n,m,k Z ,m,k 0,s ,u arbitrary terms }

Now, dene the set G of generating space-time terms as follows(i) P (c) G(ii) P (x) G for each variable x(iii) For each term t, and each g G and g , the term K (t, ) G.(iv) G is the smallest set of terms satisfying (i), (ii) and (iii).Further, dene = g(g G).The members of g for g G are called the space-time terms in normal

form with respect to g and the g is called the space-time component of g,in particular g g.

For space-time terms in normal form with respect to g, we dene theg-time prex, g-time coordinate and g-space coordinate as follows:term in g-time prex g-time coordinate g-space coordinateS n T m (g) T m (g) m nS n T m HT k (g) T m HT k (g) m + k nS n T m N H (s,t,T k (g)) T m N H (s, t ,T k (g)) m + k + 1 n

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Proposition 2.1: There is an effective procedure which, given a space-timeterm s, produces a term t (i.e., in normal form ) such that the laws Ientail s t.

Proof. The procedure is described inductively on the complexity of terms.To begin, of course, the normal form of g G is g. If t = P (s) for some normalform space-time term s g then the normal form for t is the same as thatof s.

If t = H (s) or T (s), for some normal form space-time term s g , thenthe normal form t for t is given in the following table:

s H (s) T (s)S n T m (g) HT m (g) S n T m +1 (g)S n T m HT k (g) HT m + k (g) S n T m +1 HT k (g)S n T m N H (t1, t2, T k (g)) HT m + k +1 (g) S n T m +1 N H (t1, t2, T k (g))

If t = S (s) or S 1(s) for some normal form space-time term s then thenormal form for t is obtained from s by adding or subtracting 1 respectivelyto the space component.

If t = N H (s1, s2, s) for a normal form space-time term s, then the normalform t for t is given in the following table:

s tS n T m (g) N H (s1, s2, T m (g))S n T m HT k (g) N H (s1, s2, T m + k (g))S n T m N H (t1, t2, T k (g)) N H (s1, s2, T m + k +1 (g))

If t = K (s1, s) for a normal form space-time term s then t is in normalform.

This completes the description of the procedure.

Remark. In the ensuing development, we will always deal only with space-time elements in normal form, and when we write H (), S (), etc. for ,we will mean the normal form of H (), etc.

Proposition 2.2: For terms s, t g with time coordinates m, n respectively,if m < n then the laws V entail F (s, t ) 1F .

Proposition 2.3: For terms s, t g with the same time prex but different space coordinates, the laws V entail either R(s, t ) 1 or R(t, s ) 1.

3. NON-UNIFORM SOLVABILITY OF THE WORD PROBLEM

This section is devoted to a proof of the following:

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Theorem 3.1. V does not have uniformly solvable word problem.Proof. For any initial tape conguration

. . . eL a0 a1 a2 . . . ak eR . . .

where indicates head position) for the universal Turing machine, there is acorresponding nite presentation

P : C Q (c) q0 and C (S 1(c)) eL and C (c) a0 and . . . C (S k (c)) ak and C (S k +1 (c)) eR

We claim that the universal Turing machine, started on that congura-tion, eventually halts, if and only if E (q0) h (equivalently, 0 1) followsfrom the presentation P in the variety V . Thus, since there is no algorithmwhich determines, given an initial tape conguration, whether or not theuniversal Turing machine will halt, this establishes the fact that V does nothave uniformly solvable word problem.

( ) : This direction is clear; P together with the equations deningV entail the analogous information at each successive conguration. If themachine halts at time n then we obtain P (c) such that C Q () h andso

0 E (h) E (C Q ()) 1 E (q0)follow from P in the variety V .

( ) : Suppose the machine, started on the above conguration, neverhalts. Then we produce a model A V satisfying all the equations in P , inwhich 0 = 1 .

The set of elements of A is {}Q{S n T m (c)|n Z , m N }{n |n Z and n 1} {nF |n Z and n 1}.

The operations are dened in A as follows:(i) T,S,S 1 are dened on elements of the form S n T m (c) according to

equations I so as to yield elements again of this form; for other elements y,T (y) = T (c), S (y) = S (c), S 1(y) = S 1(c).

(ii) P maps all elements of the form S n T m (c) identically and all otherelements to c, in particular P (c) = c.

(iii) U (n) = ( n + 1) and U (nF ) = ( n + 1) F for n 0, U (1) = 1, U (1F ) =1F . U maps all other elements to .

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(iv) E (q) = 1 for all qQ, q = hE (h) = 0E maps all other elements to .

(v) R(S n T m (c), S k T j (c)) = 1 if k > nk n if k n .

R(x, y) = R(P (x), P (y)) otherwise.

(vi) F (S n T m (c), S k T j (c)) = 1F if j > m( j m)F if j mF (x, y) = F (P (x), P (y)) otherwise.(vii) K (0, S n T m (c)) = S n T m (c) = K (0F , S n T m (c))

K (1, S nT

m(c)) = c = K (1F , S

nT

m(c))

K (x, y) = c otherwise.(viii) K (d, d) = 0 for all d {0, 1, 0F , 1F , c}Q K (d, e) = 1 for all d, e as above with d = eK (k, d) = 1 for all k {n |n 1}, all d = 0 , 1K (k, d) = 1 for all k {nF |n 1}, all d = 0 F , 1F K (S n T m (c), d) = 1 for all constants d = c (including all n and n F , with

n 1)K (x, y) = else.The values of H (x), C (x), C Q (x), N (q,a,H (x)), N Q (q,a,H (x)), and

N H (q,a,H (x)) for x {S n

T m

(c)|n Z , m N } are dened by induction onm:dene H (c) = HS n (c) = cC Q (c) = C Q S n (c) = q0C (S n (c)) as in P for 1 n k + 1C (S n (c)) = B for all other values of nN Q (q,a,c), N H (q,a,c, ) and N (q,a,c) are dened as in equations II (note

that c = H (c)) .Suppose we have already dened, for all n Z ,H (S n T m (c)) = H (T m (c)) = S k T m c for some k

C Q (S n

T m

(c)) = C Q (T m

(c)) QC (S n T m (c)) and N H , N Q , N for all triples (q,a,HT m (c)), with appropriate values,

i.e., im(N ) etc.Then dene for all n ZH (S n T m +1 (c)) = N H (C Q (HT m (c)), C (T H m (c)) , HT m (c))

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C Q (S n

T m +1

(c)) = N Q (C Q (HT m

(c)), C (HT m

(c)) , HT m

(c))C (T HT m (c)) = N (C Q (HT m (c)) , C (HT m (c)) , HT m (c))C (S n T HT m (c)) = C (S n HT m (c)) for all n = 0and then dene N Q , N , N H for all triples (q,a,HT m +1 (c)) according to

the rules II.This completes the inductive denition.

Dene N H (x,y,z ) = N H (x,y,P (z)) if the latter has already been dened.Dene N H and H on all other elements to have value c.Dene C Q (y) = C Q (c) = q0 for all y not of the form S n T m (c)C (y) = C (c) = a0 , for all y not of the form S n T m (c)

N Q (x,y,z ) = N (x,y,z ) = for all values not dened above.Dene C (S n T m (c), n) = C (S

n T m +1 (c)) if n 0C (S n T m (c)) if n = 1

, C (x, y) =

otherwise.Then the resulting algebra A satises all the laws of the variety and the

equations of the presentation P , and 0 = 1 in A.

4. SOLVABLE WORD PROBLEM IN THE DEGENERATE CASEThis section and the next are devoted to proving that V has solvable

word problem.Let P be a nite presentation on a generating set X = {x1, . . . , x n } and

let P be the congruence on F X generated by the relations of P togetherwith the substitution instances of the laws dening our variety V . We mustprove that P is decidable.

Denition. P has degenerate space-time if P (t)P P (c) for all terms t. Notethat, by the laws of V if P has degenerate space-time then 0 P 1 and 0F P 1F .In fact, the laws of V allow this conclusion to be drawn from any failure of the operations S , T to behave without loops. Also the laws of VII imply thatif either 0P 1 or 0F P 1F , then space-time is degenerate.

We rst prove the the following.

Theorem 4.1 If P has degenerate space-time then the word problem for P relative to our variety V is decidable.

Proof. In this case, in the presented algebra F (X )/ P , all the operations P ,T , S , S 1, H and N H are constant with value P (c). Moreover, R and F are

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constant, with value 0 and 0 F respectively, C Q is constant with value C Q P (c)and C is constant, with value C P (c).

Now, consider the type obtained from the one with which we are working,by deleting the operations P , T , S , S 1, H , N H , R, F , C Q , C , and addingthree constants c1, c2, and c3, which will stand for P (c), C Q P (c), and C P (c)respectively. Then there is an effective procedure which, given a term of thelarger type, produces a term of the smaller type which is equivalent to itmodulo the laws for our variety, and the equations 0 1, 0F 1F , c1 P (c),c2 C Q P (c), c3 = C P (c).

Thus if we consider the variety V of this reduced type dened by thefollowing equations:

equations II for N Q and N , with space-time terms, and terms in theimage of C Q and C replaced by c1, c2, c3 respectively.

c3 C (c1, 1)U (0) U (1) 1U (0F ) U (1F ) 1F equations VII, where P (x) is replaced by c1, C Q P (x) by c2, C P (x) by

c3 and R(P (x), P (y)) and F (P (x), P (y)) by 0.E (c2) 1E (h) 0.

Then, if the terms in the presentation P are replaced by their equivalentsin the new type we obtain a presentation P which relative to the varietydescribed above, is equivalent to P relative to the original variety. Here byequivalent we mean that there is an effective translation between terms sothat P entails s t relative to the variety V if and only if P entails s trelative to the variety V . Since P is a presentation considered relative to avariety dened by nitely many laws which involve no variables, by Corol-lary 1.2 this word problem is solvable, which shows that the word problemfor P relative to our variety is solvable too.

5. SOLVABLE WORD PROBLEM IN THE NON-DEGENERATE CASE

5.1 PLAN OF THE PROOF

In this section we prove the following:

Theorem 5.1 If P has non-degenerate space-time then the word problem for P relative to our variety V is decidable.

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Proof: The proof is presented in the remaining subsections of this section.In the remainder of this subsection we will describe the strategy of the proof.We will dene by induction an increasing sequence of partial subalgebras(An , A n ). There are three points to be veried. First, for all n every instanceof the laws of the variety and the relations of P with elements of An isvalidated by n . Second, for a, b An , if a n b then the laws of the varietyand P imply aP b. The set of terms will, apart from passing to normal forms,equal An . Hence P will essentially equal n . Third, the constructionof the An and the n is uniformly effective and hence n and P aredecidable. We will give a careful denition of the An and n , but we willleave it to the reader to verify the three points mentioned above. One otherpoint which is worth mentioning is that before constructing ( A0, 0) we willdemand more information about P other than its being non-degenerate. Werst dene two auxiliary sets A and B.

5.2 DEFINITION OF A

Let BP consist of all terms appearing in the presentation P and all theirsubterms and all constants of the variety V . Let GP consist of P (c), P (x) foreach generator x of the presentation, and all terms of the form K (s, t ) B P ;thus GP is nite.

For each g GP , we let g, P be the set of all terms built from g using the

unary operations P,S,S 1, T, H , and N H (s,t, ) where s, t BP . Furtherlet g, P be the set of members of g, P which are in normal form with respectto g, and let P = g, P (g GP ).

Now, we dene A as follows: it contains(1) the terms in GP and all constants(2) the terms in P (3) U (d) for d = 0 , 1, 0F , 1F (4) R(, ), F (, ), UR(, ), UF (, ) for , P (5) N Q (q, a, HT n (g)) ,

N (q, a, HT n (g)) ,

N H (q, a, HT n

(g)) for all g GP , qQ, a N Q (qL , C HT n (g), HT n (g)) ,N Q (qR , C HT n (g), HT n (g)) for all g GP , all qQ QLR , all n 0.

(6) C (), C Q ( ), all , P (7) N (C Q HT n (g), C HT n (g), HT n (g)) ,

N Q (C Q HT n (g), C HT n (g), HT n (g)) ,

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N H (C Q HT n

(g), C HT n

(g), HT n

(g)) for all g GP , all n 0.(8) C (, R (,H )),C (, R (H, )) ,C (, 1) for P

(9) K (d, d) for constants dK (d, e) for all constants d, e with d = e.All substitution instances of terms in laws VII (i) and (ii) where P (x)

and P (y) are replaced by arbitrary , P .

(10) EC Q (), for all P and E (h)

Note that membership in A is decidable.

5.3 DEFINITION OF B

Before we can dene B, we need some preliminary results.

Lemma 5.2 For any g, P , { g, P | P } is nite.

Proof. If , g, P and P then it follows from laws of V and thenon-degeneracy of P that and have the same time coordinate. (For ex-ample, S n T k g P S m T iHT r g implies HT k g = H (S n T k g)P H (S m T iHT r g) =HT i+ r g and this yields k = i + r. ) Moreover, two terms in g, P with thesame time prex and different space coordinates cannot be congruent mod-

ulo P . Since there are only nitely many time prexes with the same timecoordinate as , this establishes the result.

Corollary 5.3 For any term t, { P | P t} is nite.

Denition. For a nite F P , the maximum time vector of F is (mg )gG Pwhere mg is the maximum g-time coordinate of elements of F g (or 0 if F g is empty). We also make an ad hoc denition and say a space-timeterm s is a right subterm of a space-time t by induction on the constructionof t. If t is Hu , Su , or S 1u for a space-time term u then s is a right subtermof t if it is either t or a right subterm of u. If t is N H (w,v,u ) where u isa space-time term then s is a right subterm of t if it is either t or a rightsubterm of u.Lemma 5.4 For any nite subset F P with maximum time vector (mg)gG Pthere is a nite F P with the same maximum time vector, such that

(i) F F (ii) if P is a right subterm of F then F

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(iii) if P and P for F then

F (iv) if P and the normal form of T belongs to F then F .

Proof. We may assume that for each g GP , T j g F for all j mg .Now, let g GP and consider the set F g g, P F which consists of all

elements of g, P F whose time coordinate relative to g is mg . Let F g g, P consist of all those g, P for which there exists F g with P . ByLemma 5.2, F g is nite.

Let k1 and k2 be the maximum and minimum, respectively, of the spacecoordinate of members of F g . Note that, since T m g (g) F , we have k2 0 k1.

Let F g consist of those members of g which are the normal forms of allterms of the form S k () where k1 k k2, and F g . Then F g isnite and contains F g . We will show

(a) F g , g , P implies F g(b) F g , g a right subterm of with time coordinate mg relative

to g implies F g .re (a): Suppose is the normal form of S k () where k1 k k2 and

F g . Then S k P and hence S k P and so the normal form of S k belongs to F g . Thus , which is the normal form of S k S k , belongs to F g .

re (b): Suppose F g; then is the normal form of a term S k () where k

1 k k

2and F

g. Let the space coordinate of be n and the

time prex of be ; then = S n + k . Moreover, all terms of the form S i for n k1 i n k2 belong to F g . Since n k1 0 n k2, it followsthat if i is any number between n + k and 0, then S i F g . Now, anynormal form subterm of with the same time component has the same timeprex and hence this shows that every right subterm of with the same timecomponent belongs to F g .

Let F = F g(g GP ); then F is nite, F F , and F satises (ii) and(iii) for any g with g-time coordinate mg . Add to F each term gwith g-time coordinate mg 1 such that the normal form of T belongs toF g . The result is still nite. Now repeat the procedure for elements of g-time coordinate mg 1 , etc. to eventually obtain the desired set F . Thiscompletes the proof.

Denition. Dene B as follows:Recall that BP consists of all terms appearing in the presentation P and

all subterms thereof, and all constants of our variety V.

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Enlarge BP as follows:(i) For each bBP , if there exists a A with aP b, add one such a, andchoose a P whenever possible.

(ii) For each g GP let mg be the maximum time coordinate of all theelements of g that we have so far, and add all g-time prexes up to timemg .

(iii) Let F consist of all elements of P that we have so far. Add the setF F given in the above lemma.

(iv) For all , F , add C Q (), C (), R(, ), F (, ), U (R(, )),U (F (, )) .

The resulting set B is nite, is closed under taking subterms, and for , P , if B and P then B.

Let B = P |B; then B is nite and hence decidable. Also B containsthe relations of P .

5.4 DEFINITION OF A0Now, dene A0 = A B; then membership in A0 is decidable. We

are going to dene a partial congruence relation 0 on A0 so that the pair(A0, 0) is a partial subalgebra such that membership in 0 as well as A0 isdecidable. In fact, 0 will be P restricted to A0, but we will dene 0 byinduction on the complexity of terms and the size of the time coordinate for

members of P .For a, b B, a 0 b if and only if a B b.For a A, bB, a 0 b if and only if there exists cA B with a 0 c

(as described below) and c 0 b, i.e., c B b. Since B is nite, we decidewhether a 0 b by searching through all c A B and checking the lattertwo conditions. Thus it is enough to describe 0 between pairs of elementsof A.

There are some members of A that we can essentially ignore, because weknow they must be in the relation 0 to other elements that we have to dealwith anyway.

Thus, to begin, we decree:U (0) 0 U (1) 0 1U (0F ) 0 U (1F ) 0 1F

and for all , p,UR(, ) 0 R(, ) where is the normal form of S ( )UF (, ) 0 F (, ) where is the normal form of T ( )

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and so we may ignore, for the purposes of dening 0 between elements of A, all those elements of A which are in the range of U.Similarly, using the appropriate terms given in laws II and III we may

ignore the elements of A in the range of N or N Q , by making them 0congruent to elements in Q or the range of C Q , and or the range of C ,respectively.

We dispose in the same way of the elements of A that are in the range of C , K or K .

Thus we need only dene 0 between pairs of elements of A that are ei-ther constants, -elements, or in the range of the operations C , C Q , F andR. Moreover, since P is non-degenerate, we know by laws VIII that the in-terpreted images of these latter four operations are disjoint from one anotherand from all the interpretations of -elements in F (X )/ P . In addition, allthe constants are pairwise distinct in F (X )/ P , and A-elements in the rangeof C , C Q , F , and R can be P -congruent to constants only if they belongto , Q, {0, 1}, {0F , 1F }, respectively.

5.5 DEFINITION OF 0 FOR ELEMENTS WITH SMALL TIME COM-PONENT

We begin by describing 0 for elements , C (), C Q (), R(, ) andF (, ) for , P with time coordinate less than or equal to the maximum

occurring in B , relative to whatever space-time component and are in.(1) For , P with time coordinate less than or equal to the maximum

in B , dene 0 if and only if S n B S n

where n is the space coordinate of .Note that S n B, and hence if P then S n P S n and hence

S n B S n . The converse is also true, of course. The point about thedenition is that, given , we know its space coordinate and so we can decide 0 because B is decidable. Moreover (and we will need this later), given, we can calculate all (there are only nitely many) P with 0 .

(2) For , P with time coordinates less than or equal to the maxi-mum in B ,(i) R(, ) 0 0 if and only if either 0 as in (1) aboveor , B and R(, ) B 0.(ii) R(, ) 0 1 if and only if either there exists n > 0 with 0 S n

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or there exists P B and n 0 with 0 S n

and R(, ) B 0or there exists P B and n > 0 with 0 S n and R(, ) B 1.Note that this is decidable: for example, to check whether there exists

n > 0 with 0 S n , it is enough to determine whether there exists n > 0with S m B S n m where m is the space coordinate of , and the latteris decidable because B is nite.

(iii) R(1, 1) 0 R(2, 2) if and only if either both are congruent to 0 or 1 by (i) or (ii)or 1 0 2 and 1 0 2or there exist 1 and 2 and n 0 with R(1, 1) B R(2, 2) and

1 0 S n 1, 2 0 S n 2.

Remark. From the above denition, we have R(, ) 0 0 for all P with time coordinate less than or equal to the maximum in B , and moreover,if R(, ) 0 0 or 1 then R(, S ( )) 0 1, and so the laws V for R and thesevalues of P (x), P (y), are satised.

(3) For , , , P with time coordinate less than or equal to themaximum in B , dene

F (, ) 0 0F if and only if F (H , H ) B 0F F (, ) 0 1F if and only if F (H,H ) B 1F F (, ) 0 F ( , ) if and only if F (H,H ) B F (H , H ).(4) For ,

P with time coordinate less than or equal to the maximum

in B , deneC Q () 0 qQ if and only if C Q (H ) B qC Q () 0 C Q ( ) if and only if either both are 0 the same qQor C Q (H ) B C Q (H ).(5) The description of when C () 0 C ( ) is somewhat more compli-

cated. First, for space-time elements , , dene to mean = T and either R(,H ) 0 1 or R(H, ) 0 1. Further, dene to mean .

Note that if then = , and if 0 then 0 .Now dene if and only if there is a nite sequence of -moves from

to , i.e., if and only if there exist 1, 2, . . . , k such that = 1 2 3 . . . k = . Similarly dene .

Now, deneC () 0 C ( )

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if and only if there exist natural numbers k n and 1, 2, . . . , n p withtime coordinates less than or equal to the maximum in B such that(i) = 1, or C () B C (1)

and (ii) i i+1 for i odd, i k i i+1 for i odd, i > k i 0 i+1 for i even

and (iii) n = or C (n ) B C ( ).Note that if such a sequence exists then the length of the shortest possible

such sequence (including the lengths of the sequences involved in the and

parts) is bounded above by twice the sum of the maximum time coordinatesof elements in B . Hence we can decide, given and , whether such asequence exists.

Further, deneC () 0 a

if and only if there exists B with C () 0 C ( ) as above and C ( ) Ba.

With this denition, the congruence 0 (up to the maximum time coor-dinate in B) satises the laws IV. The identities implied by laws II and IIIfor C are also satised because they only involve C H (), and all the H ()belong to B.

5.6 COMPLETION OF THE DEFINITION OF 0Now, we complete the denition of 0 for elements of A which are, or

which involve, space-time elements with time coordinate larger than the max-imum in B, by induction on the time coordinate.

Suppose we have described 0 as above for pairs (, ), (R(, ), R( , )),etc. whenever the P -elements , , etc. in the space-time coordinate of ghave time coordinate less than or equal kg . The following describes 0 forthose elements in the space-time coordinate of g involving time coordinatekg + 1 .

1. (i) For , A, in the same space-time component, say that of g,with time coordinate kg + 1, dene

0 if and only if either = or = T , = T and 0 or = S n HT kg +1 (g) and

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either there exist qQ, a with C Q HT kg

(g) 0 q and C HT kg

(g) 0aand = S n T for 0 S (q,a )HT kg (g)or there exists q Q with C Q HT kg (g) 0 qL and = S n T for 0

S 1HT kg (g)or there exists qQ with C Q HT kg (g) 0 qRand = S n T for 0 SH T kg (g)or = S n N H (s,t,T kg (g)) and s 0 C Q T kg (g)and t 0 C T kg (g).or = S n N H (t1, t2, T kg (g)), = S n N H (s1, s2, T kg (g))and t1 B s1, t2 B s2or vice-versa (with , switched).(ii) For , in different space-time components, say in the space-time

component of g and in the space-time component of y, with time coordi-nates kg + 1 and less than or equal ky + 1 respectively, dene 0 if andonly if one of the following holds:

Case 1 = S n T kg +1 (g) and there exists in the space-time componentof y with 0 T kg (g) and 0 S n T by the preceeding description forthe same space-time component.

Case 2 = S n T m +1 HT k (g) and there exists in the space-time compo-nent of y with 0 T m HT k (g) and 0 S n T .

Case 3 = S n HT kg +1 (g) and there exists in the space-time componentof y with 0 T kg (g) and 0 S n H T .

Case 4 = N H (t1, t2, T kg (g)) and there exists in the space-time com-ponent of y with 0 T k (g) and 0 N H (t1, t2, ).

Case 5 = S n T m +1 N H (t1, t2, T k (g)) and there exists in the space-timecomponent of y with 0 T m N H (t1, t2, T k (g)) and 0 S n T .

2 (i) Dene R(, ) 0 0 where has time coordinate kg + 1 relative tog and has time coordinate less than or equal to ky + 1 relative to y (orvice-versa) if and only if 0 as in 1.

(ii) Dene R(, ) 0 1 for , as in (i) if and only if there exists n > 0

and P with 0 S n

and R(, ) 0 0 (i.e., 0 ).(Note that we can nd all possible values for and hence can decidewhether these conditions are satised.)

(iii) Dene R(, ) 0 R( , ) (where all of , , , have time coordi-nate less than or equal to the relevant kg + 1 and one of them has that time

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coordinate) if and only if either both R(, ) and R( , ) are 0 0 or bothare 0 1 by (i) or (ii) respectively, or 0 and 0 .3. For , as in 2 we deneF (, ) 0 0F if and only if H 0 H.Further, we will dene F (, ) 0 1F if and only if F (H , H ) 0 1F , so

it is enough to consider = HT k (g), = HT m (y).If k kg and m = ky + 1 then deneF (, ) 0 1F if and only if F (T k (g), T ky (y)) 0 0F or 1F .If k = kg + 1 and m ky + 1 then deneF (, ) 0 1F if and only if there exists n with 0 < n m and HT k (g) 0

HT m n (y).Finally, dene F (, ) 0 F ( , ) if and only if either both are 0 to 0F

or both are 0 to 1F by the above, or 0 and 0 , or both and have time coordinate less than the induction step, and there exists n 0 withH = HT n + p(g), H = HT n + m (y) and F (,HT p(g)) 0 F ( , HT m (y)) .

4. For in the space-time component of g with time coordinate kg + 1,we dene C Q () 0 q Q if and only if C Q (H ) 0 q Q, and for H ,which is just HT kg +1 (g), we dene C Q (H ) 0 q Q if and only if eitherC Q HT kg (g) 0 qR or qL , or there exists q Q, a with (q , a) = q andC Q HT kg (g) 0 q and C HT kg (g) 0 a.

Then, dene C Q () 0 C Q ( ) if and only if either H 0 H or bothC Q () and C Q ( ) are 0 to some qQ by the preceding paragraph.

5. Now, for in the space-time component of g with time coordinatekg + 1, we dene

C () a if and only if Case 1 = T HT kg (g) andeither there exists qQ, b with (q, b) = a and

C (HT kg (g)) 0 b and C Q (HT kg (g)) 0 q

or a = eR and there exists qQ with C Q (H T kg (g)) 0

qRor a = eL and there exists qQ with C Q (H T kg (g)) 0 qLCase 2 There exists with = T and either R(,H ) 0 1 or R(H, ) 0

1 and in addition C ( ) 0 aCase 3 There exists in another space-time component with C ( ) 0 a

by the above two cases and 0 .

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Finally, we dene C () 0 C ( ) if and only if either both are 0some a by the preceding denition, or 0 , or, if is in the space-time component of g with time coordinate kg + 1 and is in the space-timecomponent of y with time coordinate less than or equal to ky and thereexists with = T , and either R(,H) or R(H,) 0 1 and in additionC () 0 C ( ) (or vice-versa with the roles of and reversed).

This completes the denition of ( A0, 0). As the notation suggests 0 isan equivalence relation, in fact a partial congruence. In particular transitivityis taken care of in the inductive construction. Furthermore it should be notedthat 0 is decidable.

5.7 DEFINITION OF ( An , n )Now, suppose we have dened, for each m n, a decidable partial sub-

algebra (Am , m ) such that Am Am +1 , m = m +1 |Am , m is a partialcongruence, the laws of our variety are contained in m insofar as they ap-ply to the elements of Am , and in addition

(i) Am is closed under P , H , S , S 1, T (modulo normal form) and forall a Am and any operation among P , C , C Q , R, F , if a is n -equivalentto an element in the image of the operation then it is m -equivalent to anelement in the image of that operation.Further we construct by induction algorithms which

(ii) given space-time elements, , Am , determine whether there existsk with m S k .

(iii) given space-time elements , Am , determine whether there existsk > 0 with m T k .

(iv) given a space-time element Am determine whether it is m tosome element of A0, and if so, produces such an element .

(v) given an element a Am and an operation among P , C , C Q , R, F ,determine if a is m to an element in the range of the operation.

The base case of the construction is n = 0. Note that ( A0, 0) has therst four of these ve properties: (i) follows from the denition of A0 and

the fact that P (c) A0.(ii) is seen as follows, for, A0 : we can effectively list as 1, . . . , n ,the nitely many elements in A0 to which is 0. For some k, 0 S k if and only if for some i, i and have the same time prex. The lattercondition can be effectively checked.

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(iii) Let 1, . . . , n be as above. Then there is k so that 0 T k

if andonly if some i is in the space-time component of , k0 is the difference of the time coordinates of i and , and i 0 T k 0 .

(iv) is trivial for A0.

Property (v) is a bit trickier. Since an element a is n -equivalent to anelement in the range of P if and only if a n P (a), only the other operationspresent any problems. Below we will state an inductive hypothesis on theequivalence n . The inductive hypothesis will have two uses. First it willallow us to verify property (v) and the second part of (i) by giving a completedescription of which elements are n -equivalent to an element in the range

of a non-space-time operation. Second the hypothesis will determine n , therestriction of n +1 to An (dened below). In the remarks after the inductivehypothesis for elements in the image of R, we will expand on these points.The reader will be able to observe the inductive hypotheses hold for the casen = 0 and so (v) holds as well.

Let An be An together with the image of all the elements of An under theoperations R, F , U , C Q , N Q , C , N , C , E , K .

Extend n to a partial congruence n on An by considering each opera-tion in turn. First extend it to An together with the image of R by letting itbe the unique symmetric relation extending n which satises the inductivehypothesis. Continuing, given an operation we dene n on the image of that operation, An and the operations previously considered, by letting it bethe unique symmetric relation which satises the inductive hypothesis andextends the restriction of n previously dened. Transitivity will be an easyconsequence of the denition since we will always link to a previous n .

(1) Image of RInductive HypothesisR(s, t ) n 0 if and only if either for some 1, 2 A0, P (s) n 1, P (t)

n 2 and R(1, 2) 0 0 or P (s) n P (t).R(s, t ) n 1 if and only if either for some

1,

2A

0P (s) n

1, P (t) n

2 and R(1, 2) 0 1 or there is some k > 0 such that P (t) n S k P (s).R(s, t ) n R(s , t ) if and only if either for some 1, 2, 3, 4 A0,

P (s) n 1, P (t) n 2, P (s ) n 3, P (t ) n 4 and R(1, 2) 0 R(3,4) or R(s, t) n 0 n R(s , t ) (as above) or R(s, t ) n 1 n R(s , t ) (asabove) or P (s) n P (s ) and P (t) n P (t ).

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R(u, v) n U (s) if and only if either there are t, 1, 2 B so that t n s,P (u) n 1, P (v) n 2 and U (t) 0 R(1, 2) or there are 1, 2 An sothat s n R(1, 2) and R(1, S 2) n R(u, v).

R(s, t) n u if and only if either one of the above cases holds or there isv B such that R(s, t ) n v as above, and v n u.

Remark. Note rst that every equivalence on the right hand side of theinductive hypothesis either concerns elements in An or is covered in earlierclauses. As to the decidability of the relation, at various points we needto know if there are elements with a certain property. For example in thefourth clause we ask whether there are 1, 2 so that s n R(1, 2) andR(1, S 2) n R(u, v). By property (v) we can tell if there are

1,

2so

that s n R(1, 2). Since n is a partial congruence, if there are 1, 2so that s n R(1, 2) and R(1, S 2) n R(u, v) then for all 1, 2, s nR(1, 2) implies that R(1, S 2) n R(u, v). Hence we have an algorithmfor answering the question. Similar comments apply throughout.

There remains property (v) to consider. It is enough in view of theinductive hypothesis to be able to decide when elements of the form U (s)and members of B are in the range of R. Since B is nite, we can assumewe know the answer for elements of B relative to A0. By property (i), wewill then know the answer for all An , if we can settle the case n = 0. By thefourth clause, we can reduce the question to either elements of B or elementsof A0. In A0 all the elements of the form U (s) are either in B , 0-equivalentto an element in the range of R or 0-equivalent to an element in the rangeof F . Since no element in the range of R can be 0-equivalent to an elementin the range of F , we can decide whether a given element of A0 in the rangeof U is 0-equivalent to an element in the range of R. Such considerationsrecur throughout.

(2) Image of F Inductive HypothesisF (s, t ) n 0F if and only if either there are 1, 2 A0 so that H (s) n

1, H (t) n 2 and F (1, 2) 0 0F or H (s) n H (t).F (s, t ) n 1F if and only if either there are 1, 2 A0 so that H (s) n

1, H (t) n 2 and F (1, 2) 0 1F or there is k > 0 H (t) n HT k (s).F (s, t ) n F (s , t ) if and only if either there are 1, 2, 1, 2 A0 so

that H (s) n 1, H (t) n 2, H (s ) n 1, H (t ) n 2 and F (1, 2) 0F (1, 2) or F (s, t ) n 0F n F (s , t ) (as above) or F (s, t ) n 1F n F (s , t )

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(as above) or H (s) n H (s ) and H (t) n H (t ).F (u, v) n U (s) if and only if either there are t, 1, 2 B so that t ns, H (u) n 1, H (v) n 2 and U (t) 0 F (1, 2) or there are 1, 2 so thats n F (1, 2) and F (1, T 2) n F (u, v).

F (s, t ) n u if and only if either one of the above cases holds or there isv B such that F (s, t ) n v as above, and v n u.

(3) Image of U Inductive HypothesisU (s) n 0 if and only if either for some t B, s n t and U (t) 0 0 or

there are 1, 2 An so that s n R(1, 2) and R(1, S 2) n 0U (s) n 1 if and only if either for some t B, s n t and U (t) 0 1 or

there are 1, 2 An so that s n R(1, 2) and R(1, S 2) n 1U (s) n 0F if and only if either for some t B, s n t and U (t) 0 0F

or there are 1, 2 An so that s n F (1, 2) and F (1, T 2) n 0F U (s) n 1F if and only if either for some t B, s n t and U (t) 0 1F

or there are 1, 2 An so that s n F (1, 2) and F (1, T 2) n 1F U (s) n R(u, v) if and only if either there are t, 1, 2 B, so that t n

s, P (u) n 1, P (v) n 2 and U (t) 0 R(1, 2) or there are 1, 2 Anso that s n R(1, 2) and R(1, S 2) n R(u, v).

U (s) n F (u, v) if and only if either there are t, 1, 2 B, so that t n s,

P (u) n 1, P (v) n 2 and U (t) 0 F (1, 2) or there are 1, 2 An sothat s n F (1, 2) and F (1, T 2) n F (u, v).

U (s) n U (t) if and only if either U (s) n 0 n U (t) (as above) orU (s) n 1 n U (t) or U (s) n 0F n U (t) or U (s) n 1F n U (t) or s n tor there are s , t B so that s n s , t n t and U (s ) B U (t ).

U (s) n u if and only if either one of the above cases holds or there isv B such that U (s) n v as above, and v n u.

(4) Image of C QInductive HypothesisC Q (s) n q for qQ if and only if there is A

0such that H (s) n

and C Q () 0 q.C Q (s) n C Q (t) if and only if either there are 1, 2 A0 so that H (s) n

1, H (t) n 2 and C Q (1) n C Q (2) or H (s) n H (t)C Q (s) n N Q (u, v, w) if and only if either H (w) n w, HT (w) n H (s),

C Q (w) n u and C (w) n v or for some qQ, C Q (s) n q n N Q (u,v,w )

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or there are A0, and u0, v0, w0 B so that H (s) n , u0 n u, v0 n v,w0 n w and C Q () 0 N Q (u0, v0, w0).C Q (s) n u if and only if either one of the above cases holds or there is

v B such that C Q (s) n v as above and v n u.Remark. In the third clause we have to decide whether C Q (w) n u andC (w) n v. In order that n be well dened we need to know that if N Q (u,v,w ) An then C () An where is the normal form of H (w).This can be veried by induction on n.

(5) Image of N QInductive Hypothesis

N Q (s, t ,u ) n q Q if and only if either there exist p Q and a such that s n p, t n a , H (u) n u and ( p,a) = q or s n qL or qR ,H (u) n u and t n C (u) or u n H (u), s n C Q (u), t n C (u) andC Q (T (u)) n q or there are s , t , u B so that s n s , t n t , u n uN Q (s,t,u ) B and N Q (s , t , u ) B q.

N Q (s, t ,u ) n N Q (s , t , u ) if and only if either for someqQ, N Q (s,t,u ) nq n N Q (s , t , u ) or s n s , t n t and u n u or u n H (u), u n H (u ),s n C Q (u), s n C Q (u ), t n C (u), t n C (u ) and C Q T (u) nC Q T (u ) or there exist b B b such that N Q (s, t ,u ) n b (as above) andN Q (s , t , u ) n b .

N Q

(u,v,w ) n

C Q

(s) if and only if either H (w) n

w, C Q

(w) n

u,C (w) n v and C Q T (w) n C Q (s) or for some q Q, C Q (s) n q nN Q (u,v,w ) or there is some b B so that N Q (u,v,w ) n b (as in theparagraph above) and C Q (s) n b.

N Q (s, t ,u ) n v if and only if either one of the above cases holds or thereis w B such that N Q (s, t ,u ) n w as above, and v n w.

(6) Image of C Inductive HypothesisIf R(s, H (s)) n 1 and P (s) is not equivalent to an element of A0 then

R(T (s), HT (s)) is not n 1 and dually for R(H (s), s).

C (s) n a if and only if there are A0 such that n P (s) andC () 0 a.C (s) n C (t) if and only if either there are 1, 2 A0 so that P (s) n

1, P (t) n 2 and C (1) 0 C (2) or P (s) n P (t) or P (s) n T P (t)and R(t, H (t)) n 1 or R(H (t), t ) n 1 or the last condition holds with theroles of s and t reversed.

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C (s) n N (u,v,w ) if and only if either there is a , so that C (s) na n N (u,v,w ) or u n C Q (w) and v n C (w), w n H (w) and C (s) nC (T (w)) or there are A0, u , v , w B such that P (s) n , u n u ,v n v , w n w , N (u , v , w ) B and C () n N (u , v , w ).

C (s) n C (t, u) if and only if either for some a , C (s) n a nC (t, u ) or u n R(H (t), t ) or u n R(t, H (t)) and C (s) n C (T (t)) oru n 1, t n P (t) and C (s) n C (t) or there are t , u B and A0 sothat t n t , u n u , C (t , u ) B, P (s) n and C (t , u ) 0 C ().

C (s) n u if and only if either one of the above cases holds or there isv B such that C (s) n v as above and v n u.

(7) Image of N Inductive HypothesisN (s, t ,u ) n a if and only if either there are qQ, b, such that

u n H (u), (q, b) = a and s n q, t n b or a = eL (resp. eR ) and there isq Q, such that u n H (u), s n qL (resp. qR ), t n C (u) or H (u) n u,C Q (u) n s, C (u) n t and a n C T (u) or there are s , t , u B so thats n s , t n t , u n u , N (s , t , u ) B and N (s , t , u ) B a.

N (s, t ,u ) n N (v,w,z ) if and only if either they are both n -equivalentto the same a or s n v, t n w and u n z or H (u) n u, C Q (u) n s,C (u) n t, H (z) n z, C Q (z) n v, C (z) n w and C T (u) n C T (z)or there are b

Bb B so that N

(s, t ,u )

nb and N

(v,w,z )

nb (as

above).N (u,v,w ) n C (s) if and only if either there is a , so that C (s) n

a n N (u,v,w ) or u n C Q (w) and v n C (w), w n H (w) and C (s) nC (T (w)) or there is A0, u , v , w B such that P (s) n , u n u ,v n v , w n w , N (u , v , w ) B and C () n N (u , v , w ).

N (s, t ,u ) n C (v, w) if and only if either there is a so thatC (v, w) n a n N (s, t ,u ) or there is some z so that C (v, w) n C (z) nN (s, t ,u ) or there is some b B so that N (s,t,u ) n b (as above) andb n C (v, w).

N (s, t ,u ) n v if and only if either one of the above cases holds or thereis w B such that N (s,t,u ) n w as above and v n w.

(8) Image of C

Inductive HypothesisC (u, v) n a if and only if either there is u n P (u), v n

R(u, H (u)) or R(H (u), u) and C T (u) n a or v n 1, u n P (u) and

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P (u) n a or there exist u , v B so that u n u , v n v , C

(u , v ) Band C (u , v ) B a.C (t, u ) n C (s) if and only if either for some a , C (s) n a n

C (t, u ) or u n R(H (t), t ) or u n R(t, H (t)) and C (s) n C (T (t)) oru n 1, t n P (t) and C (s) n C (t) or there are t , u B and A0 sothat t n t , u n u , C (t , u ) B, P (s) n and C (t , u ) 0 C ().

C (s, t ) n C (u, v) if and only if either both are n -equivalent to thesame a or there are 1, 2 so that C (s, t ) n C (1), C (u, v) n C (2)and C (1) n C (2) or s n u and t n v or there are b1 B b2 B sothat C (s, t ) n b1 and C (s, t ) n b2 as above.

C (v, w) n N (s,t,u ) if and only if either there is a so thatC (v, w) n a n N (s, t ,u ) or there is some z so that C (v, w) n C (z) nN (s, t ,u ) or there is some bB so that N (s, t ,u ) n b and b n C (v, w)(as above).

C (s, t ) n u if and only if either one of the above cases holds or there isv B such that C (s, t ) n v as above and u n v.

(9) Image of E Inductive HypothesisE (s) n 1 if and only if either for some , s n C Q () or there is t B

so that s n t and E (t) B 1.E (s)

n0 if and only if either s

nh or there is t B so that s

nt

and E (t) B 0.E (s) n E (t) if and only if either s n t or E (s) n 1 n E (t) or

E (s) n 0 n E (t) or there are b1 B b2 so that E (s) n b1 and E (t) n b2(as above).

E (s) n t if and only if either one of the above cases holds or there is uB such that E (s) n u (as above) and u n t.

(10) Image of K

Inductive HypothesisK (s, t ) n 0 if and only if either s n t and either s is n -equivalent

to a constant or P (s) n s or s is equivalent to an element in the imageof C , C Q , R or F or there are b1, b2 B so that s n b1, t n b2 andK (b1, b2) B 0.

K (s, t ) n 1 if and only if either P (s) n s and there is a constant d = cwith t n d or s is equivalent to an element in the range of C and there isa constant d / so that t n d or s is equivalent to an element in the range

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of C Q and there is a constant d /Q so that t n d or s is equivalent to anelement in the range of R and there is a constant d = 0 , 1 so that t n d or sis equivalent to an element in the range of F and there is a constant d = 0 F ,1F so that t n d or s is equivalent to an element in the range of one of theoperations P , C , C Q , R, F and t is equivalent to an element in the rangeof a different one of these operations or there are b1, b2 B so that s n b1,t n b2 and K (b1, b2) B 1.

K (s, t ) n K (u, v) if and only if either K (s, t ) n 0 n K (u, v) orK (s, t ) n 1 n K (u, v) or s n u and t n v or there is b1 B b2 so thatK (s, t ) n b1 and K (u, v) n b2 (as above).

K (s, t ) n u if and only if either one of the above cases holds or there isv B such that K (s, t ) n v (as above) and u n v.

5.8 DEFINITION OF ( An , n )

Dene An and n as follows:for n even:

An = An {N H (s,t,H )|s, t An , An space-time }for n odd:

An = An {K (s, )|s An , An space-time }

We will extend n to a partial congruence n on An . First, for n even,we dene N H (s,t,H ) An An to be reducible (to u) if and only if one of the following holds:

(i) there exists s , t , with u = N H (s , t , H ) B and s n s , t n tand H n H .

(ii) s n qQ, and t n a , and u = S (q,a )T H ()(iii) s n qL and t n C H () and u = S 1T H ()(iv) s n qR and t n C H (), and u = ST H ()(v) s n C Q H (), t n C H () and u = HT.

Now dene, for n even,N H (s,t,H ) n N H (s , t , H ) (for both N H (s,t,H ) and N H (s , t , H )

An

An ) if and only if either s n

s , t n

t , H H or both N H (s,t,H )and N H (s , t , H ) are reducible, to u, u respectively , and u n u

N H (s,t,H ) n v An if and only if either N H (s,t,H ) is reducible tou, and u n v, or v = N H (s , t , H ) and s n s , t n t , H n H .

Similarly, for n odd, we dene K (s, ) An An to be reducible (to u)if and only if one of the following hold

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(i) there exist s , with u = K (s , ) B and s n s, n (ii) s n 0 or 0F and u = (iii) s n 1 or 1F and u = P (c).

Now , for n odd dene K (s, ) n K (s , ) (for both K (s, ) and K (s , )and K (s , ) in An An if and only if either s n s and n or bothK (s, ) and K (s , ) are reducible to u, u respectively, and u n u .

K (s, ) n v An if and only if K (s, ) is reducible to u, and u n vor v = K (s , ) and both K (s, ) and K (s , ) are irreducible, and s n s , n .

Finally, we dene An +1 and n +1 to extend An and n as follows:For n even: An +1 = An {S n T m | An An , n Z , m 0}.For n odd:

An +1 = An {S n T m | An An , n Z , m 0}

{S n T m HT k | An An , n Z ,m,k 0}.

5.9 DEFINITION OF ( An +1 , n +1 )

We extend n to n +1 on An +1 as follows:

For n even:

S n

T m

n +1 S n

T m

if and only if n = n , m = m and n andS n T m n +1 v An if and only if is reducible, to u, and S n T m u n v.For n odd:S n T m n +1 S n T m if and only if n = n , m = m , and n .S n T m HT k n +1 S n T m HT k if and only if n = n . m = m , k = k ,

and = .Finally, S n T m n +1 v An if and only if reduces to u and S n T m u n

v, and similarly with S n T m HT k .

This completes the denition of ( An +1 , n +1 ). It is straightforward to checkthat ( An +1 , n +1 ) is a decidable partial congruence satisfying (i), (ii), (iii),

(iv) above. Property (v) can be veried in the other cases as it was after thedenition of the inductive hypothesis on R. Also, as discussed in subsection5.1, n 0 n is precisely P restricted to n 0 An . The An and n areuniformly decidable in n. Together this means that n 0 An and n 0 n aredecidable. Since n 0 An contains all terms, modulo (effectively) reducingspace-time terms to their normal forms, this completes the proof.

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If we combine Theoerems 3.1, 4.1 and 5.1, we obtain the following theo-rem.

Theorem 5.5 There is a nitely based variety of nite type which has solvablebut not uniformly solvable word problem.

6. A RECURSIVELY BASED VARIETY DEFINED BY LAWS INVOLV-ING NO VARIABLES.

6.1 DEFINITION OF THE VARIETY

In this section we will describe a recursively based variety of nite type,dened by laws which involve no variables, which has solvable but not uni-formly solvable word problem.

The variety is a modication of the nitely based variety dened in thepreceding sections. The use of innitely many axioms allows us to use asimpler picture of space-time.

The operations are the same, except that P , C , and U are deleted andK and K are identied. Specically, the operations are:

constants: c, all a , all qQ, 0, 1, 0F , 1F unary: T , S , S 1, H , C , C Q , E binary: F , R, K

ternary: N H , N Q , N .Dene, for each k, H k = HT k (c), and let = {S n T m (H k )|n Z ,m,k

N }. These will be the space-time elements.The laws dening the variety are as follows:

I. H (c) c,T (S n T m (H k )) S n T m +1 (H k ),S (S n T m (H k )) S n +1 T m (H k ),S 1(S n T m (H k )) S n 1T m (H k ),H (S n T m (H k )) H m + k for all k, m 0 and n Z .

II. N Q (q,a,H k ) (q, a) for all qQ, a , k 0,N H (q,a,H k) S (q,a ) T (H k ) for all qQ, a , k 0,N (q,a,H k ) (q, a) for all qQ QLR , a , k 0.N Q (qL , C (H k ), H k ) q N Q (qR , C (H k ), H k ),N H (qL , C (H k ), H k ) ST (H k ),N H (qR , C (H k ), H k ) S 1T (H k ),

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N (qL , C (H k ), H k ) eL ,N (qR , C (H k ), H k ) eR , for all qQ \ QLR and k 0.

III. C T (H k ) N (C Q (H k ), C (H k ), H k ),C Q T (H k ) N Q (C Q (H k ), C (H k ), H k ),H k +1 N H (C Q (H k ), C (H k ), H k ) for all k 0.

IV. C Q () C Q H () for all .

V. R(, ) 0 for all ,R(, S k ()) 1 for all k > 0, and ,F (, ) 0F for all ,

F (, T k

()) 1F for all and k > 0,F (, ) F (H,H ( )) for all , .

VI. C (S n T H ()) C (S n H ()) for all and n = 0.

VII. K (0, ) for all ,K (1, ) c for all ,K (d, d) 0 for all constants d,K (d, e) 1 for all constants d = e = c,K (, d ) 1, for all and constants d = c,K (C (), d) 1 for all constants d /, and ,K (C Q (), d) 1 for all constants d /Q, and ,K (R(, ), d) 1 for all , , and d = 0 , 1,K (F (, ), d) 1 for all , , and d = 0 F , 1F .

K (t, t ) = 0 and K (s, t ) = 1 for all s, t where s, t belong to differentmembers of the following list of sets: , {C ()| }, {C Q ()| },{R(, )|, }, {F (, )|, }.

VIII. EC Q () 1 for all E (h) 0.

Note that every term generated from c by the operations S , S 1, T andH is equivalent modulo the above laws I to an element of ; in fact thereis an effective procedure which, given such a term t, produces with tequivalent (modulo I) to . Thus we may, and will, ignore all such terms texcept those in .

6.2 NON-UNIFORM SOLVABILITY OF THE WORD PROBLEM

Proposition 6.1 V does not have uniformly solvable word problem.

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Proof. The proof is analogous to the proof of Theorem 3.1: for an initial tapeconguration as described there, we have associated the same presentation P and prove that the universal Turing machine, started on that conguration,eventually halts if and only if 0 P 1.

As in the proof of Theorem 3.1, the only if part is clear.For the if direction, if the machine does not halt, we again produce a

model A V satisfying all the P equations, in which 0 = 1 .The underlying set of A is as in the proof of Theorem 3.1, and the oper-

ations are as dened there, with the following changes:(ii), (iii) and the denition of C are deleted (since we have deleted the

operations P , U and C ).In (v) R(x, y) = unless both x, y .In (vi) F (x, y) = unless both x, y .In (vii) K (x, y) = for x, y not in the form of the rst two lines.In (viii) replace K by K .

6.3 SOLVABILITY OF THE WORD PROBLEM

The proof that V has solvable word problem is somewhat different thanthe proof in section 5.

We again differentiate two cases: whether or not P has degenerate space-time, which in this case means P c for all .

In the degenerate case, we have ( , c) P for all , and hence theequations dening our variety are equivalent (modulo P ) to nitely manyequations, namely S (c) = T (c) = S 1(c) = c together with all instances of the equations dening the variety with c substituted for the arbitrary which appear. Thus P is nitely generated relative to the variety of allalgebras, and hence is decidable by Corollary 1.2.

In the non-degenerate case we proceed, at rst, similarly to section 5,bearing in mind that there are fewer space-time elements (see above denitionof ), but time coordinates, time prexes, and space prexes are dened asbefore, except that all these notions are always relative to c = H (c), i.e. theonly space-time component is that of c.

The proof of the next lemma is essentially the same as the proofs of Lemmas 5.2 and 5.3.

Lemma 6.2 For any nite subset F , with maximum time coordinate m,there is a nite F with the same maximum time coordinate such that

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(i) F F (ii) if is a right subterm of F then F (iii) if and P for F then F (iv) if and T F then F .

Denition of A:A consists of all terms appearing in the equations dening the variety,

and all subterms thereof, plus all elements C (), and K (, ) for , ,(the elements C Q () are already included).

Let A be the restriction to A of the congruence dening our variety,then, because we have no non-trivial information about C and C Q , rules

III cannot be applied in a non-trivial way, and hence A is decidable, asin condition (3) of Proposition 1.1. Thus ( A, A ) is a partial subalgebrasatisfying the hypotheses of Proposition 1.1.

Denition of B :Let BP consist of all terms appearing in the presentation P , and all

subterms thereof, and all constants.Enlarge BP as follows:(i) For each bBP , if there exists a A with aP b, add one such a, and

choose a whenever possible.(ii) Let F consist of all the elements of we have so far, and add the set

F of Lemma 6.2.(iii) For all , F , add C Q (), C (), R(, ), F (, ).

The resulting set is B . It is closed under taking subterms, and for , , if B and P then B.

Let B be P |B, then B and B are both nite, and hence decidable.

Denition of A0:Let A0 = AB, and let 0 be the partial congruence on AB generated

by A B ; we are going to show that (A0, 0) is a partial subalgebrasatisfying the hypothesis of Proposition 1. Since P is generated by A Band hence by 0, the decidability of P will then follow from Proposition 1.1,the proof of which is deferred to the next subsection.

Since membership in A is decidable, and B is nite, we know that mem-bership in A0 is decidable.

Next, we need to establish that 0 is decidable; this, however, can beproved analogously to the proof in section 5 that 0 (as dened there) is

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decidable, deleting from that proof consideration of elements which we donot have in this example, such as elements in the image of U or C andspace-time elements except those in the presently dened .

It remains to check that ( A0, 0) satises hypothesis (3) of Proposition1.1, i.e., that there is an algorithm which, given an operation of arity n,and a1, . . . , a n A0, determines whether there exist b1, . . . , bn A0 witha i 0 bi and (b1, . . . , bn ) A.

Now, B is nite, and hence we can check all elements of the form (b1, . . . , bn ) B, and decide whether a i 0 bi .

Thus it is enough to decide whether there exist b1, . . . , bn A with(b1, . . . , bn ) A and a i 0 bi . Moreover, if some a i B A then a i BP and so if there exists bi A with a i 0 bi then there exists ci A B witha i 0 ci and we may replace a i by ci . Thus we may assume without loss of generality that all a i A.

Hence we have reduced the problem to the following:

given a1, . . . , a n A do there exist b1, . . . , bn A with a i 0 biand (b1, . . . , bn ) A?

There is an effective procedure which, given a A, produces d A suchthat d A a and either d , d is a constant, or d C Q (), d C (),

d R(, ), or d F (, ). Thus we may assume that each a i is already of this form.

We consider the operations in turnT : For a A, if there exists bA with b 0 a and T (b) A then b

hence a . Thus there is such a b if and only if a .S, S 1 and H are the same as T.R: For a1, a2 A, if there exist bi 0 a i with bi A and R(b1, b2) A

then bi and hence a i , conversely if a i then R(a1, a2) A.F : Is the same as R.C Q : For bA, C Q (b) A if and only if b, hence there exists b 0 a

with C Q (b) A if and only if a C : same as C Q .N Q : If a1, a2, a3 A and there exist bi 0 a i with N Q (b1, b2, b3) A,

then b3 = H k for some k, then because a3 0 b3 we must have that the timecomponent of a3 is k so we can just check whether a3 H k . If the answeris affirmative, then for b1 and b2 we have N Q (b1, b2, H k ) A if and only if

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either a1 0 q Q and a2 0 a , or a1 0 qL or qL or C Q (H k ) anda2 0 C (H k ); there are only nitely many cases to check.N , N H : The argument is the same as for N Q .E : For a A, if there exists bA with E (b) A then either a 0 h or

a 0 C () for some . The latter occurs if and only if either a im(C )or a 0 to some d or a 0 C (b) B; these nitely many cases can bechecked.

K : If a1 0 b1 and a2 0 b2 and K (b1, b2) A then there are variouspossibilities:

(1) b2 , hence a2 , and a1 0 0 or a1 0 1 or a1 is in or in theimage of C , C Q , R or F , or a1 0 an element in the image of one of theseoperations in B . These nitely many cases can be checked.

(2) a1 and a2 are each 0 some constant (possibly different ones).(3) a1 and a2 is either 0 some constant not equal to c, or a2 , or

a2 is in the image of C , C Q , R or F , or a2 is 0 to an element in the imageof one of these operations in B.

(4) a1 is in the image of C or is 0 an element of B which is in the imageof C , and a2 0 d, a constant /.

(5) Similar to (4), with C replaced by C Q , or R, or F respectively, withthe appropriate constraint on the constant d 0 a2.This completes the proof.

Applying Proposition 1.1, we have proved the following theorem.

Theorem 6.4 There is a variety in a nite language dened by a recursive set of laws involving only constants which has solvable but not uniformly solvableword problem.

6.4 PROOF OF PROPOSITION 1.1

We complete this section with the promised proof of Proposition 1.1.

Proof (of Proposition 1.1). We rst produce a partial subalgebra ( B, B )satisfying (1) to (3), such that A B, and A B , which has the feature

that if a i B bi for 1 i n and if (a1, . . . , a n ) B then (b1,... , bn ) B.B and B are dened by induction on the complexity of terms, as follows.Let B0 = A and 0 be A . For each natural number k, let

Bk +1 = Bk {(b1, . . . , bn )| , bi Bk and there exists a i k biwith (a1, . . . , a n ) A},

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Rk+1 = {((b1, . . . , bn ), a)| , bi Bk , a A and there exists a i kbi with a = (a1, . . . , a n )}.Let k+1 = A ( A Rk+1 ) (R 1k +1 A ) (R

1k +1 A Rk +1 ).

Dene B = Bk (k ) and dene B = k (k ).Note that each Bk is closed under subterms.We will prove the following by induction on k:(i) k k +1 .(ii) If bBk and b k a, b k c for a, c A then a A c.(iii) k |B i = i for all i < k.(iv) k is transitive.

(v) k is a partial congruence on Bk .k = 0: trivial.Induction Step: Suppose we have (i) to (v) for k.(i) Then Rk +1 Rk+2 and hence k +1 k +2 .(ii) Suppose bBk+1 and b k +1 a, b k+1 c for a, c A. Note that

Rk+1 |A A , and hence if bA then b A a and b A c so a A c.Assume b /A. Then b = (b1, . . . , bn ) and there exist a i k bi with

(a1, . . . , a n ) A and (a1, . . . , a n ) A a. Similarly there exist ci k bi with(c1,...,cn ) A and (c1,...,cn ) A c. But by the induction hypothesis weget a i A ci and hence (a1,...,a n ) A (c1,...,cn ) so a A c.

(iii) It is enough to prove that k +1

|Bk

= k

, and for this it is enoughto prove that Rk+1 |Bk Rk (or A if k = 0). However, if ( (b1, . . . , bn ), a) Rk+1 and b = (b1, . . . , bn ) Bk then there exist a1, . . . , a n A with bi k a iand a = (a1, . . . , a n ). If k = 0 then we have b A and hence (b, a) A .If k > 0 then b BK implies that b1, . . . , bn Bk 1 and so the inductionhypothesis yields bi k 1 a i and hence (b, a) k .

(iv) is an direct consequence of (ii).(v) For k = 0 this is just the hypothesis on ( A, A ), since we have assumed

that A is a partial congruence on A. Suppose bi k +1 di and (b1, . . . , bn ) Bk+1 and (d1, . . . , d n ) Bk+1 . Then bi , di Bk for 1 i n and henceby (iii), bi k di . Also, there exist a i , ci A, 1 i n with bi k a iand di k ci and (a1, . . . , a n ), (c1, . . . , c n ) A. By (iv) and (ii) we obtaina i A ci and hence (a1, . . . , a n ) A (c1, . . . , c n ) and thus (b1, . . . , bn ) k+1(d1, . . . , d n ).

It remains to verify (1), (2), and (3) for ( B, B ).

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Documents

Alan H. Mekler, Evelyn Nelson and Saharon Shelah- A Variety with Solvable, but not Uniformly Solvable, Word Problem