recursive inseparability for residual bounds of finite algebras

Recursive Inseparability for Residual Bounds of Finite AlgebrasAuthor(s): Ralph McKenzieSource: The Journal of Symbolic Logic, Vol. 65, No. 4 (Dec., 2000), pp. 1863-1880Published by: Association for Symbolic LogicStable URL: http://www.jstor.org/stable/2695083 .

Accessed: 14/06/2014 09:53

Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp

.JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact [email protected].

.

Association for Symbolic Logic is collaborating with JSTOR to digitize, preserve and extend access to TheJournal of Symbolic Logic.

http://www.jstor.org

This content downloaded from 195.34.79.223 on Sat, 14 Jun 2014 09:53:54 AMAll use subject to JSTOR Terms and Conditions

http://www.jstor.org/action/showPublisher?publisherCode=asl

http://www.jstor.org/stable/2695083?origin=JSTOR-pdf

http://www.jstor.org/page/info/about/policies/terms.jsp


THE JOURNAL OF SYMBOLIC LOGIC

Volume 65. Number 4. Dec. 2000

RECURSIVE INSEPARABILITY FOR RESIDUAL BOUNDS OF FINITE ALGEBRAS

RALPH McKENZIE

Abstract. We exhibit a construction which produces for every Turing machine T with two halting states ,uo and u - I, an algebra B(T) (finite and of finite type) with the property that the variety generated by B(T) is residually large if T halts in state u I while if T halts in state ao then this variety is residually bounded by a finite cardinal.

For a variety V, we denote by resb(V) the least cardinal i such that every subdirectly irreducible algebra in V has a cardinality < i, if there is such a cardinal, and we write resb(V) = oc if there is not. For an algebra A we write resb(A) in place of resb(V) where V = V (A) is the variety generated by A, and we call this entity the residual bound of A. It was proved in W Taylor [8] and R. McKenzie, S. Shelah [5] that there are just countably many possible values for the residual bound of a finite algebra. In fact, it always belongs to the set

{O ,3, 4,. .. , co, col, (2'),}.

Here, co denotes the least infinite cardinal (or the set of all natural numbers), c91 = co+ is the next larger cardinal after co, and (2w)+ is the successor cardinal to the cardinal of the continuum.

A finite algebra of finite type can be described, up to isomorphism, as a finite list, in which some integers are given (the elements of the algebra), and the tuples making up the basic operations are then listed, in some appropriately readable fashion. Such a list, or algebra-description, is a suitable input for an algorithm which one might hope to design, the purpose of which would be to compute the residual bound of the algebra described by the input. In R. McKenzie [3], it was proved that no such algorithm exists; in fact, the properties resb(A) < co and resb(A) < co are undecidable. C. Latting [1] modified this to show that resb(A) < co, and resb(A) < 2' are undecidable. The main result of this paper is that there exists no recursive set of finite algebras (or algebra-descriptions) which includes all algebras having a finite residual bound and includes no algebra having 00 as its residual bound. This result implies those earlier results of McKenzie and Latting, and also establishes that the property resb(A) = o0 is undecidable.

Received March 23, 1999; revised June 30, 1999. Research supported by NSF Grant No. DMS-9403187.

? 2000. Association for Symbolic Logic 0022-48 12/00/6504-0025/$2.80

1863



1864 RALPH McKENZIE

The author proved that resb(A) = oc is undecidable in May 1995. The proof used Turing machines with one halting state, as had been done for all of the earlier undecidability results for residual bounds. Before this result was written up for publication, R. Willard observed that by using a Turing machine with two halting states and modifying the argument in minor ways, one should get the result on recursive inseparability that we prove in this paper. We are indebted to him for this beautiful idea.

?1. Turing machines. For the purposes of this paper, we need Turing machines with two distinct halting states. Turing machine descriptions will be built out of the following alphabet of symbols. The "tape symbols" 0, 1, the "motion symbols" L and R, and the "internal state symbols" u,, (-1 < n < co). The halting states will be jio and u - 1. Our machines perform operations on a two-way infinite tape ruled into squares on each of which is printed a 0 or a 1. The tape serves both as scratch-paper and memory area for the computation. At any point in the computing process, the next operation to be performed is determined by two things: the current internal state of the machine, and the symbol (0 or 1) that appears on the square currently being scanned by the machine's reading head. An operation consists in changing or leaving fixed the symbol on the scanned square and then moving the reading head one square to the left or to the right, and at the same time changing the internal state.

A "tape" is a function t from Z {0, ? 1, ?2,.. . } to {0, 1}. The symbol "printed" on the n-th square of the tape is t(n). By the blank tape we mean the constant function to(x) = 0.

An "instruction" is any pentuple of the form uirsTy where {r, s} C {0, 1}, T C {L, R}, Hi is any internal state symbol other than j'o and gu-l, and y is any internal state symbol, not excluding juo or u-1. The instruction ,uirsTy "instructs" the machine that if it finds itself in internal state i reading a square with the tape symbol r printed on it, then it must: first, replace r by s on the current square, then if T = L move the reading head one square to the left, and if T = R move the reading head one square to the right, and finally, change its internal state to y.

DEFINITION 1.1. A Turing Machine T is a finite set of machine instructions which, for some k > 1 satisfies the following:

1. T has, for each 1 < i < k and for each r E {0, 1 } precisely one instruction ,uirs Ty.

2. No symbol uj, j > k occurs in the instructions of T.

A T-configuration is a triple Q = (t, n, y) in which t is a tape, n E Z and y is one of the internal state symbols u j, -1 < i < k. A configuration describes the situation where T is in internal state y reading the nth square of the tape t. If the line yt(n)sTy' occurs among the list of instructions of T, then the machine is obligated to move into state y' after first converting the tape t to t' with t'(n) = s and t'(k) = t(k) for all k 74 n, and moving the reading head one square to the left or right depending as T = L or T = R. In the two cases T = L or T = R, the resulting situation is respectively described by the new configuration Q' = (t', n - 1, y') or (t', n + 1, y'). We say that our machine produces Q' when presented with Q, and we write T(Q) = Q'.



RECURSIVE INSEPARABILITY FOR RESIDUAL BOUNDS OF FINITE ALGEBRAS 1865

We are interested in the computation that results when we place a machine on the zero-th square of the blank tape and set its internal state to u l. By the "initial configuration" we mean Qo = (to, 0, , 1-) with to the blank tape. Assume that T is a Turing machine. By placing the machine in the initial configuration Qo, a sequence of successive configurations

Qo,0Q, , Qi.Qm .

is generated, in which each Q,,1?+ is the configuration produced by T from Q,,1. This sequence of configurations may be finite or infinite. Recall that according to our definition of machine instruction, there is no instruction of the form uorsTy or ,u-lrsTy. From this together with the other restrictions we placed on T, this sequence of configurations produced by T will be finite if Q,.1 = (t, n, uo) or

- (t, n, ,u1) for some m, t, n, in which case no succeeding configuration Q,,z+i can be produced. We say that T halts iff the sequence of configurations it produces is finite; and we say that T halts in state y if Q,7, = (t, n, y) with y C {,uo, uj - }, for some m, t, n.

A basic result of recursion theory is the existence of a recursively inseparable pair of disjoint recursively enumerable sets. From this, it is easy to see that for Turing machines with two halting states, defined as above, the set of machines that halt in state guo is recursively inseparable from the set of machines that halt in state Pu

?2. The result. THEOREM. The class offinite algebras offinite type having finite residual bound is

recursively inseparablefrom the class offinite algebras offinite type having residual bound infinity.

PROOF. In the next section, we define an algebra B(T), finite and of finite type, corresponding to any Turing machine with the two halting states j'o and -l1 as described in ?1. In ?5, we show that if T halts in state ,u-I then resb(B((T)) = oo. In ?6, we show that if T halts in state ,o then resb (B(T)) is finite. This theorem is thus a corollary of the observation detailed at the end of the previous section. -

?3. Algebras. Given a machine description 7T that uses the non-halting states 1 . ..., 1k and the halting states u- 1 and uo as prescribed in ?1, Iwe wish to define

an algebra B(T). This algebra B(T) is closely related to the algebra A(T) defined in R. McKenzie [3]. Its universe will be the disjoint union of a 20k + 40-element set V and a five-element set {0, 1, 2, H. Q}.

Notation for the elements of B (T) and their natural sub-groupings is introduced below.

U= {1,2,H}5

V {C,5D,,Mi, C itD,,M[} for-i < i < k and {r,s} C {0, 1},

Vi= V,,? U Vi' , Vi Vio U Vi , and V = Vi,

B(T) {0, Q} U U U V

We shall make essential use of the permutation bar of order two defined on V so that ( Ci,',) = C,,., (C. s) = Cs,5, etc.



1866 RALPH McKENZIE

We now define the operations of B(T). To begin with, we have the "semilattice operation" x A y. This is the binary greatest lower bound operation for the ordered set (B(T), <) where we define < so that x < y iffx = 0 or x = y. This means that x A x = x and x A y = 0 if x 74 y.

Further operations J, J', So, S1, S, T are defined on B (T) as follows.

J(x,y,z)=x if x=y

=xAz ifx = C V

= 0 otherwise.

J'(x,y,z) = x Az if x = y

= x ifx = y C V

= 0 otherwise.

So(u, x, y, z) = (x A y) V (x A z) if u C Vo or {x, y, z} C {0, Q} = 0 otherwise.

SI (u, x, y, z) = (x A y) V (x A z) if u C { 1, 2} or {x, y, z} C {0, Q} = 0 otherwise.

S2(UVXYZ)= (xAy)V(xAz) ifu = v C Vor{x,y,z} C {0,Q} =0 otherwise.

In the above definitions, for any x, y, z c B(T), (x A y) V (x A z) denotes the least upper bound < x of the two elements x A y and x A z. This element exists since for any x, the interval from 0 to x is a chain.

For {x, y, z} C B(T) we denote by t(x, y, z) the element w such that w z if x = y and w x if x 74 y. This is just the ternary discriminator operation. Now we define T.

T(x, y, z) = t(x, y, z) if {x, y, z} C {0, Q} or {x, y, z} C {v, v} for some v CV1

= 0 else.

Operations derived from T. Our machine operations (operations derived from T) have the function of permitting machine computations to be modelled algebraically Many of our machine operations will have companion operations which allow us to model "reversed" computations. The first operation will allow us in certain situations to produce an element of an algebra B(T)x which represents the blank tape with one square selected for the reading head of our machine.

I(x) C=?o ifx =1

=M if x = H

=D% if x = 2

0 else.

For each instruction of the formui rsLu,, in T and for each t c {0, 1 } we have the operation Li,.t, which will allow us to mimic the operation of T when it is reading r




and in state iu and if it happens that the square just to the left of the one the reading head sits on has t printed on it.

Li,.t (x, y, u) C,", if x = y = 1, u= C,:, for some s'

=Mltn if x = H. y = l, it = Cit,.

-Ds, if x = 2, y = H, = Mi[ =Ds if x = y = 2, u = Ds. for some s'

int it

-v if ucVandLi,.,(x,y,i)=vcV according to the preceding lines

=-Q if x = y = 1 and u Q

- 0 else.

The collection of all the operations Li,.t we write as L. For each instruction of the form uirsL u,.. in T with m 74 -1 and for each

t C {0, 1 } we have the operation Lir, t, which is the companion operation of Li,,. It is the unique operation satisfying

Li,.t (x, y, u) = v when Lit (x, y, v) = u 74 0

= 0 otherwise.

The collection of all the operations Lo., we write as Lo. For each instruction of the form ,uirsRu,.1 in T and for each t C {0, I} we have

an operation Ri,.t, and we shall denote the collection of all these operations by Rz.

Ri,.t(x,Y, U) Cm, if x=Y 1 forsomes'

= C'snt if x = H. y = 1, it = Mi"

= M,1t if x = 25y = H, i = D~t =Ds if x = y = 2, u = Ds.

for some s'

-= v if u C V and Ri,., (x, y, i) = v C V according to the preceding lines

=-Q ifx=y 1 and u=Q - 0 else.

For each instruction of the form juirsRu,... in T with m 74 -1 and for each t C {0, 1} we have the companion operation Rot of the operation Ri,.t. We shall denote the collection of all these operations by RJZ. R?., is the unique operation satisfying

R, (x, yu) v when Ri,., (x, y, v) = u 74 0

0 otherwise.

To introduce the final set of operations, let F1, . F. , F, be a list of all the operations in the set L U Lo U R U R0.

A certain binary relation -< on the set U is integral to our construction. We define x < y to mean that x - y = 2, or x = 2 and y = H, or x = H andy = 1. or x y = 1. Observe that for 1 < i < c, Fi(x,y,z) is0exceptwhen {x,y} C U and x -y.



1868 RALPH McKENZIE

Now for 1 < i < c we define operations U/I and U2.

Ui'(x, y, z,u) = Fi (x, y, u) if x y, x z, y 74 z, Fj (x, y, u) CV

= Fi (xyu) if x yy z

= 0 otherwise.

Ui2(x, y,z, u) = Fj (y,ZI, u) if x CZ, Y Z, x 74 y, Fi (y, z, u) CV

= Fi(y, z, u) if x y, y z

= 0 otherwise.

We define B(T) to be the algebra whose universe is B (T) and whose operations are the constant 0 and the ternary operation T, and all operations of the set

A {A,J,J ,SO,SI,S2}

and all operations of the set

M {I} U L U Lo U JZ U -R

and all operations of the set

a-{Uil, U2: I < i < c}.

?4. Algebraic encoding of machine computations. The configuration algebra of a Turing machine T is the set of all T-configurations, defined in ? 1, together with the partial function which takes any configuration Q whose state is not a halting state to the configuration Q' = T(Q) produced from Q by T. Let A = B(T)x where X is some nonvoid set. The operations of B(T) are designed specifically to allow us to "encode" in A certain subsets of the configuration algebra together with the production function restricted (as a partial mapping) to a subset. If X is infinite, we can encode any connected subset of the configuration algebra in A. In this section, we shall examine the encoding idea.

To accomplish an encoding, one must select certain elements of A which will serve as "markers" for some individual tape squares, and other elements which will serve to represent individual configurations. The production function will be represented variously, depending on the configuration it is being applied to, through the application of certain of the operations of A to an ordered triple consisting of two marker elements and a configuration element. Recall that when F(x, y, z) is any one of the operations

Lirt (x, y, z), Li,, (X, y, z), Rist(, izRt(xyz

derived from T, then F(u,v,x) 74 0 implies {u,v} C U, x E V U {Q}, and u -< v where -< is the binary relation on U defined in ?3. Our marker elements will be U-valued functions and our configuration elements will be V U { Q}-valued functions.

DEFINITION4.1. If f, f' E UX then we write f -< f to denote that f(x) -

f'(x) for all x E X. A subset F of A will be called sequentiable provided that F C UX, f-'({H}) 7 0 for all f E F, and moreover, the set can be written as F =f{fn n E N} where N is an interval (convex subset) of Z (the set of integers), in such a way that f-< fn+I whenever {n, n + 1 } C N.




Any sequentiable subset of A can serve as a set of markers for encoding computations. Let = {lf, n E N} be a sequentiable set in A. For each n E N we denote by Xn the nonvoid set f -1 ({H}) and we put

XL nfl ({i}) nEN

XR =nf -

({21). nEN

Then {XL, XR} U {Xn: n E N} is a partitioning of X into pairwise disjoint sets. Choose any function q EE {O, I}X and any set X' C XL.

Now consider an arbitrary configuration Q = (t, n, ii) in which n E N. For such a configuration, we define ,B =,(Q) in A like this: we put /3(x) Q for x E X' and

/3(x) = Ch() when x E XL- X' it (n)

= ct (j when x(EX1, = Ct((n) whnx iX, j < n

= Mt(') when x E X,,

= D'tW when x E Xn<j it(17) j

= D5'(x) whenxEXR. it (n)

Thus /3 encodes t restricted to N, as t(j) is the superscript of /B(x) when x E X1. We have also arranged that for all x outside of X', /3(x) encodes pii and t(n), and moreover n is encoded as the unique j such that /3(x) is an M-symbol when x E Xi.

It is convenient to use Ao to denote the set of all functions h E A such that 0 E h(X).

The next lemma shows in what sense we have encoded computation by represent- ing configurations Q by the functions /3 ( Q).

LEMMA 4.1.

(I) Assume that {j, j'} C N and Q (t, n, ui) is a configuration with n E N.

(1) Let uir's'Lun' be an instruction of T and e E {0, 1}. Then Li,,., (fj, f1',B3(Q)) c A -Ao iffi' = i, j' = n, j = n - 1, r' = t(n) and e = t(n - 1). If these equations hold, then LiirI (fi - f,il, 3( Q)) /3 (T(Q)).

(2) Let 1ui'r's'R1,u, be an instruction of T and e c {0, 1}. Then Rine E(f,

fj1,/3(Q)) c A-AO iff i' = i, j = n, j' = n + 1, r' = t(n) ande = t(n + 1). If these equations hold, then Rie,., (fn' fn+i, 3(Q)) = 6(T(Q)).

(II) Assume that { j, j'} C N and Q = (t, n, /m) is a configuration with n c N. (1) Suppose that piur's'Lp,u71 is an instruction ofT with m' 7 -1 and E {O, 1}.

Then Li',?, (Vj' fjl, P(Q)) c A- AO iff m/ = m, j = n, j' = n + 1, e = t(n) and s' = t(n + 1). If these equations hold, then Li?,,E (fn, ftn+i, /(Q)) = W(Q')

for a configuration Q' which satisfies Lilrie (f,7, fn+l, /3(Q')) /3(Q) (and is uniquely determined by this equation).

(2) Suppose that pi/r's'RIuml is an instruction ofT with m' I -1 and {0, 1}. Then R,?, V(f1 ,f1, f3(Q)) c A - A0 iff m' = m, j = n - 1, ' = n, e = t(n) ands' = t(n - 1). If these equations hold, then Rril,., (frn-1, fn, PM(Q)) = (Q)



1870 RALPH McKENZIE

for a configuration Q' which satisfies Riu, (fn -, fl, /3(Q')) = /3(Q) (and is uniquely determined by this equation).

?5. If T halts in state p-uI. THEOREM 5. 1. If the Turing machine T halts in state j'-1 then resb (B(T)) = oc. Throughout this section, we assume that T is a Turing machine and the sequence

of configurations produced by T starting on the initial configuration Qo = (to, 0, gui) is Qo, ..., Q, where Qj = (tj,nj, y) for j < m and y,,, = p u-. We put N = { no,... ,n,, 1}. This is an interval in the set of integers and we write it also as N = [b, c] where b and c are the extreme points of the interval.

To prove the theorem, let i be any non-zero cardinal and let YO be a set of cardinality i disjoint from N. We are going to construct a subdirectly irreducible algebra K/( of cardinality > 2i, with K a subalgebra of the algebra A = B(T) Y. where Y = Yo U N.

For n E N, let an be the element of A defined by a,1(n) = H, an (x) = 1 for X E [b, n- 1] U Yo, and a,1(x) = 2 for x E [n + 1, c]. Put

S = {ab, * * * , a,}.

Let qO E A be the function such that qo(x) Q for x E Yo, qo(x) = Co for x E [b,-1],qo(O) = MO and qo(x) = Do forx E [1,c]. Put

AO {f E B(T)Y :f(YO) C {0, Q} andf N = qo N}-

Now we define K to be the subuniverse of A generated by S U Ao, and K to be the subalgebra of A with that universe.

DEFINITION 5.1. Let Q = (t, j, y) be any configuration of T, focussed at j and in state y. We say that Q is properly in N provided j E N and t (i) = 0 for all f E Z-N.

We use CfgN to denote the set of all configurations properly in N. We write P <N Q to denote that there is a finite sequence Po =Q P1Q r . .. Pt = P consisting of configurations properly in N such that Pi+1 = T(Pi) for all i < i.

Thus the halting computation Qo,... , Q... of T consists of configurations be- longing to CfgN and we have Qm <?N QO.

DEFINITION 5.2. We put Cfg= {Q E Cfg Q= Qin or Q,1-I <?N Q}. This is the same as the smallest set C of configurations containing Qo and such that T(Q) = Q' with Q, Q' E CfgN always implies Q' E C if Q E C, and implies Q E C when Q' E C and Q' 74 Q,11.

We take Cfg2 to be the smallest set C of configurations which includes every "initial" configuration (to, j, pu) with j E N (where to is the blank tape), and has the property that T(Q) = Q' with Q, Q' C Cfg1 always implies Q' C C if Q C C, and implies Q C C when Q' C C and the state of Q' is not pu

REMARK 5.1. For each Q C Cfg2 there is a configuration P C CfgN such that P <N Q and P <N (toj,,uI) for some j C N. From this it follows that for Q C Cfg2, the state of Q is not 'o. Moreover, for Q C Cfg2 , if the state of Q is the other halting state Ju - then it is forced that Q = Q,,. (This is because the consecutive stretch of tape covered by the addresses in N contains not enough tape




to model the halting computation starting at (to, j, ,u ) unless ] = 0.) Consequently, we have Cfg' C Cfg2.

DEFINITION 5.3. Corresponding to any Q C CfgN we define two elements /= fl(Q) and i =(Q) of A. Assume that Q (t, n, H i) and that t(n) = r.

,6(x) = Co if x E Yo

= C't.(') if x (E [b,n n-1]

=M, if x n

= Dt(x) if x C[n +1, c] ir

i(x) = fl(x) if x E N

= Q if x E Yo.

Also, for 0 < j < m we put qj = X(Qj). Note that this is consistent with the earlier definition of qo.

DEFINITION 5.4. For Q E Cfg9 we put

A(Q) = f EB(T) Y of (YO) C {0, Q} and f IN i(Q) N}

We define A U{A(Q): Q E Cfg9 }; and we putF {fl(Q): Q E Cfg2}. Notice that Ao-part of the generating set for K-is identical with A(Qo).

By Ko we denote the set of all functions f c K such that 0 c f (N).

LEMMA 5.1. We have that

K = Ko U S U F U A

C ({0} U U)Y U ({O} U V)Y U ({o, Q} Yo x ({O} U V) N).

PROOF. The elements of A U F are generated from S U Ao by applying the operations of the set M repeatedly (See Lemma 4.1.) This gives K D Ko U S U A U F. To get the reverse inclusion, it is useful to first prove that K C A' where

A' = ({0} U U)y U ({0} U V)y U ({O, Q}Yo x ({o} U V)N).

This can be done by straightforward reasoning, showing that A' is closed under each of the operations of the algebra A.

The proof is then completed by showing that the set Ko US UAUF is a subuniverse. We can mention that proving the closure under the operation So is facilitated by Remark 5.1. 1

DEFINITION 5.5. We define T to be the equivalence relation on K such that fTg if f= g or {f,g} C G where

G = Ko U {f E A(Q,11): 0 E Jf(Yo)}.

LEMMA 5.2. T is a congruence of K.

PROOF. To prove this, one must check that T has the substitution property for each of the operations of K. This is trivial for the operations So, SI, S9 because Remark 5.1 and Lemma 5.1 imply that the range of each of these operations in K is included in the set Ko. It also follows from Lemma 5.1 that if {f o, f 1, f2} C K



1872 RALPH McKENZIE

and T(fo,f 1, f2) f Ko then none of the fi belongs to Ko or to A(Q,,7); thus T trivially has the substitution property for the operation T.

It is rather straightforward to show that A, J, J' preserve T. This leaves the operations of the set H U M to be checked. In fact, when any one

of these operations takes one of its input variables from the set G, the operation outputs an element of Ko: so these operations trivially have the desired property. -

We take 0 to be a congruence of K maximal among all congruences that include T and do not contain the pair (q,,,, 0).

The next lemma ends our proof of Theorem 5.1. LEMMA 5.3. 0 identifies no two elements of AO. PROOF. Let f =4 g, f, g E Ao. The congruence generated by the pair (f, g) in K

identifies the three elements T(f, g, qo) = q , T(g, f, qo) = q 2 and T(f, f, qo) T(g, g, qo) = qo. There is a term t(i, y) (built from the operations of L U R) such that t (ab, a, qO) = q,,, and t (ah, a, qO) = q' where q'1, agrees with qO on Yo. Here one of the two elements qn7, q;.. belongs to G, so that if we did have (f, g) E 0 then we would also have q, Oq' 00 for either i = 1 or i = 2. -1

?6. If T halts in state /uo. THEOREM 6.1. If the Turing machine T halts in state Ho then resb (B(T)) < co. In this section, we assume that the sequence of configurations produced by T

starting on the initial configuration QO = (to, 0, Iul) is Qo . . . , Qn, where Qj =

(tj, nj, yX) for j < m and Yi = Io. We put N = {no.... , n,,1;} and we define 7T(T) =

IN 1. Theorem 6.1 will be proved by modifying the analysis in R. McKenzie [3]. We begin with a definition from [3].

DEFINITION 6.1. When A and B are algebras and B E V (A), we define the dimension of B with respect to A, written dimA(B), to be the least cardinal K such that B E HS(AX) for some set X with I X - . We say that S is a large si in V(A) if S E V (A) is subdirectly irreducible and dimA (S) > 1.

ASSUMPTION I. In Lemmas 6.1-6.13, we assume that C is a finite large si in V (B(T)), that ICI > 2, and that C = B/O where B C B(T) Y, with I Y equal to the dimension of C with respect to B(T). Thus 1 < I YJ < co. We denote by 0 the unique cover of 0 in Con B.

LEMMA 6.1. There exist elements /, /' E B and to E Y satisfying (i) For any x, y E B we have (x, y) E 0 ifffor every polynomial f of B,

f W = fi +- f (Y) = A

(ii) The entire interval [0, /3] in the semilattice (BQ(T), A) Y is included in B andfor all s E Y, /(s) O.

(iii) /3' < /3, (/3', /3) E 0, and /3'(to) = 0 while /3'(s) = /3(s) for all s E Y - {to} (iv) For all x, y E B we have f/Ox -* f = x while

x < f, y < /3, (x,y) E - * (x,y) E Q.

In particular, x < f/ implies (x, x A /3') E 0. (v) Let f be any polynomial operation of B. If u < f/ and f (u) = /3 then also

f (u A nlot = l




PROOF. Except for a slight change in the formulation of (v), this is Lemma 5.1 of [3]. The proof carries over verbatim. -1

ASSUMPTION II. We choose and hold fixed elements /3, /3' E B and to E Y satisfying the properties listed above. We put Yo /3-({Q}), Yi = Y - Yo, and we define Bo {f E B: E f (Y1)}.

LEMMA 6.2.

(i) There exists no element f E B with f (Y1) C Vo or f (Y1) C {1,2}; thus Yi #4 0.

(ii) There do not exist f, g E B satisfying f (t) = g(t) E V for all t E Yi. (iii) Iff G- B andfor all t E Y, f (t) = f(t) or f (t) = fl(t) E V, then f = /.

PROOF. Suppose that f E B and f ( Yi) C Vo. Then restricted to the interval [0, /3] in B, the operation D (x, y, z) = So(f, x, y, z) is just (x A y) V (x A z) (acting co-ordinatewise). Choose t E Y - {to} and let /3" be the member of B which is 0 at t and agrees with /3 elsewhere (just as fl' is 0 at to and agrees with /3 elsewhere). Let F (x) be the polynomial operation f (x) =So(f, 3, /', x). Then a simple calculation shows that f (f/") = /, while f (f/" A /3') = /3'. This contradicts Lemma 6.1 (v).

A similar contradiction is obtained, using the operation Si, if f ( Y1) C { 1, 2}. If f(t) g= (t) for all t E Y1 (f, g E B), use the operation S2.

Now suppose that f E B and f (t) = /(t) or f (t) = /(t) for all t E Y. Assume that also f 4 /3, in order to reach a contradiction. By (ii), we must have f (t) / 3(t) for some t, as well as f (t) =- /(t) for some t. Choose t1 E Y so that f (t1) /3(t1)

iff f (to) = ,B(to). Let 3,i be the function that agrees with /3 except at ti, where it takes the value 0. By Lemma 6. 1, /3, E B. There are two cases. If f (to) = /(to), define F(x) = J'(3, f, x). If f (t1) = /3(t1), define F(x) = J(3, f, x). In each case, we calculate that F (/3i) = f/ and F (/3i A /3') = /', yielding the usual contradiction. -1

LEMMA 6.3. Suppose that f/ = T(fo, f 1, f2) where {fo, f 1, f2} C B.

(i) We have /3 (Y,) C V1?O and Yo C {to}. If g E B and g(t) E {f/(t), /(t)} for all t (E Y,, it follows that g I y, = fl I Y, .

(ii) Whenever f/ = T(go, g1, g2) with {go, g1, g2} C B, we have that gi j y, = fly, and gi ( Yo) C {0, Q} for all i E {0, 1, 2}.

PROOF. It is obvious that our assumption implies /3( Y1) C V1?O. Statement (ii) is a consequence of statement (i). Let us prove (i). If Yo = 0, all parts of (i) follow from Lemma 6.2 (iii). Thus we can assume that Yo #4 0. To prove that Yo = {to}, assume otherwise. Then we can choose t1 E Yo - {to}. Let /3, be the function that takes the value 0 at t1 and equals /3 elsewhere. Consider the polynomial function F (x) = T(x, 3i, /). We have F(/3l) = 3 while F (/' A /3i) = /', contradicting Lemma 6.1 (v). Thus Yo {to}= _o_

Now assume that g E B and g(t) E {f/(t), /(t)} for all t E Y,, while g(t') /3(t') for at least one t' E Y,. Replacing g by T(g,/f,g), we can assume that g( Yo) C {0, Q}. Now we calculate that g' = T(g, /3', /3) agrees with g on Y, and satisfies gI(to) = Q. Then Lemma 6.2 (iii) implies that g' /3, and this entails that g I y, = fy I y, as desired. -1

LEMMA6.4. Wehave/3(Y,) C Viforsomei e{-1,0,...,k}and{r {,1}.



1874 RALPH McKENZIE

PROOF. This is essentially Lemma 5.3 of [3]. We have two cases. First, assume that whenever f is a non-constant polynomial of B, u E B and f (u) = f6 then u = ,. Then it follows from Lemma 6.1 (i), (iv) that 0 = (B - {f})2 U idB. But we have assumed that ICI > 2 so this case is impossible.

So, now, we do have a non-constant polynomial f with f (it) =f where u is some element of B distinct from fl. We use the concept of the degree of a polynomial g, which is defined so that constant polynomials and the identity function have degree 0, and in general, if g is non-constant and non-identity, then the degree of g, or deg(g), is m + 1 where m is as small as possible such that g can be expressed as a fundamental operation of the algebra applied to a sequence of polynomials each of which has degree < m.

We can assume that f is a non-constant polynomial satisfying f- ({f/}) g {f1} and that for every non-constant polynomial g with deg(g) < deg(f) we have g1 ({I6}) c {I6}. Now f is not of degree 0, so we can express f as the result of applying one of the non-constant basic operations F of B to a system of polynomials each of which is of a degree less than the degree off . The argument breaks into many cases, corresponding to the choice of F. It is easy to see that F = A is impossible. Also, by Lemma 6.2 (i), (ii), F cannot be SO, SI, or S2. If F is any one of the other fundamental operations besides J or J' then the desired conclusion immediately follows. The cases where F = J or F = J' are handled using Lemma 6.2 (iii). We leave this argument to the reader. -d

DEFINITION 6.2. We define B, to be the least subset X of B including 16 such that whenever F is an n-ary member of M and {fo, * * *, fn - C B (n = 1 or n = 3) and F(fo,... fn-) E Xthen{Ifo. f fM-I} C X.

LEMMA 6.5. (1) Iff E B, then either f E U' and f(Yo) C {1}, or else f(Yo) C {Q} and

f (Y1) C Vi, for some i andr. (2) Iff E B, and g E B andfor all t E Y holds g(t) f (t) E V or g(t) f (t)

then g = f. (3) Let F(fo, fi, g) h E B, where F E M - {I}. Then fo - fi; moreover, if

f c B then fo - f impliesfI = f, while f -< fI and f(Yo) C {1} imply fo = f

PROOF. See the proof of Lemma 5.7 in [3]. (The proof of (2) uses Lemma 6.2 (iii).) -d

DEFINITION 6.3. We write E for the set B, n U Y and Q for the set B, - E. For f E Q there exists, by Lemma 6.5 (1), a unique integer i E [-1, k] with f (YI) C Vi. We call i the state of f.

We recall that the sequentiable subsets of B were defined in Definition 4.1 and Zt(T) was defined at the beginning of this section.

LEMMA 6.6. Every sequentiable subset of B has fewer than r( T) elements.

PROOF. This is Lemma 5.4 of [3]. The same proof goes through here without change. If B contained a sequentiable set of size 7T(T), then using the operations in M (including I), we could produce an element in B n v7Y. But this would contradict Lemma 6.2 (i). -d




LEMMA 6.7. ThesetY issequentiable. Wehave XE1 < 1(7Q)and QJ < 211 (k+1) EJ.

PROOF. These are the same assertions expressed in Lemmas 5.8, 5.9 and 5.10 of[3] and can be proved in precisely the same way. All the functions in B, are constant on Yo (either Q or 1), and B, projects injectively onto a subset of the algebra B('T) Y'.

The new operations of B(T), not existing in A(T), which play a role in this proof are those of ?2 U R0, and they are handled exactly as the operations of L U R. (Also see the remarks in the proof Lemma 6.13 below.) -A

LEMMA 6.8. If Yo = 0 or B* n ({Q} U V1?O) Y = 0 then B* is identical with the set of all u E B such that f (u) = /3 for some non-constant polynomial function f of B. In this case,

JCJ = JB/O0 = 1 + JB*J = I + XE1 + IQI < (27(')(k + 1) + 1)7(T).

PROOF. We assume that Yo = 0 or B* n ({Q} U V10o)Y= 0. First we claim that if f is any non-constant polynomial function of B then f' (B*) C B,. This claim will imply that when u E B - B* then no non-constant polynomial operation f can have f (u) = ,. It should be clear that when u E B* then f (u) = /3 for a non-constant polynomial f that can be obtained by composing some operations of the set M.

Suppose that the claim fails. Let f be a non-constant polynomial of minimal degree among all those with the property f - (B*) g B,. Let u E B - B* with f (u) E B,. We write f as a non-constant basic operation F of B applied to some polynomials of lesser degree than the degree of f. The proof of the claim consists in showing that whatever F is, a contradiction can be obtained. The various statements in Lemma 6.5 will be very helpful. With their help, it is easy to see that F cannot be J or J'. Clearly F cannot be A. It cannot be So, SI or S2, since by Lemma 6.2, these operations take values only in Bo, while by Lemma 6.5 (i), B* is disjoint from Bo.

Suppose that F = T, say f = T(fo, f, f2) where f i are polynomials of degree less than deg(f). We have that T(fo(u), fI(u) f2(u)) = v E B,. Hence by the definition of T and Lemma 6.5, it follows that v E B* n ({Q} U Vo) Y. Then our initial assumption gives that Yo = 0. Now each of f X (u) is equal to v or to v3 at every t E Y, = Y-by definition of T and since v does not take any values in {0, Q}. Hence by Lemma 6.5 (ii), we have f i (u) = v. Now the minimality of deg(f) implies that each f X is constant. Hence so is f . This is a contradiction.

If F E M it is pretty trivial to find the contradiction. Suppose finally that F Uil. (The case F = Ui2 yields to the same argument.) Say f = Uil (fo, fi, f2, P 3) where f X are of lesser degree. Now f (u) -v E B* implies that F (fVo(u), f I (),

f 3 ()) (t) E {v(t), v(t)} for all t E Y. Then by Lemma 6.5 (ii), v = Fi (fo(u), f I (), f 3 ()) . This gives that {fo(u), f I(), f 3 ()} C B* (definition of B*) and hence fo, f l, f 3 are constant by minimality of deg(f). The fact that

Ui' (fo u), f I ), f 2(u, f 3(u) = Fi (fo u), f I ), 3 W))

and these functions are everywhere non-zero implies that f2(u) = fi (u) EB* (definition of U11). Thus also f2 must be constant, implying that f is constant- the desired contradiction.



1876 RALPH McKENZIE

We have now established the claim. From the characterization of B, thus arrived at, and from the characterization of 0 in Lemma 6.1 (i), we find that all elements of B - B, are 0-equivalent. Our estimate for an upper bound of ISI then follows from Lemma 6.7. -

LEMMA 6.9. Assume that Yo 7 0 and B, n ({Q} u v0) Y 7 0. Then Yo = {to}. Choose f E B, with f (to) = Q and f (YI) C Vl7o, and let f ' be the function that agrees with f on Yi and has f '(to) = 0. Then the pair of elements f, f ' have the same properties, expressed in Lemma 6.1, as the pair /3, /3'.

PROOF. Since T does not halt in state pu-1, then Q cannot contain any element in state -1 (see Definition 6.3).

Briefly, here is the reasoning behind this assertion: The proof of Lemma 6.7, for which we referred the reader to [3], shows that-ignoring the difference between bar'ed and un-bar'ed members of V-the elements of Q encode configurations, in the sense of ?4, relative to the sequentiable set S. The chosen element f encodes a configuration P that has the machine in state /uI reading a blank tape. The set of configurations encoded in Q is connected under the partial function induced by T, in the configuration algebra. If Q contains an element encoding a configuration Q in state ,u - 1, then since P and Q are connected under T and Q is not in the domain of T, it follows that En (p) = Q for some n. But then T started on a blank tape in state ,u I halts in state u - 1, contradicting our assumption that it halts in state puo. Thus Q contains no element in state p- 1.

Since Q contains no element in state -1, then whenever F E M - {I } and F(x, y, u) = v with {x, y, u, v} C B*, it follows that M contains the reverse (or companion) operation F' of F and we have F'(x, y, v) = u. Thus if we let E = {ao,... an }, then for any u,v E Q, there are terms z(x0,.X. ,Xn,y) and .'(X, . . ., yn, Y) such that -c(ao, . . .a, u) = v and z'(ao, . . .a, v) = u. These

terms are built out of operations in M. In particular, we have polynomials z0(-, y), zI (a, y) such that zo(-, /3) = f and zI (a, f ) = /3 where f is the chosen element.

Our proof that Yo {to} is just an elaboration of our proof for Lemma 6.3 (i). Assuming that Yo =4 {to} then we can choose t1 E Yo - {to}. Let fil be the function in B(T)y such that /3l(t1) = 0 and /3l(t) /3(t) for all t E Y - {ti}. Now p = zo(-, il) equals f except at t1 where p(ti) 0; and p' =zo(-,& l A /3') agrees with p except at to, where it is 0. In fact, we have f' = /3(a,,f') and p' = p A f'. We define a polynomial function H(x) by

H(x) = IQ a-, T (oG(a, x),p ,) )

It can be verified that H(/31) = /3 and H(/31 A /3') /3'. This contradicts Lemma 6.1 (iv).

To see that f, f ', to have the same properties expressed in Lemma 6.1 as do fl/, to, observe that the polynomial functions Hi(x) = zi(-, x) (i E {0, 1}) used

in the previous paragraph induce A-preserving bijections between the intervals [0, fli] and [0, f ]. Then if, for example, we have a polynomial H(x) such that H(u) = f and H(v) =4 f where u, v E B, we can form the polynomial H'(x) = zi (a, H(x) A f ) which satisfies H'(u) /3, H'(v) 7 /3. The remaining details of this proof are left to the reader. -A




In view of Lemmas 6.8 and 6.9, it will suffice to bound ICI under the following assumption, which we now adopt on top of the earlier Assumption I and Assump- tion II.

ASSUMPTION III. Henceforth we assume that Yo {to} and /3( Y1) C VTO.

DEFINITION 6.4. We denote by p,, the projection homomorphism of B onto a subalgebra B" of B(T) Y', and by q we denote the kernel of p,1, a congruence of B. For f E B, we shall generally write simply f " for p,1 (f ). For any polynomial operation F (xo, . . . , x ) of B, we denote by F'" the polynomial operation of B" which satisfies F(f0 ,...,fni)= f 1) for all {fo f * * i} C B

DEFINITION 6.5. We define B,' just as B, was defined, but with reference to /311 in B" in place of f, in B. Namely, B'5 is the smallest subset X of B" which includes /3" and is such that whenever F is an n-ary member of M and { fo, . f - l } C B7

(n = I orn = 3) and F(fo,... , fn-1) E X then {fo. f * ,J-11} C XI

LEMMA 6.10.

(1) Iff E B*' then either f (YI) C U or else f (YI) C Vi, for some i and r. (2) Iff E B47 andg E B" andfor all t E Yi holds g (t) = f (t) E V or g (t) = f (t)

then g = f. (3) Let F(fo,fi,g) =h e B47where F EM-{I}. Thenfo -<fi;moreover,if

f E B" then fo < f implies f I = f, while f -< f i implies fo = f .

PROOF. For (1), just note that the set U ' U Ui,. Vi]' contains /3" and has the closure properties required of B$.

For (2), we note that at least the element f = f7 E B*7 has the required property, since Assumption III contains the hypothesis of Lemma 6.3, namely we have ,6 = T(/3,/3,/3). Now suppose that f E B'1 and g E B" and for all t E Yi we have f (t) = g (t) or f (t) = g (t) E V. Iff E U Y" then it follows automatically that f = g. The other case, by (1), is when f E V Y . In this case, note that it follows by an easy induction that there is a polynomial

F(y) = t(fo,. f , f y, ,)

where t is a term built out of operations of M and {fo, . . fn} C B1 n U Y1, with the property that /3" = F(f); and this polynomial is such that the element F(g) will be equal to /35 at those t where g = f and will be equal to /3" at those t where g f . Now we have that F(g) must be equal to /3', by Lemma 6.3; and hence g =ft -

For (3), let h =F(fo,fi,g) e B$' where F =Fi E M - {I}. We see that fo -< f I by examining the definition of F1 (since h (t) is non-zero for all t). Suppose that fo -< f . Let h' = Uil (fo, f l, f, g). Now h' = h at precisely those t where f = f l, and h' = h elsewhere. Hence (2), already proved, implies that f = f l, since h E B*. The argument to get f C< f I fo = f uses the operation U? in an analogous fashion. -A

LEMMA 6.1 1. For g E B we have that go e B' iff the algebra B has a polynomial function F(x) such that F(g) = /3 and F" is non-constant on B".

PROOF. The proof is parallel to our proof of Lemma 6.8. Write B*' for the set p -I (B*4). First, let g be any member of B*'. As we remarked in the proof of



1878 RALPH McKENZIE

Lemma 6.10, there is a term t(xo,.x. ,Xn) built out of operations in M so that t (fo, . . ., fl) = ," where g7 E {ffo, . . ., f4n C Bq, and where t(xo . . ., x,,) 0 in B(T) whenever {xo, Xn IC B (T) and at least one of the xi is 0. Let f E B be chosen to satisfy p?/ (f') f i, with g = f 'for some i. This gives us a polynomial F(x) of B so that F(g)7 f=" and F(0)?1 - 0. We now put

G (x) = T fl, T(3 F (x), fl), fl

One can verify that G (g) = /3 and G (0)" - 0, so that G" is non-constant. To complete the proof of this lemma, what remains to be shown is that F'- (Bk') C

B*' for every polynomial F(x) of B such that F" is non-constant. As usual, we prove it by assuming, to the contrary, that we have a polynomial F (x), with F" non-constant and F-1(B*') g B*', while G-1(B*') C B,' holds for every polynomial G(x) with G" non-constant and deg(G) < deg(F). Let u C B -B'

withF(u) =g E B*'. We write F(x) as a basic operation H applied to polynomials having lesser

degree. Consider what that basic operation might be. Suppose that it is J'. Then we have F(x) = J' (Fo(x), F1 (x), F2(x)) where the Fi have lesser degree. Now F(u) = g c B*' implies that g" = Fo(u)" since g is non-zero on Y1; and implies that

Fo(u)" = F1 (u)"I or Fo(u)"I = F1 (u)'1

at every t. Then by Lemma 6.10 (2),

Fo(u)" = Fi (u)"'I= g".

This implies that gt' = Fo(u)" A F2(u)"7 from the definition of J'. Thus we get also F2(u)- = g". Now since Fo(u), F1(u), and F2(u) all belong to B*', it follows that FOJ, F', F2' are constant. But then so is F(x)'", a contradiction.

The cases where H = J or H = A are similar but easier to handle. We cannot have H C {SO, SI, S2} since by Lemma 6.2, these operations take only values in the set Bo, which is disjoint from B*'.

If H= T,sayF(x)= T (Fo(x),Fi(x),F2(x)),then it is forced that g'" eV100)Y' and each of Fi (u)7 agrees with g" or g" everywhere. Hence in this case, Fi (u) =-g'

for each i by Lemma 6.10; and a contradiction is reached as before. The case where H c M is quite easy and requires no new ideas. The case where

H c H is also easy, using part (3) of Lemma 6.10 as well as part (2). We leave these final arguments to the reader. -A

LEMMA 6.12. Every sequentiable subset of B" has fewer than 7t(T) elements.

PROOF. If this fails, then Bf has an element in V Y'. (See the proof of Lemma 6.6.) This would contradict Lemma 6.2 (i). -

DEFINITION 6.6. We write X" for the set B*4 n U Y, and Qf1 for the set B*` -E.

LEMMA 6.13. IC < ((27(T)(k + 1) + 1) .7t(T) + I) (20k + 45).

PROOF. The analysis in [3] that begins with Definition 5.3 and ends with Lemma 5.10 can be applied here to show that X" is sequentiable and conclude that

IB$'I < (27f(T) (k + 1) + 1) 7T(T).




Then p7 l(B*{)l < IB*K (20k + 45) obviously, as 20k + 45 is the cardinality of B(T). Finally, Lemma 6.11 in conjunction with Lemma 6.1 (i) implies that any two members of B - p' l(B*4) which agree at to are O-equivalent. Hence we get the above-displayed strict upper bound for ICI in the case now under consideration, where Assumption III holds.

For adapting the above-cited analysis from [3] to the present situation, our Lemma 6.2 (i) adequately substitutes for Lemma 5.2 of [3], our Lemma 6.12 substitutes for Lemma 5.4 of [3], our Lemma 6.10 substitutes for Lemma 5.7 (1)-(2)-(5) of [3], and our Lemma 6.11 substitutes for Lemma 5.7 (3) of [3]. The statement of Lemma 5.7 (4) of [3] plays no role. (The analogous assertion here, which would be that no two elements of p,1 (B*a) are O-equivalent, may not be true.)

Of course, this analysis must deal with the operations of the set L' U R) which are missing from the algebra A(T) of [3]. but it should be clear that at least for this part of the argument, these new operations can be treated exactly as the operations ofLUR. -d

PROOF OF THEOREM 6.1. The number supplied in Lemma 6.13 bounding IC under Assumption III is larger than the one supplied in Lemma 6.8, and it is larger than IB(T) . Therefore, this number is a strict bound for the cardinalities of all finite subdirectly irreducible algebras in the variety V (B(T)). Since the variety consists of locally finite algebras, and all of its finite subdirectly irreducible algebras are bounded in size, it cannot have any infinite subdirectly irreducible algebras. (See R. W Quackenbush [7, Lemma 2.7].) -1

REFERENCES

[1] C. LATTING, There is no algorithm to decide whether the residual character of afinite algebra is uncountable, Ph.D. dissertation, UC Berkeley, 1995.

[2] R. MCKENZIE, The residual bounds of finite algebras, International Journal of Algebra and Coin- putation, vol. 6 (1996), pp. 1-28.

[3] I The residual bound of a finite algebra is not computable, International Journal of Algebra and Computation, vol. 6 (1996), pp. 29-48.

[4] , Tarski'sfinite basis problem is undecidable., International Journal of Algebra and Compu- tation, vol. 6 (1996), pp. 49-104.

[5] R. MCKENZiE and S. SHELAH, The cardinals oj simple models for universal theories, Proceedings of the Tarski Symposium, Symposia in Pure Mathematics, vol. 25. American Mathematical Society, 1974.

pp. 53-74. [6] R. MCKENZIE and J. WOOD, The type-set of a variety is not computable, International Journal of

Algebra and Computation, to appear. [7] R. W QUACKENBUSH, Equational classes generated by finite algebras Algebra Universalis, vol. 1

(1971), pp. 265-266. [8] W TAYLOR, Residually small varieties, Algebra Universalis. vol. 2 (1972), pp. 33-53. [9] R. WILLARD, On McKenzie's method, Periodica Mathemnatica Hungarica, vol. 32 (1996). pp. 149-

165. [10] . Tarski's finite basis problem via A(T), Transactions of the American Mathematical

Society, vol. 349 (1997), pp. 2755-2774. [11] , Three lectures on the RS problem, Algebraic model theory (B. T. Hart et al., editors),

Kluwer Academic Publishers, 1997, pp. 231-254. [12] Afinite basis theorem for residuallyfinite, congruence meet-semidistributive varieties, this

JOURNAL, to appear.



1880 RALPH McKENZIE

[13] 1 Determining whether V (A) has a model companion is undecidable, manuscript.

VANDERBILT UNIVERSITY

DEPARTMENT OF MATHEMATICS

1326 STEVENSON CENTER

NASHVILLE, TN 37240. USA

E-mail: mckenziegmath.vanderbilt.edu



recursive inseparability for residual bounds of finite algebras

Documents