Proc. Natl. Acad. Sci. USA Vol. 92, pp. 617-621, January 1995 Mathematics

Algebraic aspects of the computably enumerable degrees (computability theory/recursive function theory/computably enumerable sets/Turing computability/degrees of unsolvability) THEODORE A. SLAMAN AND ROBERT I. SOARE Department of Mathematics, University of Chicago, Chicago, IL 60637 Communicated by Robert M. Solovay, University of California, Berkeley, CA, July 19, 1994

ABSTRACT A set A of nonnegative integers is computably equivalently represented by the set of Diophantine equations enumerable (c.e.), also called recursively enumerable (r.e.), if with integer coefficients and integer solutions, the word there is a computable method to list its elements. The class of problem for finitely presented groups, and a variety of other sets B which contain the same information as A under Turing unsolvable problems in mathematics and computer science. computability (theorem is the first and easiest extension of embedding result of Q into Rt. Many of the most significant theorems giving an for the well-known partial ordering of the c.e. degrees, R = (R, algebraic insight into Rk have asserted either extension or <, 0, O'). nonextension of embeddings. We extend and unify these THEOREM 1.1 (FRIEDBERG-MUCNIK). results and their proofs to produce complete and complemen- tary criteria and techniques to analyze instances of extension (3x)[0 y & x V y = a]. work on computable functions (2) showed that undecidability (Va 0)(3x, y)[x (i.e., noncomputability) is resident in the most familiar math- Sacks (8) then developed a strong version of the infinite ematical settings, even in elementary number theory. Follow- injury priority method to prove the density theorem. ing Godel, there has been an intensive study of noncomputable THEOREM 1.3 (SACKS DENSITY THEOREM). sets arising in ordinary mathematics. The computably enumer- able (c.e.) sets are of particular interest here, because these are (Va,b)[aalgorithm to decide whether YATES). x EA when given answers to all questions ofthe form "Isy E B?" We will write A CT B to indicate that A is computable in B and (3a, b)[a b & a A b = 0]. A =TB ifA partially ordered set RIt = (R, <, 0, 0') with least element 0 = there is an x so that x < a, b and 0 < x. deg(0) and greatest element 0' = deg(K). Post then posed this The density theorem method was applied by Robinson (12, famous question (Post's problem): Does there exist a c.e. 13) and others to produce stronger extension of embedding degree a which is not computable and not complete (i.e., 0 < results, while the minimal pair method was developed by a < 0')? If not, then there would be a single Turing degree Lachlan and others to produce nonextension results and embedding of various lattices into the c.e. degrees. The next The publication costs of this article were defrayed in part by page charge significant advance was the Lachlan nonsplitting theorem (14), payment. This article must therefore be hereby marked "advertisement" in accordance with 18 U.S.C. §1734 solely to indicate this fact. Abbreviation: c.e., computably enumerable. 617 Downloaded by guest on September 25, 2021 618 Mathematics: Slaman and Soare Proc. Natl. Acad Sci. USA 92 (1995) which asserts that the Sacks splitting and density theorems order embedding. We now describe conditions for deciding cannot be combined simultaneously. whether f can be extended to an embedding g: Q -- Rk. THEOREM 1.5 (LACHLAN NONSPLITrING THEOREM). Definition 2.1: For S a subset of Q, we define -,(Va,b)[bz & x) X 3(s(z))}. sion of embedding results, or nonextension of embedding !(x)=f{zlzE results for R viewed as either a partial ordering or sometimes In Section 3 we prove that if condition [1] holds, then we can with partial lattice structure. embed P into 2k so that there is no extension to an embedding The extension of embedding question was solved for certain of Q into 2R; the proof uses a combination of elements from the related structures. For example, Fejer and Shore (17) solved it of the minimal pair and splitting theorems. In Section for the c.e. tt-degrees and wtt-degrees. Slaman and Shore (18) proofs calculated the extension of embedding theory which is com- 5 we prove that if condition [2] holds, then we can construct mon to all principal ideals in 2k for which the top point is low2 a similar counterexample to the extension of embedding; the (i.e., a' = 0"). Roughly speaking, the method of proof in these proof uses a more complicated version of the Lachlan non- results is to rule out certain extensions by using the minimal splitting technology. In Section 4 we prove that if conditions [1] pair method and then to realize those remaining extensions by and [2] both fail, then we can construct the required extension; using the methods of the Sacks density theorem. However, the the proof uses a strengthening of the Sacks density theorem minimal pair method, with its associated theorems about meet method. This proves Theorem 1.6. We present in Sections 4 and and join, is not sufficient to capture all the nonextension 5 only sketches of the proofs, which are much longer and will properties of 2k. This failure was already apparent in the proof appear elsewhere. of the Harrington-Shelah theorem. Further, there were ex- tension of embedding conjectures which computability theo- Section 3. The First Nonextension Theorem rists believed but were unable to prove by using the standard methods of Sacks and Robinson. We now solve the full THEOREM 3.1. IfP and Q satisfy condition [1] then there is an extension of embedding problem for 2k. embedding of P into 2k which cannot be extended to an embed- THEOREM 1.6. Given (P, Q), a pair offinite partially ordered ding of Q into JR. sets with 0 and 1 such that P C Q, it is a uniformly computable Proof: The conjunct ga(sd(y)) C 91(sA(J(x))) in condition [1] condition whether every embedding of P into 2 extends to an is equivalent to embedding of Q into 2R. The significance of this result is threefold. First, it answers -_(3c C P)(3d E P)[d .21 (x) & c ssi(y) & d ; c]. [3] specific open questions and conjectures. Second, it unifies the algebraic extension and nonextension results about 2k. Third, Now expression [3] is equivalent to the assertion that it is its proof unifies the principal proof methods for c.e. degrees: consistent to extend P to a distributive lattice L such that it shows that the minimal pair method and Lachlan nonsplit- ting method are complementary to the Sacks density method, W\ '(x) /X\dS(y), [4] when all are suitably generalized. Using Theorem 1.6, we can derive the following decidability where w\X3(x) represents the supremum in L of the elements result, which stands in contrast to the Harrington-Shelah b C 21(x) and /X\A(y), the infimum of the elements a C d(y). undecidability theorem. A restricted H2 sentence of the lan- We first explicitly give this extension L, and then map L into guage L(<, 0, 1) is one of the form 2k to produce c.e. degrees {a,, . . . am} satisfying P for which there is no extension in 2 satisfying Q. We define L as follows. (Vx_(3y_)[D(x_) => D#x, -y")] Let P = {ei, . . . em}. Define X,, = coi1, where ()[n] = {(X,y): y = n}. For each d C P define Sd = U{Xc : c - d}. Clearly, where D(x) is any quantifier-free formula in the variables x-, c ' d iff SC C Sd. and DI(i(,-) is any complete atomic diagram in the variablesx Case 1: 21(sL(y)) C 3(x). Define w(p) = Sp for all p C P. and y. Case 2: 2(s(y)) gt 2a(x). Let {bo,. . . , bk_ } be the maximal COROLLARY 1.7. There is a procedure to decide for any elements of 21(x). Note that k . 2 by the failure of Case 1. Let restricted 2 sentence ofthe language L(<, 0, 1) whether it is true in 2k. T= U {Sd: d E 1(s(y))}, and

Section 2. The Nonextension Conditions Tb, = Tfn {(x,y): x = i mod k}. Suppose that P and Q are finite partially ordered sets with 0 Note that T = U{Tb : i < k}, where u denotes disjoint union, and 1, such that P is a subordering of Q andf: P -- 2k is a partial and Downloaded by guest on September 25, 2021 Mathematics: Slaman and Soare Proc. NatL Acad Sci USA 92 (1995) 619 cardinality of %. For each i, 0 ' i < c, we put node a = Oi in T, assign to a the requirement, [( (bi-q ))] Pa: Va. TUa, [7] For allp P define 7n(p) = Sp U (Ub::pTb,). It is easy to check that for all q, p P, q p iff ir(q) C w(p). where (Va, Ua) is the ith pair in T, and we directly code Va into Let L be the distributive lattice generated by the sets 7r(p), Ua by putting Cx, a) into Ua exactly whenx is enumerated in V,r. p C P, and let OL and 1L be its least and greatest elements. Hence, 1. =T U[a' CT UO. We call Va the direct coder for Pa. Clearly, L satisfies inequality [4], because For the incomparability requirements Y $ X, we now let ((Fe, Xe, Ye) be an enumeration of all triples (4, X, Y) such that 4) U {Xr(b): b E a(x)} D T D U{i7(d): d E (si(Y)} is a computable partial functional, Q 1= Y $ X, and at least one ofX and Y is the Q - P. For each a, Ia 2 c, let (4)a,Xa, Ya) D nf{(d):dEsEi(Y)}. be (4e, Xe, Ye) where I a I = c + e. By hypothesis we have the Now we embed L into Wt. negation of condition [1] and hence the negation of expression Case i: Suppose W{7r(p) : p E g(x)} < 1L. Hence, by [3], namely inequality [4] ds(y) contains a member other than a 1L. By E & - & ; result of Lachlan, Lerman, and Thomason (see ref. 15, (3Ca,Da P)[Da 2 ga(Xa,) Ca 5s(Ya) Da C]. [8] theorem IX.2.2), any countable distributive lattice can be For each a E T, I a I 2 c, we assign to a the corresponding embedded into the c.e. degrees (R, ', V, A, 0, 0') by an requirement Ra: 4?xa # Ya. We now describe the a-module for embedding which preserves suprema, infima, and least ele- meeting Ra. First we build a functional T'a such that ment. Let f be such an embedding of L into the c.e. degrees (R, -, v, A, 0, 0'). Forp E L define f(p) = f(p) if < IL sXa = Y a>AXa = ca. [9] andf(p) = 0' otherwise. It is necessary to usef rather than f because Theorem 3.1 requires an embedding preserving Let p5 denote the value of parameter p at the end of stage s. greatest and least elements and the mapping f may not We drop the subscript a. Define the length function and preserve the former. However, the key point is that f and restraint functions, hence f preserve inequality [4], namely, e(s) = max{x (Vy arguments y < x. ', V, A, 0, 0') by an embedding which preserves suprema, At stage s + 1 restrain with priority a any x . r(s) from infima, and greatest element. This is the dual of Case i. entering X, or entering any Vp E Q - P such that (Vp, Up3) E ' and Up = Xa. In addition, ifx < t(s), +i5(x) , and also ps(x) Section 4. The Extension Theorem , then for the least suchx at stage s + 1 define TI(x) = Cs(x) and define qi(x) to be the least fresh z E 0k'J such that z > THEOREM 4.1. If P and Q satisfy the negations of conditions zs(x), s. If later some elementy ' Sp(x) entersX U Y, then q+(x) [1] and [2], then any embedding of P into R can be extended to becomes undefined. We may later redefine +i(x) as above, an embedding of Q into kR. maintaining the property sp5(x) < +5(x) if both are defined. If Proof: The negation of condition [2] implies the following at some later stage t, x E C,, and T(x) = 0, then at stage t + weaker hypothesis, which is equivalent to it in the presence of 1 we enumerate +'t(x) in Y, and redefine t,+,(x) = 1. the negation of condition [1], We check that the a-module succeeds. If 'I = Y, then clearly TxPY = C. Hence,X .TXe Y _T C. But VF = Y (Vx E Q - P)[2(x) 0 > 9(si(2(x))) C g(x)]. [5] implies limse(s) = 00, so limsr(s) = 00, and each x 0 X is eventually permanently restrained from X. Only 13-require- We use the notation and methods of ref. 15, chapters VIII ments for 1B < a can lift an a-restraint, so computably in and XIV. Fix an of P into For each a P fix embedding X R. wa(X) we can tell whether a given restraint is permanent, as a c.e. set A 4(a). We identify P with 7r(P), identify the in Lemma 4.2 below. Hence, X CT w\(X). Using this and the elements of Q with the c.e. sets in P and those we are first clause of expression [81, we have D _T WV (X) :T X. constructing for Q - P, and write X < Y for X CT Y ifX, Y E Hence, D .T C, contrary to the last clause of expression [8]. Q. We let A, B, C, D represent members of P and X, Y, Z, U, Hence, the a-module satisfies requirement Ra. V represent members of Q. For 9' C P let w(&) be the join Definepa = (px)[4)xa(x) # Ya(x)]. Usingpa, it is easy toFshow E3{XiX E W}, where E$ is the usual join operation. We will thatra = lim infr(a,s) < cc. From q1'a,s(x) define the total function assign to each member of Q - P a c.e. set X so that X CT Y ,s(x) = qia,s(x) if 4'a,s(x) I and = s otherwise. Note that a,s(x) iff Q J= X < Y We must meet comparability requirements Y and q'c(x) t iff = 0. From pa define da c X and incomparability requirements Y $ X The compara- p'a,s+i(x) limsq'as(x) = (px)[limS#ia,s(x) = m0]. Now Y'1] is computable. Define ca = bility requirements oftype I are those corresponding to the pairs max{iap,(y)ly < da, s E co}. Given z E w@[a], z > ca, find s such almost all integers are eventually available to the a-module for either enumeration or restraint. This motivates our tree T P = {A, B} U {0, 0'} and functions F and G below. in whichA andB are incomparable, and 0 and 0' are least and If a E T, lal 2 c, let a^(d, r) 8 Twhere d, r C o. Define greatest elements. Let Q be the extension obtained by adding the true path f E [T] as follows. For n < c, define f(n) = 0. If Z and X to P so that Z is belowA, and X is above both B and a = fin for n - c define f(n ) = (d4, r0) for d4 and ra as above. Z, Xis not aboveA, and Z is not belowB. In this example, sdl(Z) Define as usual a computable approximation {fl,sE. such that is {A, 0'}, a(si(Z)) is {A, 0}, and M(X) is {O, 0}. Since Z lim inf,tf = f. We say that s is an a-stage if s = 0 or if a C f. Define G(a, s) = min{ (3r)[f3^(d, r) C if > c X and Qa(X) Z g(si(Z)), !(X) is {Z}. Then, sit((X) U I,sI(d) I al} Ia I 0, 5(X)) is {0'}, g(dsi(I(X) U ga(X))) is not contained in ga(X), and = s otherwise. Let s' be the greatest a-stage < s, if there and Theorem 5.1 applies. is one and s' = 0 otherwise. Define The construction used to prove Theorem 5.1 is impossible to if = y < G(a, s) describe as succinctly as the one in the previous section. We {.X;s(x) ,s(x)X will discuss a simplified construction for the example and the 4>aX' s(x) = & X.s Y = IIy, ingredients which appear in the general case. X',S We construct a partial order embedding of P into a which T otherwise, cannot be extended to one of Q. Let (4)e, Ze), e E (o, be an enumeration of all pairs (4), Z) such that 4) is a computable < a & 2c- & [X=Ya V E]} partial functional and Z is a c.e. set and let 0e be an 9;(a) ={3I (Ya,XP) enumeration of all computable partial functionals. For all e E and F(a, s) = max{ r(,3, s) 1,B C 9;(a)}. w, we meet the following requirements. Next we define e (a, s), mii(a, s), and r(a, s) as above but with Pe :0,B A, Fa,s(x) in place of 'Fa,s(x). Hence, r(a, s) < G(a, s). Notice = that r(a, s) can increase only at an a-stage. The modified Qe e = Ze => eeB =0 or (3e*)[A4. Ze], a-strategy is roughly the same as above except that a can define aIic,s+I(i) = z only ifa Cfs+1, but if i 8 Da,s and 4i,s(i) =Z > where Fe and Ae* are computable partial functionals which we F(a, s), then a will enumeratez in Ya.s+1 immediately (whether construct. Satisfying the above requirements is enough to or not a C fs+ ) and initialize every fy E T such that either a prove any instance of 0 A B, using the fact that RIt is dense. computable set for its exclusive use. for B B For H C w define The prototypical strategy to satisfy Pe* appears in the every 8 Oa(Xa), I[r = B, [r. Friedberg-Mucnik papers: pick a number w which is not inA; OL(H)={y E H V [y =(, a) & x E Va nl(H) temporarily, prohibit w from entering A; if there is a stage s during which 09eB3s(w) is equal to 0, then enumerate w into A & (Va, Ua) E C ]}. and prohibit any number less than O',s(w) from entering B. Thus, w enters A if and only if 0B predicts that w is not in A. LEMMA 4.2 (CORRECTNESS LEMMA). Suppose z < s, z c r(a, s), To satisfy Qe, we first enumerate a functional Fe attempting rf(a, s) is 9J(Xa)-correct, z 0 Xa, and z 0 OU(Eass), where to make rez[B equal to 0'. Subsequently, we may find that making rezFeB total interferes with our satisfying an incompa- E.,, = {tf, ,,(i) P s, or at some stage t > s we In enumerating Fe, we observe several conventions. We only initialized a (and hence all y - a). In either case z § Xa. enumerate computations into Fe which have uses belonging to LEMMA 4.3 (INCOMPARABILITY LEMMA). All the incompa- its assigned computable set. During stage s, let 4e(s) be the rability requirements Y $ X in Q are satisified. greatest 1 such that for allx less than 1, Ces(x) is equal to Ze,s(X). LEMMA 4.4 (RECOVERY LEMMA). For every Z E Q, Z CT W We act for the sake of re during stages when ee achieves a value which is larger than ever before; we call these Se-expansionary The proof uses that if (Z, Xp) E % then Z ¢ !(Xa),so by the stages. Action means the following. For each x such that definition of !(X) in condition [2].!(X3)) C a(si(Z)). Thus, FeZ,j¶y 5(x) is defined and not equal to 0s'(x), we enumerate the the procedure of Lemma 4.2 can be done computably in use ye eBS(x) into B. Then, for each x less than ee(s) such that ga(A(Z)) to determine whether q, = z is permanently F1's5i3(x) is not defined, we define rz-eZ$Bs(x) to equal 0s(x) restrained by 3 < a, enumerated in Xa, or Ii,,q(i) # x at t > s. and define -y7s @Bs(x) greater than the maximum of yz, s/Bs(x - LEMMA 4.5 (COMPARABILITY LEMMA). All the comparability 1) and e's(x). In other words, for each such x we enumerate requirements V _ U are satisified. a computation into Fe which makes use ofZe,s EDBs yze'9BI(x) The only remaining pairs to be considered are those of the and assigns the value of 0s'(x) to argument x. form V E Q - P and V E !(U). In this case by condition [5] There is a clear conflict between the strategy to enumerate we have 9((si((U))) C §h(U), and thus in particular, g(si1(V)) Fe and the Friedberg-Mucnik strategy. Once rze'sBs(x) is C gJ(U). But by Lemma 4.4, we have V 5T Wa(sl(V)), and defined it is impossible to prohibit y7-.sBs(x) from being clearly W§h(U) 5T U, SO V 5T Ua enumerated into B. This enumeration could be necessary to reflect a change in 0'. Section S. The Second Nonextension Theorem In the tradition of Lachlan (14), we resolve the conflict as follows. Consider the case in which there is exactly one active THEOREM 5.1. If P and Q satisfy condition [2], then there is strategy of higher priority and it enumerates re. We modify the an embedding of P into a which cannot be extended to an Friedberg-Mucnik strategy for Pe*: pick a number wo; pick a embedding of Q into '?I. number wwhich is not inA; temporarily, prohibit w from entering Downloaded by guest on September 25, 2021 Mathematics: Slaman and Soare Proc. Natl. Acad. Sci. USA 92 (1995) 621 A; if there is a stage s such that 0e,s (w) is equal to 0, then There are two other families of incomparability strategies to enumerate w into A, prohibit any lower priority strategy from satisfy instances of Oc # D. If C is not above QP(X), then it is enumerating numbers less than OeB'S (w) into B; at the next possible to preserve finite initial segments of C by diverting thi (De*-ex ansionary stage t, if y1eef9B'(wo) is not defined or is greater feedback from the Ie and A4* strategies to a component of B than Oe*s,s (w), require that all subsequent values of -ye are greater which is not below C. Thus a direct variation on the Friedberg- than OBss (w) and maintain the restraint onB, otherwise, drop all Mucnik strategy can be used to satisfy such requirements. If C restraint on B, enumerate ye,1Bt(wo) into B, require that all is above A(X) and above 2(X), then C is also above H. Then, subsequent values of Ye are greater than t, and begin again with C is above all the sets which drive the Qe strategies. We a larger value for w. Additionally, begin again with a larger value introduce a family of incomparability strategies which either for w whenever a number less than wo is enumerated into 0'. preserve a finite initial segment of C or have an asymptotic There are several possible asymptotic behaviors for this behavior from which we may conclude that some component strategy. Similarly to the Friedberg-Mucnik strategy, it could of Ze fails to be computable in C in the way already preserved establish an inequality between Oe0*(w) and A(w) in finitely by$e. many steps. It could also act during infinitely many stages t to The strategies are combined in a H13-priority construction. enumerate zZ,eiEDBt(Wo) into B. In the infinitary case, rZe/Bt(wo) The combinatorics are similar to those in the solution to a goes to infinity as t does. In addition, Ze exhibits a behavior childhood puzzle, the Tower of Hanoi. which is compatible with its being computable in B: once a computation of Be*,s (w) is preserved, Ze cannot change below T.A.S. was supported by National Science Foundation Grants DMS YZeSsBS(wo) unless B first changes below OBss(w). Moreover, if 89-10312 and DMS-92-12022-A01, and R.I.S., by National Science B is not changed below OS (w) before the next (e- Foundation Grants DMS 91-06714 and DMS 94-00825. expansionary stage, then Ze can be permanently prohibited from changing (on a finite set) by preserving the computation 1. Godel, K. (1931) Monatsh. Math. Phys. 38, 173-198. of Ze from A. Thus, in the infinitary case, we can enumerate 2. Godel, K. (1965) in The Undecidable: Basic Papers on Undecidable a functional Ae*, establish ABe = Ze, and recover the ability to Propositions, Unsolvable Problems and Computable Functions, ed. put numbers into A while preserving B. Davis, M. (Raven, New York), pp. 39-71. 3. Turing, A. M. (1939) Proc. London Math. Soc. 45, 161-228. In the general case, we let B enumerate the elements of 4. Post, E. L. (1944) Bull. Am. Math. Soc. 50, 284-316. A(X). We let H be an element of R(s4(2(X) U @(X))) - 5. Friedberg, R. M. (1957) Proc. Natl. Acad. Sci. USA 43, 236-238. @(X). We use H in the role of 0' to be coded into the possible 6. Mu6nik, A. A. (1956) Dokl. Akad. Nauk SSSR N.S. Math. 108, images for !(X) D @(X). There are three types of require- 194-197. ments: comparability, incomparability, and the general version 7. Sacks, G. E. (1963) Ann. Math. 77, 211-231. of Qe. To formulate the general Qe, let n be number of Zi in 8. Sacks, G. E. (1963) Ann. Math. 80, 300-312. 2(X) and for each i less than n let mi be the number of 9. Shoenfield, J. R. (1965) in Symposium on the Theory of Models, eds. Addison, J. W., Henkin, L. & Tarski, A. (North-Holland, Aij in A(Zj). Let Ze, (De be an enumeration of all possible Amsterdam), pp. 359-363. values for 2(X) and computable functionals relative to theA,Aj. 10. Lachlan, A. H. (1966) Proc. London Math. Soc. 16, 537-569. 11. Yates, C. E. M. (1966) J. Symbolic Logic 31, 159-168. Qe (Vi < n)(Vj < mi)(Q,V,I;) = Ze,i) > 12. Robinson, R. W. (1971) Ann. Math. 93, 285-314. 13. Robinson, R. W. (1971) Ann. Math. 93, 586-596. = H or (3C)(3i < n)(3e*)[C 0 d(Zi) & Ac = Ze,] 14. Lachlan, A. H. (1975) Ann. Math. Logic 9, 307-365. 15. Soare, R. I. (1987) Recursively Enumerable Sets and Degrees: A The comparability requirements C 2T D are satisfied by Study of Computable Functions and Computably Generated Sets directly coding D into C. (Springer, Heidelberg). The Qe requirements are satisfied by first building a Fe and 16. Harrington, L. A. & Shelah, S. (1980) Bull. Am. Math. Soc. 6, 79-80. attempting to establish Fe7d = H. Making Ie total may 17. Fejer, P. A. & Shore, R. A. (1985) in Recursion Theory Week, eds. interfere with the satisfaction of an incomparability require- Ebbinghaus, H. D., Muller, G. H. & Sacks, G. E. (Springer, ment Oc D for some C which is above GA(X) and not above Heidelberg), pp. 121-140. 2(X). In that case, we let i(c) be the least index i such that Z1 18. Slaman, T. A. & Shore, R. A. (1991) Ann. Pure Applied Logic 52, ; C, make Fe partial, and build Ae* to compute Ze,i(c) from C. 1-25. These are the general Pe-strategies based on the example. 19. Ambos-Spies, K (1980) Ph.D. thesis (Univ. of Munich, Munich). Downloaded by guest on September 25, 2021