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Chapter 6 The Ehrenfeucht-Fraïxsséa Game
CHAPTER 6 THE EHRENFEUCHT-FRAISSE GAME
$1.
THE PLAY OF THE GAME
Suppose we are given two linear orderings A and B. Two players, called and PLAYER 11, play the following game. First, PLAYER I picks an element from either A or B and PLAYER 11 picks an element from the other one. This procedure is repeated a total of three times, so that when the game is completed, three (not necessarily distinct) elements of each linear ordering have been selected. Let ail,aiz,ui, be the three elements of A chosen in rounds 1, 2, and 3, respectively; and let b,,, b,,, b,, be the three elements of B chosen in rounds 1, 2, and 3, respectively. For example, suppose that A has six elements {ail 1 I i 5 6) in their natural order and that B has seven elements {bjl 1 5 j 5 7) in their natural order. Suppose that PLAYER I chooses b, and PLAYER 11 chooses a4 in round 1, then PLAYER I chooses a2 and PLAYER 11 chooses b , in round 2, and finally, PLAYER I chooses a3 and PLAYER II chooses b, in round 3. We have marked, in the diagram below, each point chosen with the player who chose it and the round in which it was chosen.
PLAYER I
u2
a3
a4
. o o o I,
I1,2
11,
2
1,
3
3
0
0
bl
h2
11,
a5
a6
1 151
. h3
. a * . b.4
b5
b6
b7
Note that a i l , u i 2 ,ai, are u 4 , ci2, a 3 , respectively, and that b,,, b,,, b,, are b,, b,, b,, respectively. We say that PLAYER II has won a particular play of the game if, as the result of his strategic selections, the elements a i l ,ai2,ui3are in the same order in A as the elements b,,, b,,, b,, are in B. If PLAYER 11 has failed, then we say that PLAYER I has won the play of the game. 93
94
6.
THE EHRENFEUCHT-FRA’ISSE
GAME
In the particular play of the game described above, since a,, a 2 , a, are in the same order in A as b, ,b , ,b, are in B, PLAYER 11 has won. If PLAYER I1 had foolishly selected b6 or b, in round 3, then PLAYER I would have won; on the other hand, if he had selected b, or b, in round 3, then he would still have won. (PLAYER II could also have selected 6 , or b, in round 3, but he would then also have lost.) Going back one step, we ask whether PLAYER I could have done better on his third move, for the move he did make allowed PLAYER 11 to win. If PLAYER I had selected a, or a6 as his third move, then PLAYER 11 could have responded with either b6 or b, and won; or if PLAYER I had chosen b6 or b,, then PLAYER 11 would have chosen a, or a6 and won; or, again, if PLAYER I had chosen b,, b,, or b,, then PLAYER 11 could have won by choosing a,. If, however, PLAYER I had chosen a, on his third move, PLAYER 11 would be in a real quandary; for whichever element b,, of B he chose, b,, b , , bj, would not match up with a,, a 2 ,a,. Thus, if on his third turn, PLAYER I had chosen a , , PLAYER 11 would not have been able to prevent him from winning. Thus PLAYER 11’s second move permitted PLAYER I to guarantee a win for himself. Could PLAYER II have selected a different element on his second turn and, by so doing, forced a win? The reader can verify that he could have done so by choosing either b, or b, on his second turn (instead of bl). The reader should also verify that if PLAYER 11 chose b, on his second move, then he would be no better off than he was with b,. (Note also that if PLAYER 11 chose b6 or b, at his second move, then he would have already lost since b,, b, would not match up with a4, a 2 .) Since PLAYER 11, by playing better at move 2, could have won the game, we are naturally led to question PLAYER 1’s second move. Could PLAYER I, by making a different move on his second turn, have forced a win? The reader should verify that, in fact, he couldn’t; to make this verification, the reader should find PLAYER 11’sproper response to each possible second move Of PLAYER I.
Backtracking one more step, we see that PLAYER 11 made a correct first move-correct in the sense that, no matter what PLAYER I did on his second and third moves, PLAYER II could, by moving properly, always win the game. Thus we may say that, once PLAYER I chose a, on his first turn, PLAYER 11 then had a winning strategy. If PLAYER 11 had a winning strategy in response to every possible first move of PLAYER I, then we could reasonably say that PLAYER 11 had a winning strategy for the entire game. On the other hand, if PLAYER I had a first move in response to which PLAYER 11 had no winning strategy, then no matter what PLAYER 11 did on his first move, PLAYER I could respond in such a way that PLAYER II would have no winning strategy at that point; and pursuing
1.
THE PLAY OF THE GAME
95
this chain of thought, we see that PLAYER I could be said to have a winning strategy for the entire game. Thus, at the very outset, either PLAYER I or PLAYER 11 can force a win in this game. (This result, of course, follows from the general theorem of game theory that given any finite two-person game with complete information such that each play of the game ends in a victory for one of the two players, either the first player has a winning strategy or the second player has a winning strategy.) Returning to our game above, we ask whether PLAYER I or PLAYER I1 has a winning strategy. If PLAYEK I has a winning strategy, then his first move was terrible; if, on the other hand, PLAYER 11 has a winning strategy, then whichever first move PLAYER I substituted for his actual move, the result, assuming best play by PLAYER 11, would have been the same. The reader should try to determine for himself whether PLAYER I or PLAYER 11 has a winning strategy before reading on. It seems fairly evident that PLAYER 11 has a winning strategy. For PLAYER I1 can essentially mimic PLAYER 1’s moves, choosing an endpoint whenever PLAYER I chooses an endpoint, a point closer to the center whenever PLAYER I chooses one closer to the center. and a central point whenever PLAYER I chooses a central point. The details are easily verified. We now proceed to a more general discussion. But first we should point out that the paragraph above is incorrect; it is intended to deceive the reader who did not really try to find out who has the winning strategy and to provide a hint to the reader who tried unsuccessfully. For it may lead him to see that if the first player chooses the central point of B, then the second player cannot really choose a central point of A . If, for example, he chooses a 3 , then the first player can choose b, with devastating effect. Thus the first player actually has the winning strategy in this game. EXERCISE 6.1: In each of the following cases, determine who has a winning strategy and what that strategy is.
(1) A = 7 , B = 8 . (2) A = 0 , B = q. ( 3 ) A 5 w B 5 [. (4) A 5 w, B 5 w + w. (5) A - w + 2 + w * , B 2 . 0 - t I+U*. (6) A w + 2 + a*,B = UJ -t 3 w * .
+
In general, of course, there is no need to restrict ourselves t o the game with three moves. Thus for any fixed natural number yt 2 1 and any linear orderings A and B we have the game G,,(A,B), which is played as above with
6.
96
THE EHRENFEUCHT-FRA‘ISSE
GAME
n rounds instead of three rounds. These games (and other similar games discussed in Chapter 13) were first formulated by Ehrenfeucht [l], who used them in his analysis of the logical properties of the arithmetic of ordinals. A similar notion, without the game-theoretic terminology, had been formulated independently by Fraisse [3]. (Ehrenfeucht notes that, for the case of the ordering of the ordinals, the condition he presents is “only a new formulation of the condition given by Fraissk.”) We will discuss the applications of these games to the ordering of ordinals at the end of this chapter, and we will return to the more elaborate games in Chapter 13. For a contemporary discussion of their work, see Feferman [2]. The following formal definitions are motivated by the discussion above. DEFINITION 6.2: Let A and B be linear orderings and let n 2 1 be a fixed natural number. A play of the game G,(A,B) consists of an ordered sequence of n repetitions of the following: PLAYER I chooses an element of A or B and PLAYER II chooses an element of the other. The element of A selected at the rth turn is denoted a, and the element of B selected at the tth turn is denoted b,. We say that PLAYER 11 has won the play of the game if for each s,t I n, a, < A a, if and only if b, < B b,; otherwise we say that PLAYER I has won the play of the game. We say that PLAYER 11 has a winning strategy in G,(A,B) if there are functions f i , f i , . . . ,f, such that (i) the domain o f f , is the set of all ordered t-tuples of elements of
A v B;
(ii) given c l r c 2 , . . . , c, E A u B (representing the choices of at the first t turns)
PLAYER I
(iii) if c , , c 2 , . . . , c , ~ A u B a n d i f w e d e f i n e a, =
b, =
f,(cl, . . . ,c,)
if c, E A if c, E B,
f,(c,, . . . ,ct)
if c, E B if c, E A
iCi
for every t 5 n, then for every s, t I n, a, < A a,
if and only if
b, < b,.
Otherwise we say that PLAYER I has a winning strategy. Given the technical and notational complexity of these definitions, it is not surprising that, in the proofs of the various claims below, we avoid
1.
97
THE PLAY OF THE GAME
using the formal definitions and instead rely on intuitive arguments. The disbelieving reader may, if he wishes, convert our intuitive arguments into more formal ones. NOTATION : If PLAYER 11 has a winning strategy in the game G,(A,B), then we will write G,(A,B ) E 11; otherwise we write G,,(A,B) E I. The following lemmas are easily verified. LEMMA 6.3: (1) G,(A,B ) E I1 if and only if G,(B,A ) E 11. (2) !f A = B, then G,(A,B ) E 11. A LEMMA 6.4:
I f G,(A,B)E I1 arid 1 i m < n, then G J A , B ) E 11.
LEMMA 6.5:
(1) I f G , ( A 1 ,B , ) E I1 and G,(A,,B,)
&(A,
11, then
+ .4,,B, + B2) E 11.
(2) I f G , ( A , , B , )E I1 for each
E
E
A
I E
J , then G,(C{A,Ii E J } , c ( B i l i E J } )
11, where J is an arbitrary linear ordering.
(3) I f A has a largest (smallest)clement and B does not, then G,(A,B ) E I. (4) I f A is dense and B is not, then G , ( A , B ) E I. A
We will soon show that the binary relation on linear orderings defined by G,(A,B ) E I1 is actually an equivalence relation. The lemmas above should thus be read with each G,(A,B ) E I1 replaced by the corresponding A -,B. Theorem 6.6, which we prove below, is a basic result which allows us to resolve games with n 1 moves into games with n moves, and thereby allows us to perform induction arguments.
+
THEOREM 6.6: G,, l(A, B ) E I1 If and only
if
(i) .for euery a E A there is a h E B such that
G,(A'",B'*)
E
I1
and
G,(A'",B'b)
E
11; and
(ii) for every b E B there is an a E A such that
G,(A'a,B'b)
E
I1
and
G,(A",Bib)
E
11.
Proof : (+) Suppose that PLAYER II has a winning strategy in G,, l(A,B ) and suppose that a E A . Using the strategy, we can find an element b E B
98
6.
THE EHRENFEUCHT-FRA'I'SS~
GAME
such that if PLAYER I chooses a at his first turn, then PLAYER 11 should choose h. There are now n turns left in the game G,+,(A,B).Whenever PLAYER 1 chooses an element in A'" or B < b ,the strategy, since it produces a win for PLAYER 11, will always choose an element in B'b or A'". Thus PLAYER I ~ S winning strategy for G,+ l(A, B) includes within it a winning strategy for G,(A<", BCb),and similarly it includes a winning strategy for G,(A'", B > b ) . This proves (i). By symmetry, (ii) is also correct. ( G ) Assuming that (i) and (ii) are correct, we describe a winning strategy for PLAYER 11 in G , , l(A, B). If PLAYER I chooses an element a E A on his first turn, then PLAYER 11 uses (i) to find an element b E B. Thereafter, whenever PLAYER I chooses an element of A > " or ' B b, PLAYER 11 uses his winning strategy in G,(A'",B'b) to respond; and similarly, whenever PLAYER I chooses an element of A'" or Bcb, PLAYER 11 uses his winning strategy in G,(A'",BCb) to reply. Since there are only n subsequent moves in the game, PLAYER I can choose no more than n times from A'" or B e b (and from A'" or B z b ) .Hence PLAYER 11's winning strategies in G,(A'",B'b) and G,(A<", F b provide ) him with moves in all contingencies. If, on the other hand, PLAYER I chooses an element b E B,then PLAYER 11 uses (ii) to find his correct first move and then proceeds analogously to the above. Thus PLAYER 11 has a winning strategy in en+l ( A , B ) . COROLLARY 6.7: Let A , B, and C be linear orderings and let n 2 1 be a fixed natural number. If PLAYER II has a winning strategy in G,(A,B ) and PLAYER II has a winning strategy in G,(B,C ) ,then PLAYER 11 has a winning strategy in G,(A,C). Proof: We prove this by induction on n. For n = 1 the hypotheses imply that A and B are either both empty or both non-empty, and that B and C are either both empty or non-empty; from this it follows that A and C are either both empty or both non-empty, and hence that G,(A, C) E 11. Assume that the statement is true for n, and assume that G,+,(A,B) E I1 and G,+ ,(B, C) E 11. We will show that G , , l(A, C) E I1 using Theorem 6.6. Let a E A . Then by Theorem 6.6, there is an element b E B such that G,(A <", B < b )E I1 and G,(A>",B S b )E 11. Since G , , l(B, C ) E 11, by Theorem 6.6, there is an element c E C such that G,(BCbb, C<') E I1 and G,(B>b, C") E 11. By the induction hypothesis it follows that G,(A'", C < ' ) E I1 and G,(A'", C") E 11. Thus if a E A , there is a c E C such that G,(A'", C<') E I1 and G,(A'", C") E 11, verifying (i) of Theorem 6.6. Clause (ii) is similarly verified. Hence, by Theorem 6.6, G , , l(A, C) E 11. 4
We are now ready to make the following definition.
1. THE
99
PLAY OF THE GAME
DEFINITION 6.8: We say that A is G,-equivalent to B, written A -,, B, if PLAYER 11 has a winning strategy in G,(A,B). We say that A is G-equivalent to B, written A B, if A -,, B for every n E N . By Lemma 6.3 and Corollary 6.7, these are equivalence relations. Intuitively, if A -,, B, then PLAYER I cannot demonstrate any difference between A and B in the n moves at his disposal.
-
We now discuss various implications of Theorem 6.6.
COROLLARY 6.9: Let A and B be ,finite linear orderings of cardinality at least 2" - 1. Then PLAYER 11 hay ti winning strategy in G,(A,B). Proof: We prove the corollary by induction on n. Assume that n = 1. Then both A and B are non-empty. Hence clearly PLAYER II has a winning strategy in G , ( A ,B). Assume the corollary true for n and let A and B be finite linear orderings of cardinality at least 2"' - 1. To show that PLAYER 11 has a winning strategy in G , + , ( A , B ) it suffices to show that (i) and (ii) both hold. We will deal only with (i), since (ii) is analogous. Let a E A. Suppose that IA'"I = nL and that IA'"I = n R . Since IAl 2 2"" - 1, if nL < 2" - 1 and nR < 2" - 1, then
[ A (= 1 + nL + nR < 1 + (3" - 1)
+ (2"-
1) = 2"+l - 1,
contrary to hypothesis. Hence u cannot be both one of the first 2" - 1 elements of A and one of the last 2" - 1 elements of A . Similarly, if b E B, then b cannot be both one of the first 2" - 1 elements of B and one of the last 2" - 1 elements of B. If now the given element a E A is one of the first 2" - 1 elements of A, we let b be the corresponding element of B ; for example, if a is the third element of A , then b is the third element of B. Then A'" 2: B'], IFb] 2 2" - 1, so that by induction hypothesis G,(A'", B'b) E 11. If the given element a E A is one of the last 2" - 1 elements of A , we let b be the corresponding element of B and proceed as above. If the given element a E A has at least 2" - 1 predecessors in A and at least 2" - 1 successors in A , then we choose b to be any element of B with at least 2" - 1 successors and at least 2" - 1 predecessors; since IBI 2 2"" - 1, such an element exists. Then by induction hypothesis, G,(A '", E I1 and G,(A'", B ' b ) E 11. Hence (i), and similarly (ii), holds. Thus, by Theorem 6.6 B ) if A and B each have above, PLAYER 11 has a winning strategy in G , , cardinality 2 2"+ - 1. This completes the proof. W
100
6.
THE EHRENFEUCHT-FRAYSS~
GAME
Note that a special case of this corollary is that PLAYER 11 has a winning strategy in G3(7,8)-which is included in Exercise 6.1. That no further improvement is possible in case n = 3 is seen from the fact that PLAYER I has a winning strategy in G3(6,7). EXERCISE 6.10: (1) Show that PLAYER I has a winning strategy in G,(h,k)ifk<2"- l a n d h 2 2 " - 1. (2) Show that if k < 2" - 1 and PLAYER II has a winning strategy in G,(A,k), then A 2 k. EXERCISE 6.11 : Let A be a finite linear ordering of cardinality at least 2" - 1. Then PLAYER 11 has a winning strategy in G,(A,B) where B 2: w t; . a + w* for any order type a. [Hint: Note the fact that if b E B, then either B > b or B C b(or both) is again of the form w + t; . a' + w* for some order type a'; and see the proofs of the corollary above and the corollary below. Note, however, that, since Corollary 6.12 uses the result of this exercise, it would be quite inappropriate to use Corollary 6.12 here.]
+
COROLLARY 6.12: PLAYER 11 hus a winning strategy in G,(w,w + ( . a ) jbr any order type a and any n 2 1.
Proof: We proceed by induction on n to show that G,(w, w + t; . a ) E I1 for every a. The case where n = 1 is obvious. Assume that the statement is true for n and proceed, as in Corollary 6.9, to use Theorem 6.6. Thus suppose that a E w = A ; then let b be the corresponding element of w + 4 . a = B. Then, since A'" c B < b ,G,(A'",B'b) E 11; and, by induction hypothesis, since B'b 2: w + t; . a, G,(A'",B'b) E 11. Suppose, on the other hand, that we are given b E B. If b is in the initial w of B, then choose a to be the corresponding element of A and proceed as above. If, on the other hand, b is not in the initial w of B, then B'b 2: w + 4 . a' and B'b c w + t; . a" + w*, where a' and a" are particular order types. Let a be the 2"th element of A . Then IA'"I = 2" - 1, so that, by the exercise above, G,(A'", B < b )E 11; and A'" z w, so that, by the induction hypothesis, G,(A'",B>b) E 11. Hence conditions (i) and (ii) of Theorem 6.6 are satisfied, so that PLAYER 11 has a winning strategy in G,(o,w + ( . E) for every a and for every n.
+
In particular, PLAYER II has a winning strategy in G,(w,w t;) for every n. Thus no difference between w and w t; can be discerned through the games.
+
1.
101
THE PLAY OF THE GAME
These games were used by Ehrenfeucht in his investigations of the logical equivalence (or elementary equivalence) of mathematical structures and, more specifically, of well-orderings. Informally, two mathematical structures are said to be logically equivalent if there is no statement in the first-order predicate calculus which distinguishes them. Ehrenfeucht showed that two structures are logically equivalent if and only if they are Gequivalent, and used this characterization of logical equivalence to obtain interesting results about well-orderings. Similar results were obtained by Fraisse, using different terminology. Thus the logical interpretation of Corollary 6.12 is that o is logically equivalent to w 5 . c( for any order type CI.Thus, for example, no statement of the first-order predicate calculus is true for w but false for w (. One consequence of this fact is that no first-order statement and, more generally, no set of first-order statements, can be true precisely for those linear orderings which are well-orderings. This is usually expressed by saying that the notion of well-ordering is not first-order definable or that the class of wellorderings is not an axiomatizable class. We will return to well-orderings shortly. The logical uses of the Ehrenfeucht games will be further discussed at the end of this chapter and in Chapter 13. The equivalence relation -"partitions the class of linear orderings into a number of equivalence classes. We will now prove that for each n there are only a finite number of equivalence classes; this contrasts strongly with the equivalence relation -, which, as we will see, has infinitely many (and, in fact, has c ) distinct equivalence classes.
+
+
THEOREM 6.1 3 : For each n there are only a Jinite number of equivalence classes modulo -,,.
Proof: We proceed by induction on n to show that there are only a finite number f ( n ) of equivalence classes modulo n. For n = 1, as we already know, there are but two equivalence classes. Assume the claim for n. Let A be a linear ordering. Then each element a E A determines an ordered pair ( [ A ' " ] , [A'"]) of equivalence classes modulo -,,, where [B] is the -,,equivalence class of B. Let I ( A ) = { ( [ A ' " ] , [A'"])la E A } . We note that Theorem 6.6 can be understood as saying that A - n + l B if and only if I(A) = I(B). Hence the number of equivalence classes modulo is at most the number of subsets of the set of ordered pairs of equivalence classes modulo -,; but this number is 2J(")'s(").Hence f ( n + 1) 5 2f""rcn), and in particular f ( n 1) is finite.
+
102
6.
THE EHRENFEUCHT-FRA~’SS~GAME
The reader might naturally inquire at this point what the values of f ( n ) actually are, and he may wonder whether the function f actually attains the upper bound given in the proof and therefore satisfies the recursion f(1) = 2, f ( n + 1) = 2f(“)
The value for f ( 2 )would then, for example, be 16. In point of fact, however, .fV) = 7. EXERCISE 6.14: (1) Show that f ( 2 ) = 7 by giving a complete set of representatives for the equivalence relation m 2 . (2) Find f ( 3 ) . [Caution: This is not an exercise, but a major project.]
The fact that there are only a finite number of G,-equivalence classes for each n suggests the possibility that there is a “nice” countable set A of order types such that for every n E N and every linear ordering A there is an M E .4 such that A -, M. We will see in the next chapter that this is indeed the case. THEOREM 6.15: ulo -.
There are at least c different equivalence classes mod-
We define, as in the proof of Proposition 1.48, for each subset X G N a linear ordering L ( X ) such that if X I # X , , then L ( X , ) is not C-equivalent to L ( X 2 ) . We define A , ( X ) to be of order type q + (a + 2) if a E X and of order type 1 if a $ X ; and we let L ( X ) = C { A , ( X ) l a< w ) . Let X , Y G N such that X # Y . Let a be an element in X - Y (or, alternatively, in Y - X ) . We claim that PLAYER I has a winning strategy in C,+.(L(X),L ( Y ) ) .Indeed, on his first a + 2 turns, he chooses in order the final a + 2 elements of A,(X). He then considers PLAYER 11’sfirst a + 2 choices b , , . . . , b,+2 in L( Y ) . If, for some i. there is an element of L ( Y ) between bi and bi+ then PLAYER I chooses that element on his (a + 3)rd turn and forces PLAYER 11’s resignation. Otherwise, since there is no maximal discrete sequence of exactly a + 2 elements in L( Y ) ,either there is an element b beyond b a + 2in L( Y )with nothing of L( Y )in between, or there is an element b before b , in L( Y )with nothing of L( Y )in between. PLAYER I chooses such an element for his ( a + 3)rd turn; suppose that he chooses such a b and suppose, without loss of generality, that b,, < L l y ) b. Then PLAYER II presumably chooses an element c , + ~of L ( X ) after the last element c of A , ( X ) in L ( X ) . But then Proof:
1.
103
THE PLAY OF THE GAME
PLAYER I chooses an element between c and c,+ ,-and since PLAYER II cannot find an element between and b, he loses. Hence PLAYER I has a winning strategy in G,+,(L(X), L ( Y ) ) .Thus if X and Y are different subsets of N , then L ( X ) is not G-equivalent to L,(Y). Hence there are at least c distinct equivalence classes of countable linear ordering modulo .
-
-
Can there be more than c diflerent equivalence classes modulo ? We will show first that every G-equivalence class contains a countable representative. (This is the combinatorial version of the Lowenheim-Skolem Theorem for the predicate calculus; this method of proof was suggested by J. Schmerl.) Since there are but c distinct countable order types, this shows that there are exactly c different G-equivalence classes.
-
THEOREM 6.16: (Lowenheim-Skolem) Let A be a linear ordering. Then there is a countable linear ordering B A such that B A .
Proof: We first note that, as in the proof of Theorem 6.13, given an interval C E A and given nE N , each element U E Cdetermines an ordered pair S(C,a, n) = ([C'"],, [ C > " ] , )of equivalence classes modulo -,, where [D], is the n-equivalence class of D. Furthermore, by Theorem 6.13, for each n there are but a finite number of possible S(C, a, n). Thus it is possible to select a finite subset S(C, n) E C so that for each U E C there is a unique b E S ( C , n) with S(C, a, n) = S(C, b, n). The countable subset B of A IS defined to be the union of an ascending sequence
$8 = B, s B , G B ,
S.-._C B,G
B,+l E ' . .
of finite subsets of A, defined by induction on n. Assume that B, has already been defined; the k = k(n) points of B, partition A - B, into k + 1 intervals C,, C , , . . . , C, (some of which may be empty.) Define
B,+l= B, u U{S(Ci,n)IO5 i C:" B
I k}.
Thus for any b E A and any n, either b E B, or b E C iand for some a E Ci, -,C
-To conclude that B A.
A, it suffices to show by induction on n that B - , A . As is often the case, a stronger induction hypothesis turns out to be necessary. We will show, by induction on n, that for any b E B , B X b-,A ' b and B > b-,A'b and that if b < c are in B, then ( b , ~-, ) (~ b , ~ )(Note ~ . that this implies that B -,A.) For n = 1 this is clearly correct. Turning to the induction step, we will confine our attention to showing that (b,c ) -,+~ (b,c ) for ~
104
6.
THE
EHRENFEUCHT-FRAYSSI?
GAME
b < c elements of B ; the proof that B’b A’b and -,+ A < b is similar. We may restrict ourselves further to the case where b and c are consecutive elements of some B,,, with m 2 n. [For if b < c in B, we can find an m 2 n such that b,c E B, and we can find c , , c 2 , . . . , c, in B , such that b = c , < c2 < . . . < c,- < c, = c are consecutive elements of B , ; but if ( c i ,el+J A for each i, it follows that we have shown that (cl,c, + I)B -,+ (b,c), -,,+ ( b , ~ ) ~Since . ] b and c are consecutive elements of B,, (b,c), is one of the k(m) intervals C used to define B,, in particular, for each x E (b,c ) ~ there , is a y E (b,c ) such ~ that (b,x ) ~, (b,y ) , and ( x ,c), -, ( y ,c ) ~ . In the game G,+ l((b, c ) ~ (h, , c ) ~ )if, PLAYER I chooses an element y E (b,c ) on ~ his first move, PLAYER II chooses y also, and by the induction hypothesis, since y E B,
(b,Y ) , -,, (b. Y ) ~
and
( Y ,c ) A
-,, ( Y , c ) B ,
so that PLAYER 11 will always be able to win. If, on the other hand, PLAYER I chooses a n element x E (b, c ) on ~ his first move, then PLAYER 11 can choose y ~ ( bc, ) so ~ that (b, x ) ~, (6, and (x,c ) ~-,, ( y , c ) ~ Then, . since Y E B, the induction hypothesis applies to the intervals (b,y) and ( y , c); that is, (b,Y ) A -,(b,Y ) B and (.v, -,, ( Y , c)B. Since m 2 n, (b, -,(b,yIB and (x, c), -,( y , c ) ~by , transitivity. Thus in this case also, PLAYER 11 will always , c ) ~E)11; thus be able to win. Hence, by Theorem 6.6, G , , (b, c ) ~ (b, ( b , ~ and ) ~ the induction step is complete. (b,c ) ~
COROLLARY 6.17: modulo -. Proof:
There are exactly c d i e r e n t equivalence classes
By Theorem 6.15 and Theorem 6.16.
The Lowenheim-Skolem Theorem, as formulated here, is a special case of a more general theorem which applies to arbitrary structures and has a stronger conclusion. The logical interpretation of Theorem 6.16 is that any linear ordering is logically equivalent to a countable linear ordering; that is, given a linear ordering A , there is a countable linear ordering B (which is actually a subordering of A ) which cannot be distinguished from A using statements of the predicate calculus. This will be discussed further in Chapter 13. One further comment before concluding this section. Ehrenfeucht games can, of course, be played with other structures, not just with linear orderings. For example, if A and B are groups, then, in a play of the game G,(A, B), PLAYER 11 wins if the selections b,, b,, . . . , b, from B match the selections a,, a,, . . . , a , from A in that ai x A a j = a, if and only if bi x E bj = b, for
2.
105
GAMES AND ORDINALS
each i, j , and k . The formal development for games on such structures will appear in Chapter 13. For now we leave it as an exercise to find winning strategies for either player in the games Gz((Q, + ),
G d(Rf, .), (Q+, .)),
(Q',
.))
and
G 3 ( ( R + , (Q,
+)I,
where R+ and Q' denote the positive reals and rationals, respectively.
2. GAMES AND ORDINALS
In this section, we present a complete answer to the question, Under what circumstances are two ordinals G-equivalent? By our earlier remarks, we get the same answer to the question, Under what circumstances are two ordinals logically equivalent? THEOREM 6.18:
For any naturtrl number n > 1,
(1) o" w" . p ,for any ordinul p > 0, and (2) w" * 2 n + w" B fbr uny ordinal /3 > 0.
+
Proof: Note first that (2) implies in particular that on+ z n + I wn . fl for any ordinal /? > 0 and hence that the result in (1) is best possible. We prove these two claims simultaneously by induction on n. For n = 1, we need to show that G,(w, w . p) E I1 for any /3 > 0 and that G,(w,o + p) E I for any p < 0. The first is obvious, and for the second, PLAYER I has a rather simple strategy. He chooses the point w in o + p and, if PLAYER II chooses m + 1 in u, PLAYER I chooses m in o and then a point in w + fl between PLAYER 11's response and o. Assume now that the claims are true for all k < n. We first show that PLAYER 11 has a winning strategy in GZn(w",w" . p) for any ordinal p > 0. We will describe PLAYER 11's responses to PLAYER 1's first two moves. He will be able to respond in such a way that if a < a' are the first two selections from wnand if b < b' are the first two selections from o". p, then, if A is any one of the three intervals formed in 0)" and B is the corresponding interval in on. p, either A 5 y and B u y for some y < on- or A 'v (con-' . PI) + y and B N (a"-' . p2) + y for some y < (on-' and some PI, p2 > 0. From this it will follow, using the induction hypothesis, Lemma 6.3.2, and Lemma 6.5.2 on sums of 2(n - 1)-equivalent orderings, that Gz,(w", o". p) E 11. But it is clear that PLAYER 11 can respond so that the above conditions are met by taking care that, if PLAYER I chooses a point of the form (on-'. Po) + y where y < a n - 'then , he chooses a point of similar form; and, in particular, if Po
6.
106
THE
EHRENFEUCHT-FRAYS&
GAME
is finite, then PLAYER 11 chooses the same point in the other ordering. (Note that every tail of wkhas order type w k ;this is a special case of Exercise 3.44.3.) ~ n . P)EI, we will describe the first two moves To show that G 2 n + l ( (LO" of PLAYER 1's strategy. On his first move he picks the point w" in wn p. Then, no matter which point a in (0" is picked by PLAYER 11, PLAYER I at his second turn picks a point b in w" so that (b, a) 'v wk for some k < n (unless a is a successor, in which case PLAYER I picks its predecessor.) Then, no matter which point d < w" in o"+ [I is picked by PLAYER 11, (d,0")= 0".But, by the induction hypothesis, PLAYER I has a winning strategy in Gzfl-' ( w k ,o")for any k < n. Hence Gz,+l(co",ion /?)€I for any p > 0.
+
+
+
COROLLARY 6.19: For any natural number n, if TX * 2 n + 2 w"+' and u - 2 n + 3 p.
ci
< w""
< p, then
+
Proof: Write ci = ~ " .l u , wn2. a2 + . . . + conk. uk in Cantor Normal Form. In the game Gzn+2 ( x , (fin+l), PLAYER I first picks a point a of ci which determines a tail of order type conk.Then no matter which point PLAYER 11 selects in a"+ ',the tail it determines has order type a"+ But, by the theorem, PLAYER I has a winning strategy in the game G2nk+l(wnk, w"+'), hence also ni PLAYER I has a winning strategy in in the game C Z n + l ( ~ n k , f').r JThus
'.
G 2 , + 2 ( c i , ~ 1 " + 1 ) i f< c iw " + ' .
In the game G2,r+3(ci, p) where ci < LO"+' < p, PLAYER I first picks the point w"+' in p. Then no matter which point a in a is selected by PLAYER 11, its predecessorshave order type < a"+ ',so PLAYER I can continue his strategy as in the preceding paragraph to obtain a win in Gzn+ 3(ci, p). The Exercise below shows, by example, that the results of Corollary 6.19 are the best possible. EXERCISE 6.20:
Prove each of the following for all n:
(1) W " . 2 - 2 , w f l + 1 + to". ( 2 ) W".2-2,+l w " + ' w". (3) w" . 2 *2"+2 Q"+ -tw". (4) W " . 2 - 2 n + 1 o"+I.
'
(5) (6)
W"'3h.2,+1
OJ"+'
+
+On.
3 * 2 n + 2 or"' + w" for n > 0. 1 w". (7) on 4 - 2 n + 2 O". '
+
Note that (4) and (7) show that the results of Corollary 6.19 are the best possible; this proof of that fact is due to P. Mulhall.
2.
107
GAMES A N D ORDINALS
(8) Show that every ordinal is G2,-equivalent to some ordinal in the finite set (w" . a,
+ w n - l . a,- + 1
c(j"-2
. an-,
+
'
'.
+0
'
a1
+ ao:
where each ai < 22" and a, I 1. THEOREM 6.21 : Let a and fi he ordinals. Write a = ow. a, + a2 and p = cu"' . fi, + p2 where a, < o" and [j2 < (d".Then c1 fl if and only if a2 = f12 and either r l and p1 are both 0 or are both bigger than 0.
-
-
-
Proof : Since Q"'= wn . ow,by Exercise 3.48.4, it follows from Theorem 6.18 that coo . r l ou. ,8,, for any a l and fil which are non-zero. Hence a p if the conditions are met. Now suppose that a, = 0 but PI > 0. Let a2 < on. If PLAYER I chooses . in j? at his first turn, then the initial segments a. and Po of u and p determined by the first turn satisfy a. < W" < Po, so by Corollary 6.19, PLAYER I can win in 2n + 1 further turns. Hence G2,,+ ,(a,p) E 1. Thus if u p it follows that a1 and p1 are both 0 or both non-zero. Suppose now that c1 < p < (0'". Write p in Cantor Normal Form as w n l. a , (on* . a2 . . . mnk . a k . We claim that PLAYER I has a winning strategy in Gn(a,p) where n = (a, a2 . . . + ak)- 1 + 2n,. Indeed, at his first u 1 turns, PLAYER I chooses the points wnl. 1, wnl . 2 , . . . , w"' . a , ofp; at his . 2 , .. ., next a2 turns, he chooses the points wnl . a, + mn2 . 1, wnl . a, + on2 (d"'. a , + wf12. a,; . . . ; at his nexl ak - 1 turns, he chooses the points (1)111 . a, + . . . + g n k - 1 . ak- 1 wnk' 1 , . . . , 0"'' a1 ' ' . Unk. (ak - 1). He now surveys PLAYER 11'sfirst ( a , + a2 + . . . + ak) - 1 moves. Since /3 > a, at least one of the intervals in a must be smaller than the corresponding nl, interval in p. Since the intervals of p are all of form wm for some m I PLAYER I needs at most 2n, further moves to win. Thus G,(u,p) E I as claimed. Hence if a fi and u, p < ow,then u = p. Similarly, suppose that u = cu" . a1 + a2 and p = 0") . PI + p2 where cil and fil are non-zero and u2 < p2 < w". If a, = 0, then PLAYER I can win in G2,(r,p) by selecting a point of /3. whose successors have form okfor some k < n, and then using the strategy of Theorem 6.18.2 for his remaining 2n - 1 moves. If a2 > 0, then by picking the point ow. u1 in u he leaves PLAYER 11 with three ways of losing: Either PLAYER II picks a point of p which is less than o0 . p,, in which case the successor intervals are a, and ow. -y p2 so, by Corollary 6.19, at most 2n 1 further moves are necessary; or PLAYER 11 picks the first point of p2, in which case the successor intervals are a2 and p 2 , so that PLAYER 1 can win in at most t = (a, a, + . . . + ak)- 1 2n,
-
+
+
+
+ +
+
+
+
-
+
+
+
+
108
6.
THE EHRENFEUCHT-FRAYS&
GAME
+
more moves where fl = m"' . a , . . . + wflk. ak,as in the preceding case; or, finally, PLAYER II picks a point beyond the first point of p 2 , in which case the predecessor intervals are owand w" y where 0 < y < m", and in this case PLAYER I can win in at most 2n + 2 additional moves. Thus in the case where 0 < ci2 < f12, PLAYER I can win in the game G,+,(u,fl) where m = max(2n + 3, t ) where n and t are bounds depending only on /?.This completes the proof. H
+
Theorem 6.21 enables us to conclude that the set of all ordinals less than
UP . 2 forms a complete and irredundant set of representatives of the Gequivalence classes of well-orderings. By way of contrast, although any wellordering is G-equivalent to one of a countable number of well-orderings, the analogous statement is false for arbitrary linear ordering, as we verified in Theorem 6.15. The logical interpretation of Theorem 6.21 is that the set of all ordinals less than LO"' . 2 forms a complete and irredundant set of representatives of the logical equivalence classes of well-ordering ; this fact was noted earlier by Mostowski and Tarski [4]. The first application made of the EhrenfeuchtFrai'sse analysis is the following result of Ehrenfeucht [11 and Fraisse [3].
-
THEOREM 6.22: Let O N denote the collection of all ordinal numbers, with the usual ordering. Then O N d". Proof : Since m" m Z f lo") for each n, by Theorem 6.18, it suffices to show that (d'- 2 n ON for each n. This is proved, by induction on n, making only notational changes in the proof of Theorem 6.18.1. H
Theorem 6.22 has the following logical interpretation : Given a first-order statement about the ordering of the ordinal numbers, that statement will be true if and only if it is true about the ordinal numbers below LO'). As we will see later, this interpretation makes it possible to conclude that the theory of the ordering of the ordinal numbers is decidable; that is to say, that there is an effective procedure which, when presented with any first-order statement about the ordering of the ordinal numbers, will determine whether or not that statement is true. This result was also proved by Mostowski and Tarski [4]. Ehrenfeucht [11 also proved that a first-order statement about addition of ordinal numbers is true if and only if it is true about the ordinal numbers below IO")'"; or, in the language of games, ( O N , <,
+) - (w""', <, +)
109
REFERENCES
Furthermore, going one step further, ( O N , <, +;)
-
<, +;>.
( ~ l u w w ,
For these results, Ehrenfeucht used a more elaborate game, which we will discuss in Chapter 15. REFERENCES Ehrenfeucht, A,, An application of games to the completeness problem for formalized theories, Fund. Math. 49 (1961), 129-141. [ M R 23, #3666] 121 Feferman, S., Some recent work of Ehrenfeucht and Fraisse, Summer Institute for Symbolic Logic, Cornell University. 1957, Summaries, pp. 201 -209. [3] Fraisse, R., Sur quelques classifications des systemes de relations, Publ. Sci. Univ. Alger. Ser. A 1 (1954), 35-1 82. [Abstract: .Ipplicutions scientifiques de la logique mathimatique, Paris and Louvain, 1954,851 141 Mostowski, A,, and Tarski, A., Arithmetically definable classes and types of well-ordered systems, Bull. Amer. Math. Soc. 55 (1949), 65 and 1192.
[I]
$1.
THE PLAY OF THE GAME
Suppose we are given two linear orderings A and B. Two players, called and PLAYER 11, play the following game. First, PLAYER I picks an element from either A or B and PLAYER 11 picks an element from the other one. This procedure is repeated a total of three times, so that when the game is completed, three (not necessarily distinct) elements of each linear ordering have been selected. Let ail,aiz,ui, be the three elements of A chosen in rounds 1, 2, and 3, respectively; and let b,,, b,,, b,, be the three elements of B chosen in rounds 1, 2, and 3, respectively. For example, suppose that A has six elements {ail 1 I i 5 6) in their natural order and that B has seven elements {bjl 1 5 j 5 7) in their natural order. Suppose that PLAYER I chooses b, and PLAYER 11 chooses a4 in round 1, then PLAYER I chooses a2 and PLAYER 11 chooses b , in round 2, and finally, PLAYER I chooses a3 and PLAYER II chooses b, in round 3. We have marked, in the diagram below, each point chosen with the player who chose it and the round in which it was chosen.
PLAYER I
u2
a3
a4
. o o o I,
I1,2
11,
2
1,
3
3
0
0
bl
h2
11,
a5
a6
1 151
. h3
. a * . b.4
b5
b6
b7
Note that a i l , u i 2 ,ai, are u 4 , ci2, a 3 , respectively, and that b,,, b,,, b,, are b,, b,, b,, respectively. We say that PLAYER II has won a particular play of the game if, as the result of his strategic selections, the elements a i l ,ai2,ui3are in the same order in A as the elements b,,, b,,, b,, are in B. If PLAYER 11 has failed, then we say that PLAYER I has won the play of the game. 93
94
6.
THE EHRENFEUCHT-FRA’ISSE
GAME
In the particular play of the game described above, since a,, a 2 , a, are in the same order in A as b, ,b , ,b, are in B, PLAYER 11 has won. If PLAYER I1 had foolishly selected b6 or b, in round 3, then PLAYER I would have won; on the other hand, if he had selected b, or b, in round 3, then he would still have won. (PLAYER II could also have selected 6 , or b, in round 3, but he would then also have lost.) Going back one step, we ask whether PLAYER I could have done better on his third move, for the move he did make allowed PLAYER 11 to win. If PLAYER I had selected a, or a6 as his third move, then PLAYER 11 could have responded with either b6 or b, and won; or if PLAYER I had chosen b6 or b,, then PLAYER 11 would have chosen a, or a6 and won; or, again, if PLAYER I had chosen b,, b,, or b,, then PLAYER 11 could have won by choosing a,. If, however, PLAYER I had chosen a, on his third move, PLAYER 11 would be in a real quandary; for whichever element b,, of B he chose, b,, b , , bj, would not match up with a,, a 2 ,a,. Thus, if on his third turn, PLAYER I had chosen a , , PLAYER 11 would not have been able to prevent him from winning. Thus PLAYER 11’s second move permitted PLAYER I to guarantee a win for himself. Could PLAYER II have selected a different element on his second turn and, by so doing, forced a win? The reader can verify that he could have done so by choosing either b, or b, on his second turn (instead of bl). The reader should also verify that if PLAYER 11 chose b, on his second move, then he would be no better off than he was with b,. (Note also that if PLAYER 11 chose b6 or b, at his second move, then he would have already lost since b,, b, would not match up with a4, a 2 .) Since PLAYER 11, by playing better at move 2, could have won the game, we are naturally led to question PLAYER 1’s second move. Could PLAYER I, by making a different move on his second turn, have forced a win? The reader should verify that, in fact, he couldn’t; to make this verification, the reader should find PLAYER 11’sproper response to each possible second move Of PLAYER I.
Backtracking one more step, we see that PLAYER 11 made a correct first move-correct in the sense that, no matter what PLAYER I did on his second and third moves, PLAYER II could, by moving properly, always win the game. Thus we may say that, once PLAYER I chose a, on his first turn, PLAYER 11 then had a winning strategy. If PLAYER 11 had a winning strategy in response to every possible first move of PLAYER I, then we could reasonably say that PLAYER 11 had a winning strategy for the entire game. On the other hand, if PLAYER I had a first move in response to which PLAYER 11 had no winning strategy, then no matter what PLAYER 11 did on his first move, PLAYER I could respond in such a way that PLAYER II would have no winning strategy at that point; and pursuing
1.
THE PLAY OF THE GAME
95
this chain of thought, we see that PLAYER I could be said to have a winning strategy for the entire game. Thus, at the very outset, either PLAYER I or PLAYER 11 can force a win in this game. (This result, of course, follows from the general theorem of game theory that given any finite two-person game with complete information such that each play of the game ends in a victory for one of the two players, either the first player has a winning strategy or the second player has a winning strategy.) Returning to our game above, we ask whether PLAYER I or PLAYER I1 has a winning strategy. If PLAYEK I has a winning strategy, then his first move was terrible; if, on the other hand, PLAYER 11 has a winning strategy, then whichever first move PLAYER I substituted for his actual move, the result, assuming best play by PLAYER 11, would have been the same. The reader should try to determine for himself whether PLAYER I or PLAYER 11 has a winning strategy before reading on. It seems fairly evident that PLAYER 11 has a winning strategy. For PLAYER I1 can essentially mimic PLAYER 1’s moves, choosing an endpoint whenever PLAYER I chooses an endpoint, a point closer to the center whenever PLAYER I chooses one closer to the center. and a central point whenever PLAYER I chooses a central point. The details are easily verified. We now proceed to a more general discussion. But first we should point out that the paragraph above is incorrect; it is intended to deceive the reader who did not really try to find out who has the winning strategy and to provide a hint to the reader who tried unsuccessfully. For it may lead him to see that if the first player chooses the central point of B, then the second player cannot really choose a central point of A . If, for example, he chooses a 3 , then the first player can choose b, with devastating effect. Thus the first player actually has the winning strategy in this game. EXERCISE 6.1: In each of the following cases, determine who has a winning strategy and what that strategy is.
(1) A = 7 , B = 8 . (2) A = 0 , B = q. ( 3 ) A 5 w B 5 [. (4) A 5 w, B 5 w + w. (5) A - w + 2 + w * , B 2 . 0 - t I+U*. (6) A w + 2 + a*,B = UJ -t 3 w * .
+
In general, of course, there is no need to restrict ourselves t o the game with three moves. Thus for any fixed natural number yt 2 1 and any linear orderings A and B we have the game G,,(A,B), which is played as above with
6.
96
THE EHRENFEUCHT-FRA‘ISSE
GAME
n rounds instead of three rounds. These games (and other similar games discussed in Chapter 13) were first formulated by Ehrenfeucht [l], who used them in his analysis of the logical properties of the arithmetic of ordinals. A similar notion, without the game-theoretic terminology, had been formulated independently by Fraisse [3]. (Ehrenfeucht notes that, for the case of the ordering of the ordinals, the condition he presents is “only a new formulation of the condition given by Fraissk.”) We will discuss the applications of these games to the ordering of ordinals at the end of this chapter, and we will return to the more elaborate games in Chapter 13. For a contemporary discussion of their work, see Feferman [2]. The following formal definitions are motivated by the discussion above. DEFINITION 6.2: Let A and B be linear orderings and let n 2 1 be a fixed natural number. A play of the game G,(A,B) consists of an ordered sequence of n repetitions of the following: PLAYER I chooses an element of A or B and PLAYER II chooses an element of the other. The element of A selected at the rth turn is denoted a, and the element of B selected at the tth turn is denoted b,. We say that PLAYER 11 has won the play of the game if for each s,t I n, a, < A a, if and only if b, < B b,; otherwise we say that PLAYER I has won the play of the game. We say that PLAYER 11 has a winning strategy in G,(A,B) if there are functions f i , f i , . . . ,f, such that (i) the domain o f f , is the set of all ordered t-tuples of elements of
A v B;
(ii) given c l r c 2 , . . . , c, E A u B (representing the choices of at the first t turns)
PLAYER I
(iii) if c , , c 2 , . . . , c , ~ A u B a n d i f w e d e f i n e a, =
b, =
f,(cl, . . . ,c,)
if c, E A if c, E B,
f,(c,, . . . ,ct)
if c, E B if c, E A
iCi
for every t 5 n, then for every s, t I n, a, < A a,
if and only if
b, < b,.
Otherwise we say that PLAYER I has a winning strategy. Given the technical and notational complexity of these definitions, it is not surprising that, in the proofs of the various claims below, we avoid
1.
97
THE PLAY OF THE GAME
using the formal definitions and instead rely on intuitive arguments. The disbelieving reader may, if he wishes, convert our intuitive arguments into more formal ones. NOTATION : If PLAYER 11 has a winning strategy in the game G,(A,B), then we will write G,(A,B ) E 11; otherwise we write G,,(A,B) E I. The following lemmas are easily verified. LEMMA 6.3: (1) G,(A,B ) E I1 if and only if G,(B,A ) E 11. (2) !f A = B, then G,(A,B ) E 11. A LEMMA 6.4:
I f G,(A,B)E I1 arid 1 i m < n, then G J A , B ) E 11.
LEMMA 6.5:
(1) I f G , ( A 1 ,B , ) E I1 and G,(A,,B,)
&(A,
11, then
+ .4,,B, + B2) E 11.
(2) I f G , ( A , , B , )E I1 for each
E
E
A
I E
J , then G,(C{A,Ii E J } , c ( B i l i E J } )
11, where J is an arbitrary linear ordering.
(3) I f A has a largest (smallest)clement and B does not, then G,(A,B ) E I. (4) I f A is dense and B is not, then G , ( A , B ) E I. A
We will soon show that the binary relation on linear orderings defined by G,(A,B ) E I1 is actually an equivalence relation. The lemmas above should thus be read with each G,(A,B ) E I1 replaced by the corresponding A -,B. Theorem 6.6, which we prove below, is a basic result which allows us to resolve games with n 1 moves into games with n moves, and thereby allows us to perform induction arguments.
+
THEOREM 6.6: G,, l(A, B ) E I1 If and only
if
(i) .for euery a E A there is a h E B such that
G,(A'",B'*)
E
I1
and
G,(A'",B'b)
E
11; and
(ii) for every b E B there is an a E A such that
G,(A'a,B'b)
E
I1
and
G,(A",Bib)
E
11.
Proof : (+) Suppose that PLAYER II has a winning strategy in G,, l(A,B ) and suppose that a E A . Using the strategy, we can find an element b E B
98
6.
THE EHRENFEUCHT-FRA'I'SS~
GAME
such that if PLAYER I chooses a at his first turn, then PLAYER 11 should choose h. There are now n turns left in the game G,+,(A,B).Whenever PLAYER 1 chooses an element in A'" or B < b ,the strategy, since it produces a win for PLAYER 11, will always choose an element in B'b or A'". Thus PLAYER I ~ S winning strategy for G,+ l(A, B) includes within it a winning strategy for G,(A<", BCb),and similarly it includes a winning strategy for G,(A'", B > b ) . This proves (i). By symmetry, (ii) is also correct. ( G ) Assuming that (i) and (ii) are correct, we describe a winning strategy for PLAYER 11 in G , , l(A, B). If PLAYER I chooses an element a E A on his first turn, then PLAYER 11 uses (i) to find an element b E B. Thereafter, whenever PLAYER I chooses an element of A > " or ' B b, PLAYER 11 uses his winning strategy in G,(A'",B'b) to respond; and similarly, whenever PLAYER I chooses an element of A'" or Bcb, PLAYER 11 uses his winning strategy in G,(A'",BCb) to reply. Since there are only n subsequent moves in the game, PLAYER I can choose no more than n times from A'" or B e b (and from A'" or B z b ) .Hence PLAYER 11's winning strategies in G,(A'",B'b) and G,(A<", F b provide ) him with moves in all contingencies. If, on the other hand, PLAYER I chooses an element b E B,then PLAYER 11 uses (ii) to find his correct first move and then proceeds analogously to the above. Thus PLAYER 11 has a winning strategy in en+l ( A , B ) . COROLLARY 6.7: Let A , B, and C be linear orderings and let n 2 1 be a fixed natural number. If PLAYER II has a winning strategy in G,(A,B ) and PLAYER II has a winning strategy in G,(B,C ) ,then PLAYER 11 has a winning strategy in G,(A,C). Proof: We prove this by induction on n. For n = 1 the hypotheses imply that A and B are either both empty or both non-empty, and that B and C are either both empty or non-empty; from this it follows that A and C are either both empty or both non-empty, and hence that G,(A, C) E 11. Assume that the statement is true for n, and assume that G,+,(A,B) E I1 and G,+ ,(B, C) E 11. We will show that G , , l(A, C) E I1 using Theorem 6.6. Let a E A . Then by Theorem 6.6, there is an element b E B such that G,(A <", B < b )E I1 and G,(A>",B S b )E 11. Since G , , l(B, C ) E 11, by Theorem 6.6, there is an element c E C such that G,(BCbb, C<') E I1 and G,(B>b, C") E 11. By the induction hypothesis it follows that G,(A'", C < ' ) E I1 and G,(A'", C") E 11. Thus if a E A , there is a c E C such that G,(A'", C<') E I1 and G,(A'", C") E 11, verifying (i) of Theorem 6.6. Clause (ii) is similarly verified. Hence, by Theorem 6.6, G , , l(A, C) E 11. 4
We are now ready to make the following definition.
1. THE
99
PLAY OF THE GAME
DEFINITION 6.8: We say that A is G,-equivalent to B, written A -,, B, if PLAYER 11 has a winning strategy in G,(A,B). We say that A is G-equivalent to B, written A B, if A -,, B for every n E N . By Lemma 6.3 and Corollary 6.7, these are equivalence relations. Intuitively, if A -,, B, then PLAYER I cannot demonstrate any difference between A and B in the n moves at his disposal.
-
We now discuss various implications of Theorem 6.6.
COROLLARY 6.9: Let A and B be ,finite linear orderings of cardinality at least 2" - 1. Then PLAYER 11 hay ti winning strategy in G,(A,B). Proof: We prove the corollary by induction on n. Assume that n = 1. Then both A and B are non-empty. Hence clearly PLAYER II has a winning strategy in G , ( A ,B). Assume the corollary true for n and let A and B be finite linear orderings of cardinality at least 2"' - 1. To show that PLAYER 11 has a winning strategy in G , + , ( A , B ) it suffices to show that (i) and (ii) both hold. We will deal only with (i), since (ii) is analogous. Let a E A. Suppose that IA'"I = nL and that IA'"I = n R . Since IAl 2 2"" - 1, if nL < 2" - 1 and nR < 2" - 1, then
[ A (= 1 + nL + nR < 1 + (3" - 1)
+ (2"-
1) = 2"+l - 1,
contrary to hypothesis. Hence u cannot be both one of the first 2" - 1 elements of A and one of the last 2" - 1 elements of A . Similarly, if b E B, then b cannot be both one of the first 2" - 1 elements of B and one of the last 2" - 1 elements of B. If now the given element a E A is one of the first 2" - 1 elements of A, we let b be the corresponding element of B ; for example, if a is the third element of A , then b is the third element of B. Then A'" 2: B
100
6.
THE EHRENFEUCHT-FRAYSS~
GAME
Note that a special case of this corollary is that PLAYER 11 has a winning strategy in G3(7,8)-which is included in Exercise 6.1. That no further improvement is possible in case n = 3 is seen from the fact that PLAYER I has a winning strategy in G3(6,7). EXERCISE 6.10: (1) Show that PLAYER I has a winning strategy in G,(h,k)ifk<2"- l a n d h 2 2 " - 1. (2) Show that if k < 2" - 1 and PLAYER II has a winning strategy in G,(A,k), then A 2 k. EXERCISE 6.11 : Let A be a finite linear ordering of cardinality at least 2" - 1. Then PLAYER 11 has a winning strategy in G,(A,B) where B 2: w t; . a + w* for any order type a. [Hint: Note the fact that if b E B, then either B > b or B C b(or both) is again of the form w + t; . a' + w* for some order type a'; and see the proofs of the corollary above and the corollary below. Note, however, that, since Corollary 6.12 uses the result of this exercise, it would be quite inappropriate to use Corollary 6.12 here.]
+
COROLLARY 6.12: PLAYER 11 hus a winning strategy in G,(w,w + ( . a ) jbr any order type a and any n 2 1.
Proof: We proceed by induction on n to show that G,(w, w + t; . a ) E I1 for every a. The case where n = 1 is obvious. Assume that the statement is true for n and proceed, as in Corollary 6.9, to use Theorem 6.6. Thus suppose that a E w = A ; then let b be the corresponding element of w + 4 . a = B. Then, since A'" c B < b ,G,(A'",B'b) E 11; and, by induction hypothesis, since B'b 2: w + t; . a, G,(A'",B'b) E 11. Suppose, on the other hand, that we are given b E B. If b is in the initial w of B, then choose a to be the corresponding element of A and proceed as above. If, on the other hand, b is not in the initial w of B, then B'b 2: w + 4 . a' and B'b c w + t; . a" + w*, where a' and a" are particular order types. Let a be the 2"th element of A . Then IA'"I = 2" - 1, so that, by the exercise above, G,(A'", B < b )E 11; and A'" z w, so that, by the induction hypothesis, G,(A'",B>b) E 11. Hence conditions (i) and (ii) of Theorem 6.6 are satisfied, so that PLAYER 11 has a winning strategy in G,(o,w + ( . E) for every a and for every n.
+
In particular, PLAYER II has a winning strategy in G,(w,w t;) for every n. Thus no difference between w and w t; can be discerned through the games.
+
1.
101
THE PLAY OF THE GAME
These games were used by Ehrenfeucht in his investigations of the logical equivalence (or elementary equivalence) of mathematical structures and, more specifically, of well-orderings. Informally, two mathematical structures are said to be logically equivalent if there is no statement in the first-order predicate calculus which distinguishes them. Ehrenfeucht showed that two structures are logically equivalent if and only if they are Gequivalent, and used this characterization of logical equivalence to obtain interesting results about well-orderings. Similar results were obtained by Fraisse, using different terminology. Thus the logical interpretation of Corollary 6.12 is that o is logically equivalent to w 5 . c( for any order type CI.Thus, for example, no statement of the first-order predicate calculus is true for w but false for w (. One consequence of this fact is that no first-order statement and, more generally, no set of first-order statements, can be true precisely for those linear orderings which are well-orderings. This is usually expressed by saying that the notion of well-ordering is not first-order definable or that the class of wellorderings is not an axiomatizable class. We will return to well-orderings shortly. The logical uses of the Ehrenfeucht games will be further discussed at the end of this chapter and in Chapter 13. The equivalence relation -"partitions the class of linear orderings into a number of equivalence classes. We will now prove that for each n there are only a finite number of equivalence classes; this contrasts strongly with the equivalence relation -, which, as we will see, has infinitely many (and, in fact, has c ) distinct equivalence classes.
+
+
THEOREM 6.1 3 : For each n there are only a Jinite number of equivalence classes modulo -,,.
Proof: We proceed by induction on n to show that there are only a finite number f ( n ) of equivalence classes modulo n. For n = 1, as we already know, there are but two equivalence classes. Assume the claim for n. Let A be a linear ordering. Then each element a E A determines an ordered pair ( [ A ' " ] , [A'"]) of equivalence classes modulo -,,, where [B] is the -,,equivalence class of B. Let I ( A ) = { ( [ A ' " ] , [A'"])la E A } . We note that Theorem 6.6 can be understood as saying that A - n + l B if and only if I(A) = I(B). Hence the number of equivalence classes modulo is at most the number of subsets of the set of ordered pairs of equivalence classes modulo -,; but this number is 2J(")'s(").Hence f ( n + 1) 5 2f""rcn), and in particular f ( n 1) is finite.
+
102
6.
THE EHRENFEUCHT-FRA~’SS~GAME
The reader might naturally inquire at this point what the values of f ( n ) actually are, and he may wonder whether the function f actually attains the upper bound given in the proof and therefore satisfies the recursion f(1) = 2, f ( n + 1) = 2f(“)
The value for f ( 2 )would then, for example, be 16. In point of fact, however, .fV) = 7. EXERCISE 6.14: (1) Show that f ( 2 ) = 7 by giving a complete set of representatives for the equivalence relation m 2 . (2) Find f ( 3 ) . [Caution: This is not an exercise, but a major project.]
The fact that there are only a finite number of G,-equivalence classes for each n suggests the possibility that there is a “nice” countable set A of order types such that for every n E N and every linear ordering A there is an M E .4 such that A -, M. We will see in the next chapter that this is indeed the case. THEOREM 6.15: ulo -.
There are at least c different equivalence classes mod-
We define, as in the proof of Proposition 1.48, for each subset X G N a linear ordering L ( X ) such that if X I # X , , then L ( X , ) is not C-equivalent to L ( X 2 ) . We define A , ( X ) to be of order type q + (a + 2) if a E X and of order type 1 if a $ X ; and we let L ( X ) = C { A , ( X ) l a< w ) . Let X , Y G N such that X # Y . Let a be an element in X - Y (or, alternatively, in Y - X ) . We claim that PLAYER I has a winning strategy in C,+.(L(X),L ( Y ) ) .Indeed, on his first a + 2 turns, he chooses in order the final a + 2 elements of A,(X). He then considers PLAYER 11’sfirst a + 2 choices b , , . . . , b,+2 in L( Y ) . If, for some i. there is an element of L ( Y ) between bi and bi+ then PLAYER I chooses that element on his (a + 3)rd turn and forces PLAYER 11’s resignation. Otherwise, since there is no maximal discrete sequence of exactly a + 2 elements in L( Y ) ,either there is an element b beyond b a + 2in L( Y )with nothing of L( Y )in between, or there is an element b before b , in L( Y )with nothing of L( Y )in between. PLAYER I chooses such an element for his ( a + 3)rd turn; suppose that he chooses such a b and suppose, without loss of generality, that b,, < L l y ) b. Then PLAYER II presumably chooses an element c , + ~of L ( X ) after the last element c of A , ( X ) in L ( X ) . But then Proof:
1.
103
THE PLAY OF THE GAME
PLAYER I chooses an element between c and c,+ ,-and since PLAYER II cannot find an element between and b, he loses. Hence PLAYER I has a winning strategy in G,+,(L(X), L ( Y ) ) .Thus if X and Y are different subsets of N , then L ( X ) is not G-equivalent to L,(Y). Hence there are at least c distinct equivalence classes of countable linear ordering modulo .
-
-
Can there be more than c diflerent equivalence classes modulo ? We will show first that every G-equivalence class contains a countable representative. (This is the combinatorial version of the Lowenheim-Skolem Theorem for the predicate calculus; this method of proof was suggested by J. Schmerl.) Since there are but c distinct countable order types, this shows that there are exactly c different G-equivalence classes.
-
THEOREM 6.16: (Lowenheim-Skolem) Let A be a linear ordering. Then there is a countable linear ordering B A such that B A .
Proof: We first note that, as in the proof of Theorem 6.13, given an interval C E A and given nE N , each element U E Cdetermines an ordered pair S(C,a, n) = ([C'"],, [ C > " ] , )of equivalence classes modulo -,, where [D], is the n-equivalence class of D. Furthermore, by Theorem 6.13, for each n there are but a finite number of possible S(C, a, n). Thus it is possible to select a finite subset S(C, n) E C so that for each U E C there is a unique b E S ( C , n) with S(C, a, n) = S(C, b, n). The countable subset B of A IS defined to be the union of an ascending sequence
$8 = B, s B , G B ,
S.-._C B,G
B,+l E ' . .
of finite subsets of A, defined by induction on n. Assume that B, has already been defined; the k = k(n) points of B, partition A - B, into k + 1 intervals C,, C , , . . . , C, (some of which may be empty.) Define
B,+l= B, u U{S(Ci,n)IO5 i C:" B
I k}.
Thus for any b E A and any n, either b E B, or b E C iand for some a E Ci, -,C
-To conclude that B A.
A, it suffices to show by induction on n that B - , A . As is often the case, a stronger induction hypothesis turns out to be necessary. We will show, by induction on n, that for any b E B , B X b-,A ' b and B > b-,A'b and that if b < c are in B, then ( b , ~-, ) (~ b , ~ )(Note ~ . that this implies that B -,A.) For n = 1 this is clearly correct. Turning to the induction step, we will confine our attention to showing that (b,c ) -,+~ (b,c ) for ~
104
6.
THE
EHRENFEUCHT-FRAYSSI?
GAME
b < c elements of B ; the proof that B’b A’b and -,+ A < b is similar. We may restrict ourselves further to the case where b and c are consecutive elements of some B,,, with m 2 n. [For if b < c in B, we can find an m 2 n such that b,c E B, and we can find c , , c 2 , . . . , c, in B , such that b = c , < c2 < . . . < c,- < c, = c are consecutive elements of B , ; but if ( c i ,el+J A for each i, it follows that we have shown that (cl,c, + I)B -,+ (b,c), -,,+ ( b , ~ ) ~Since . ] b and c are consecutive elements of B,, (b,c), is one of the k(m) intervals C used to define B,, in particular, for each x E (b,c ) ~ there , is a y E (b,c ) such ~ that (b,x ) ~, (b,y ) , and ( x ,c), -, ( y ,c ) ~ . In the game G,+ l((b, c ) ~ (h, , c ) ~ )if, PLAYER I chooses an element y E (b,c ) on ~ his first move, PLAYER II chooses y also, and by the induction hypothesis, since y E B,
(b,Y ) , -,, (b. Y ) ~
and
( Y ,c ) A
-,, ( Y , c ) B ,
so that PLAYER 11 will always be able to win. If, on the other hand, PLAYER I chooses a n element x E (b, c ) on ~ his first move, then PLAYER 11 can choose y ~ ( bc, ) so ~ that (b, x ) ~, (6, and (x,c ) ~-,, ( y , c ) ~ Then, . since Y E B, the induction hypothesis applies to the intervals (b,y) and ( y , c); that is, (b,Y ) A -,(b,Y ) B and (.v, -,, ( Y , c)B. Since m 2 n, (b, -,(b,yIB and (x, c), -,( y , c ) ~by , transitivity. Thus in this case also, PLAYER 11 will always , c ) ~E)11; thus be able to win. Hence, by Theorem 6.6, G , , (b, c ) ~ (b, ( b , ~ and ) ~ the induction step is complete. (b,c ) ~
COROLLARY 6.17: modulo -. Proof:
There are exactly c d i e r e n t equivalence classes
By Theorem 6.15 and Theorem 6.16.
The Lowenheim-Skolem Theorem, as formulated here, is a special case of a more general theorem which applies to arbitrary structures and has a stronger conclusion. The logical interpretation of Theorem 6.16 is that any linear ordering is logically equivalent to a countable linear ordering; that is, given a linear ordering A , there is a countable linear ordering B (which is actually a subordering of A ) which cannot be distinguished from A using statements of the predicate calculus. This will be discussed further in Chapter 13. One further comment before concluding this section. Ehrenfeucht games can, of course, be played with other structures, not just with linear orderings. For example, if A and B are groups, then, in a play of the game G,(A, B), PLAYER 11 wins if the selections b,, b,, . . . , b, from B match the selections a,, a,, . . . , a , from A in that ai x A a j = a, if and only if bi x E bj = b, for
2.
105
GAMES AND ORDINALS
each i, j , and k . The formal development for games on such structures will appear in Chapter 13. For now we leave it as an exercise to find winning strategies for either player in the games Gz((Q, + ),
G d(Rf, .), (Q+, .)),
(Q',
.))
and
G 3 ( ( R + , (Q,
+)I,
where R+ and Q' denote the positive reals and rationals, respectively.
2. GAMES AND ORDINALS
In this section, we present a complete answer to the question, Under what circumstances are two ordinals G-equivalent? By our earlier remarks, we get the same answer to the question, Under what circumstances are two ordinals logically equivalent? THEOREM 6.18:
For any naturtrl number n > 1,
(1) o" w" . p ,for any ordinul p > 0, and (2) w" * 2 n + w" B fbr uny ordinal /3 > 0.
+
Proof: Note first that (2) implies in particular that on+ z n + I wn . fl for any ordinal /? > 0 and hence that the result in (1) is best possible. We prove these two claims simultaneously by induction on n. For n = 1, we need to show that G,(w, w . p) E I1 for any /3 > 0 and that G,(w,o + p) E I for any p < 0. The first is obvious, and for the second, PLAYER I has a rather simple strategy. He chooses the point w in o + p and, if PLAYER II chooses m + 1 in u, PLAYER I chooses m in o and then a point in w + fl between PLAYER 11's response and o. Assume now that the claims are true for all k < n. We first show that PLAYER 11 has a winning strategy in GZn(w",w" . p) for any ordinal p > 0. We will describe PLAYER 11's responses to PLAYER 1's first two moves. He will be able to respond in such a way that if a < a' are the first two selections from wnand if b < b' are the first two selections from o". p, then, if A is any one of the three intervals formed in 0)" and B is the corresponding interval in on. p, either A 5 y and B u y for some y < on- or A 'v (con-' . PI) + y and B N (a"-' . p2) + y for some y < (on-' and some PI, p2 > 0. From this it will follow, using the induction hypothesis, Lemma 6.3.2, and Lemma 6.5.2 on sums of 2(n - 1)-equivalent orderings, that Gz,(w", o". p) E 11. But it is clear that PLAYER 11 can respond so that the above conditions are met by taking care that, if PLAYER I chooses a point of the form (on-'. Po) + y where y < a n - 'then , he chooses a point of similar form; and, in particular, if Po
6.
106
THE
EHRENFEUCHT-FRAYS&
GAME
is finite, then PLAYER 11 chooses the same point in the other ordering. (Note that every tail of wkhas order type w k ;this is a special case of Exercise 3.44.3.) ~ n . P)EI, we will describe the first two moves To show that G 2 n + l ( (LO" of PLAYER 1's strategy. On his first move he picks the point w" in wn p. Then, no matter which point a in (0" is picked by PLAYER 11, PLAYER I at his second turn picks a point b in w" so that (b, a) 'v wk for some k < n (unless a is a successor, in which case PLAYER I picks its predecessor.) Then, no matter which point d < w" in o"+ [I is picked by PLAYER 11, (d,0")= 0".But, by the induction hypothesis, PLAYER I has a winning strategy in Gzfl-' ( w k ,o")for any k < n. Hence Gz,+l(co",ion /?)€I for any p > 0.
+
+
+
COROLLARY 6.19: For any natural number n, if TX * 2 n + 2 w"+' and u - 2 n + 3 p.
ci
< w""
< p, then
+
Proof: Write ci = ~ " .l u , wn2. a2 + . . . + conk. uk in Cantor Normal Form. In the game Gzn+2 ( x , (fin+l), PLAYER I first picks a point a of ci which determines a tail of order type conk.Then no matter which point PLAYER 11 selects in a"+ ',the tail it determines has order type a"+ But, by the theorem, PLAYER I has a winning strategy in the game G2nk+l(wnk, w"+'), hence also ni PLAYER I has a winning strategy in in the game C Z n + l ( ~ n k , f').r JThus
'.
G 2 , + 2 ( c i , ~ 1 " + 1 ) i f< c iw " + ' .
In the game G2,r+3(ci, p) where ci < LO"+' < p, PLAYER I first picks the point w"+' in p. Then no matter which point a in a is selected by PLAYER 11, its predecessorshave order type < a"+ ',so PLAYER I can continue his strategy as in the preceding paragraph to obtain a win in Gzn+ 3(ci, p). The Exercise below shows, by example, that the results of Corollary 6.19 are the best possible. EXERCISE 6.20:
Prove each of the following for all n:
(1) W " . 2 - 2 , w f l + 1 + to". ( 2 ) W".2-2,+l w " + ' w". (3) w" . 2 *2"+2 Q"+ -tw". (4) W " . 2 - 2 n + 1 o"+I.
'
(5) (6)
W"'3h.2,+1
OJ"+'
+
+On.
3 * 2 n + 2 or"' + w" for n > 0. 1 w". (7) on 4 - 2 n + 2 O". '
+
Note that (4) and (7) show that the results of Corollary 6.19 are the best possible; this proof of that fact is due to P. Mulhall.
2.
107
GAMES A N D ORDINALS
(8) Show that every ordinal is G2,-equivalent to some ordinal in the finite set (w" . a,
+ w n - l . a,- + 1
c(j"-2
. an-,
+
'
'.
+0
'
a1
+ ao:
where each ai < 22" and a, I 1. THEOREM 6.21 : Let a and fi he ordinals. Write a = ow. a, + a2 and p = cu"' . fi, + p2 where a, < o" and [j2 < (d".Then c1 fl if and only if a2 = f12 and either r l and p1 are both 0 or are both bigger than 0.
-
-
-
Proof : Since Q"'= wn . ow,by Exercise 3.48.4, it follows from Theorem 6.18 that coo . r l ou. ,8,, for any a l and fil which are non-zero. Hence a p if the conditions are met. Now suppose that a, = 0 but PI > 0. Let a2 < on. If PLAYER I chooses . in j? at his first turn, then the initial segments a. and Po of u and p determined by the first turn satisfy a. < W" < Po, so by Corollary 6.19, PLAYER I can win in 2n + 1 further turns. Hence G2,,+ ,(a,p) E 1. Thus if u p it follows that a1 and p1 are both 0 or both non-zero. Suppose now that c1 < p < (0'". Write p in Cantor Normal Form as w n l. a , (on* . a2 . . . mnk . a k . We claim that PLAYER I has a winning strategy in Gn(a,p) where n = (a, a2 . . . + ak)- 1 + 2n,. Indeed, at his first u 1 turns, PLAYER I chooses the points wnl. 1, wnl . 2 , . . . , w"' . a , ofp; at his . 2 , .. ., next a2 turns, he chooses the points wnl . a, + mn2 . 1, wnl . a, + on2 (d"'. a , + wf12. a,; . . . ; at his nexl ak - 1 turns, he chooses the points (1)111 . a, + . . . + g n k - 1 . ak- 1 wnk' 1 , . . . , 0"'' a1 ' ' . Unk. (ak - 1). He now surveys PLAYER 11'sfirst ( a , + a2 + . . . + ak) - 1 moves. Since /3 > a, at least one of the intervals in a must be smaller than the corresponding nl, interval in p. Since the intervals of p are all of form wm for some m I PLAYER I needs at most 2n, further moves to win. Thus G,(u,p) E I as claimed. Hence if a fi and u, p < ow,then u = p. Similarly, suppose that u = cu" . a1 + a2 and p = 0") . PI + p2 where cil and fil are non-zero and u2 < p2 < w". If a, = 0, then PLAYER I can win in G2,(r,p) by selecting a point of /3. whose successors have form okfor some k < n, and then using the strategy of Theorem 6.18.2 for his remaining 2n - 1 moves. If a2 > 0, then by picking the point ow. u1 in u he leaves PLAYER 11 with three ways of losing: Either PLAYER II picks a point of p which is less than o0 . p,, in which case the successor intervals are a, and ow. -y p2 so, by Corollary 6.19, at most 2n 1 further moves are necessary; or PLAYER 11 picks the first point of p2, in which case the successor intervals are a2 and p 2 , so that PLAYER 1 can win in at most t = (a, a, + . . . + ak)- 1 2n,
-
+
+
+
+ +
+
+
+
-
+
+
+
+
108
6.
THE EHRENFEUCHT-FRAYS&
GAME
+
more moves where fl = m"' . a , . . . + wflk. ak,as in the preceding case; or, finally, PLAYER II picks a point beyond the first point of p 2 , in which case the predecessor intervals are owand w" y where 0 < y < m", and in this case PLAYER I can win in at most 2n + 2 additional moves. Thus in the case where 0 < ci2 < f12, PLAYER I can win in the game G,+,(u,fl) where m = max(2n + 3, t ) where n and t are bounds depending only on /?.This completes the proof. H
+
Theorem 6.21 enables us to conclude that the set of all ordinals less than
UP . 2 forms a complete and irredundant set of representatives of the Gequivalence classes of well-orderings. By way of contrast, although any wellordering is G-equivalent to one of a countable number of well-orderings, the analogous statement is false for arbitrary linear ordering, as we verified in Theorem 6.15. The logical interpretation of Theorem 6.21 is that the set of all ordinals less than LO"' . 2 forms a complete and irredundant set of representatives of the logical equivalence classes of well-ordering ; this fact was noted earlier by Mostowski and Tarski [4]. The first application made of the EhrenfeuchtFrai'sse analysis is the following result of Ehrenfeucht [11 and Fraisse [3].
-
THEOREM 6.22: Let O N denote the collection of all ordinal numbers, with the usual ordering. Then O N d". Proof : Since m" m Z f lo") for each n, by Theorem 6.18, it suffices to show that (d'- 2 n ON for each n. This is proved, by induction on n, making only notational changes in the proof of Theorem 6.18.1. H
Theorem 6.22 has the following logical interpretation : Given a first-order statement about the ordering of the ordinal numbers, that statement will be true if and only if it is true about the ordinal numbers below LO'). As we will see later, this interpretation makes it possible to conclude that the theory of the ordering of the ordinal numbers is decidable; that is to say, that there is an effective procedure which, when presented with any first-order statement about the ordering of the ordinal numbers, will determine whether or not that statement is true. This result was also proved by Mostowski and Tarski [4]. Ehrenfeucht [11 also proved that a first-order statement about addition of ordinal numbers is true if and only if it is true about the ordinal numbers below IO")'"; or, in the language of games, ( O N , <,
+) - (w""', <, +)
109
REFERENCES
Furthermore, going one step further, ( O N , <, +;)
-
<, +;>.
( ~ l u w w ,
For these results, Ehrenfeucht used a more elaborate game, which we will discuss in Chapter 15. REFERENCES Ehrenfeucht, A,, An application of games to the completeness problem for formalized theories, Fund. Math. 49 (1961), 129-141. [ M R 23, #3666] 121 Feferman, S., Some recent work of Ehrenfeucht and Fraisse, Summer Institute for Symbolic Logic, Cornell University. 1957, Summaries, pp. 201 -209. [3] Fraisse, R., Sur quelques classifications des systemes de relations, Publ. Sci. Univ. Alger. Ser. A 1 (1954), 35-1 82. [Abstract: .Ipplicutions scientifiques de la logique mathimatique, Paris and Louvain, 1954,851 141 Mostowski, A,, and Tarski, A., Arithmetically definable classes and types of well-ordered systems, Bull. Amer. Math. Soc. 55 (1949), 65 and 1192.
[I]