PDL Has Interpolation

PDL Has InterpolationAuthor(s): Tomasz KowalskiSource: The Journal of Symbolic Logic, Vol. 67, No. 3 (Sep., 2002), pp. 933-946Published by: Association for Symbolic LogicStable URL: http://www.jstor.org/stable/3648548 .

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THE JOURNAL OF SYMBOLIC LOGIC

Volume 67, Number 3, Sept. 2002

PDL HAS INTERPOLATION

TOMASZ KOWALSKI*

Abstract. It is proved that free dynamic algebras superamalgamate. Craig interpolation for propositional dynamic logic and superamalgamation for the variety of dynamic algebras follow.

?1. Dynamic algebras. The notion of a dynamic algebra has arisen in connection with propositional dynamic logic (PDL), "the logic of programs" as it has also been called, as an algebraic rendering of the latter. Its standard definition, unlike the one adopted here, views a dynamic algebra as a two-sorted structure, with the boolean component modelling relations between propositions, and the other component, called regular, modelling relations between programs, such as composition, choice, and iteration. The reader is asked to consult [6] for the standard definition of dynamic algebras as well as for the proofs of all facts about them stated here without proof. A full rendering of PDL requires also an algebraic counterpart of test. This is achieved by means of test algebras that may be viewed as dynamic algebras with yet more new unary operations. We know from [3], however, that PDL has interpolation iff test-free PDL has interpolation; thus, by a slight abuse of terminology we let PDL stand for the latter, unless explicitly stated otherwise.

A dynamic algebra is an algebra A = (A; V, A, -, 0, 1, f,(n E -I)), such that:

(1) (A; V, A, -, 0, 1) is a boolean algebra; (2) l is the universe of the absolutely free algebra H = (Il; o, U, *), of the type

(2, 2, 1), countably generated by the set Z = {,: n E co). (3) A satisfies:

(i) fO = 0,

(ii) f,(x V y) = fx V fy, (iii)

f•ruax = ftx V fax,

(iv) fo,,, = f f x, (v) x Vfff*.x < f*x < x V f ,*(-ix A f ,x).

The class 9 of dynamic algebras is a variety. The logic A(9) associated in the natural way with 0, is precisely (test-free ) PDL.

To simplify the notation, we will from now on identify functions f, with their indices, using lowercase Greek letters for the latter, and abbreviate 7r o a by the juxtaposition 7ra. Rewritten thus, the identities above become:

Received May 9, 2001; revised DATE NOT PROVIDED. *I am grateful to Marcus Kracht for all his encouragement and help. 'Invaluable' is in this context

the most appropriate adjective to qualify the noun 'help'.

? 2002, Association for Symbolic Logic 0022-4812/02/6703-0005/$2.40

933



934 TOMASZ KOWALSKI

(i) 70 =- 0, (ii) t(x V y) =

•tx V 7ry,

(iii) (ntU a)x = •x

V ax, (iv) (7ra)x

- n1(ax),

(v) x V 7 _*X<

it*x < xV x *(-x A 7x).

FACT 1.1 ([6] Lemma 3.1). An algebra A of the appropriate type is a dynamic algebra iffA satisfies (i)-(iv) and, moreover,for all 27 E I-, 7•*x

is the smallest element of the set {y E A: x V ny < y}.

A dynamic algebra A is called *-continuous iff 2*x = V{•cnx:

n E wco, for all 7 E H, with 7anx defined inductively by: orx

- x, 7En+1x

-- 7Mn x. A class X of

dynamic algebras is *-continuous iff all its members are.

FACT 1.2. All finite dynamic algebras are *-continuous. Ifa class 5X is *-continuous, then so is SP (X).

PROOF. The first statement is folklore. Preservation of *-continuity under direct products is trivial, as they preserve arbitrary joins. To prove preservation under subalgebras notice first that 7c*x is always an upper bound of {f7rnx: n E w}. This follows from (v), by an easy induction on n. Then, take A c B, with B *-continuous, and an a c A. We get that 7r*a is an upper bound of {f7ra: n E co} in A. We will show that it is the least upper bound. Let c E A be an upper bound of { na: n E c } in A. Clearly, c is an upper bound of {7n"a: n co w} in B as well. Then, c

_> r*a in

B, since B is *-continuous. But c belongs to A, and so does 7 *a; hence c _>

i*a in A, as desired.

Let us note on the margin that *-continuity is not preserved by either homomorphisms or ultraproducts and, thus, the class of *-continuous dynamic algebras is neither a variety nor a quasivariety.

Recall that an algebra is called residually finite iff it is a subdirect product of finite algebras.

FACT 1.3 ([6] Theorem 5.4). Free dynamic algebras are residually finite. LEMMA 1. Free dynamic algebras are *-continuous.

PROOF. From Facts 1.2 and 1.3.

?2. Amalgamation and interpolation. A quintuple (Ao, A1, A2, iO, il) is called a V- formation iff Ao, A1, A2 are similar algebras, and i : Ao --+ A1 and i2: Ao --+ A2 are embeddings. In other words, in a V-formation, A1 and A2 share an isomorphic copy of Ao as a subalgebra. Thus, dealing with V-formations we will always tacitly assume il and i2 to be identity maps, and specify the relevant V-formation as the triple (Ao, A1, A2).

A V-formation (Ao, A1,A2), with Aj E i, for j - 0, 1,2 and X a class of similar algebras, can be amalgamated in X iff there is an algebra B E XT and embeddings el: A1 - B, e2: A2 --- B, such that el o

il = e2 o i2 (or, by our

convention, el lAo = e2 Ao). Here again we will assume el, e2 to be identity maps, however, one word of caution is needed. Even if el, e2 are identity maps, their union el U e2, although always a function, in general need not be one-one, for we can have el(a) = e2(b), for an a E A1 \Ao andb E A2 \Ao. The algebra B, or sometimes the triple (B, el, e2), is called an amalgam of (Ao, A1, A2). If the



PDL HAS INTERPOLATION 935

map el U e2 is in fact one-one, B is called a strong amalgam of (Ao, A1, A2). If an amalgam (strong amalgam) of (Ao, A1, A2) exists, we say that (Ao, AI, A2) can be amalgamated (strongly amalgamated).

A class XW of similar algebras has the amalgamation property (AP) iff every V- formation from 6X can be amalgamated in 3X. The class XW has the strong amalgamation property (SAP) iff every V-formation from X can be strongly amalgamated in X.

In the vast majority of (if not all) classes of algebras related to logics, we have a natural partial ordering of the universes of these algebras. Assume that X has such an ordering. Then X is said to have the superamalgamation property (SupAP) iff for any V-formation (Ao, A1, A2), there is a B that amalgamates it, with the embeddings el : A --+ B and e2 : A2 --- B, such that we have: for any x E Ak and y E Aj, if ek(x) < ej(y), then there is a z E A0 with x < ik(z), ij (z) y, where {k,j} - { 1,2} (or, again, by our convention, simply x < z, z < y).

Clearly, SupAP is stronger than AP; thus, by obvious extensions of the definitions above we will speak of V-formations that superamalgamate, or of superamalgams of certain V-formations. Notice also, that SupAP forces el U e2 to be a one-one function. Thus, SupAP = SAP > AP. Neither of the converse implications holds.

A (propositional) logic A, with a certain reasonable implication -+, possesses Craig interpolation property iff for any formulas 0, q1 in the appropriate language, the following holds:

(t) if --A

q 0 -q, then there is a formula X, whose variables are a subset of those

that q and q/ have in common, and such that F-A -- x and F-A Z -` .

This notion has several variants and reformulations, but in our case all of these (at least all the standard ones) follow from the property defined above. In particular, this holds for so-called local and global interpolation, best known in the context of modal logic. The reader will also pardon the sloppy talk about reasonable implication-especially that in our case implication is classical.

Crucial for the result is the next lemma, which follows easily from Madaraisz's generalisation of Maximova's celebrated theorem (see [4] and [5]), and the folklore about dynamic algebras recalled above. Before we state it, we need two more definitions. Let Fx5(X) stand for the X-free algebra generated by X. We will say that

f has SupAP for free algebras iff every V-formation (Fw (X n Y), Fx (X), Fx ( Y)) has Fx (X U Y) as a superamalgam, with identity embeddings. This is essentially Definition 4.4 from [4]. Notice that we implicitly assume that all the free algebras involved exist. In particular, the subuniverse generated by X n Y must be nonempty. Then, we say that X-free algebras superamalgamate iff every V-formation above can be superamalgamated in the variety 7(1(X).

LEMMA 2. For the variety 9 of dynamic algebras the following are equivalent:

(i) 9 has SupAP, (ii) 9 has SupAP for free algebras,

(iii) free algebras from 9 superamalgamate, (iv) A(9) has Craig interpolation.

PROOF. A straightforward consequence of Lemma 4.5, Proposition 4.6 and Lem- ma 5.1, from [4, Part I]. Put 9 for K, the similarity type of dynamic algebras for



936 TOMASZ KOWALSKI

t', that of boolean algebras for t and the set {x -+ y = 1} for F in Lemma 5.1. Further, put -* for ( and {x -* y 1} for the set of equations e(x(y) = 6(x0y) in Lemma 4.5 and Proposition 4.6. Then, it is easy to verify that 9 satisfies the conditions of [4, Lemma 5.1], and A(9) those of [4, Proposition 4.6]. The condition (iii) from Lemma 5.1 is perhaps the hardest to check, but it boils down to the following: for any B C A E 9, and any congruence ideal (i.e., an ideal closed under the dynamic operators) I on B, the boolean ideal (I] C A generated by I on (the boolean reduct of) A is closed under the dynamic operators. Since (I] is simply the downward closure of I as a subset of A, the statement follows by monotonicity of dynamic operators. Obviously, in this setting, the relation <r and the property (( Craig) from [4] are, respectively, the boolean partial ordering and the usual Craig interpolation property. The equivalence of (i), (ii), and (iv) follows from the above, and, by [4, Definition 4.4], and [4, Lemma 4.5], (ii) and (iii) are also equivalent. This closes the circle.

In the main proof, instead of relying on Lemma 2 as such, we will make use of its simple but convenient fine-tuning below.

LEMMA 3. All equivalences of Lemma 2 hold also under the assumption that the generating sets of FjX(X), F~x( Y) are both finite.

PROOF. Write (ii)fi and (iii)fi, for (ii) and (iii) restricted to finitely generated free algebras. To see that (ii)n, and (iii)nf are still equivalent simply follow the original proof from [4] and check that everything still works. To see that (ii)nf implies (iv), suppose the contrary, i.e., that Craig interpolation fails. Then there are terms t(X) and s(Y) with 0 t(X) < s(Y), but no term r(Z) has 0 = t(X) < r(Z) and 9 r(Z) < s( Y); where X and Y are finite sets of variables and Z c X n Y. We then arrive at a contradiction with (ii) for finitely generated FX (X n Y), F~ (X), Fx( Y) and Fz (X U Y). The rest follows.

?3. Boolean background. Before proceeding any further with dynamic algebras, we need some control over what is going on in the boolean background. To this end, we will collect here a couple of facts about boolean algebras (BAs). The general reference is to [2], but we will occasionally point to concrete theorems or definitions the reader may consult for details.

Recall that a BA C is complete iff for all Z c C, V Z (and, by duality, A Z)

exists in C. If A c B, Z C A, and VA Z exists, then we say that B preserves VA Z

iff VB Z also exists and VA Z = VB Z. The same notion applies to an embedding e: A - B. If A is complete and e preserves all joins, e is said to be complete. Although the results in this section will be stated in terms of preservation of joins, they obviously are all equivalent to their duals stated in terms of preservation of meets.

For a BA A, A+ will denote A \ {0}. With a BA A we associate a particular topology on 4(A+), called the partial order topology, whose open sets are all downward closed subsets of A+. As usual, for Z E pc(A+), we denote its closure by clA Z and interior by intA Z; we also write rA Z for intA clA Z. A set Z is called regular iff Z = rAZ. We let MAZ stand for the downward closure of Z in A+, and write 8Aa for aA{a}.




FACT 3.1. Let X and Y be any subsets of A+, and Z a family of subsets of A+. The following hold:

(i) intA X = {a E A+: PAa C X}, clA X - {a E A+: 6Aa n X # 0}; (ii) rAX = {a E A+: Vu E A, u < a: Au n X )0};

(iii) for a E A+, 6Aa is regular in A+; (iv) if Y is open in A+, then rAX n Y C rA(X n Y); (v) 6BU Z UfBZ: z E Z}. PROOF. The statements (i)-(iv) are straightforward consequences of the definition

of partial order topology (see [2, pp. 25-26], and Proposition 4.16). Statement (v) is an easy fact about partial orders; it does not depend on any topological considerations.

The set of all regular subsets of A+ forms a boolean algebra, denoted by Ac, when endowed with the operations: X n Y, rA(X U Y), and intA(A+ \ X); with X, Y c A+. Moreover, AC is complete, with V Z - rA (U Z) and A Z - rA (n Z), for any Z C Ac. The map f: A+ - Ac, with f(a) = sAa, extends to an embedding of A into AC by putting f (0) - 0. The BA AC constructed this way is called the (minimal) completion of A. The name is justified by the next fact (see [2, Definition 4.18, and Theorem 4.19]).

FACT 3.2. For any BA A the algebra AC of regular subsets of A+ is the smallest complete BA C such that A C C and C preserves all joins that exist in A.

If A c B, then A+ is a subspace of B+, yet, it is a well-known topological phenomenon that the sets open in A+ do not have to be open in B+. Thus, for a X C A+, intA X is generally not a subset of intB X; we only have 6AX _ 6BX.

FACT 3.3. Let A C B. Then:

(i) if X and Y are open in A+, then 6BX n 6B Y 6B(X n Y); (ii) if X is regular in A+, then the restriction of [rBaBX to A+ equals X;

(iii) if B preserves all joins that exist in A and X is regular in A+, then X c rB B X.

PROOF. In (i), the inclusion from right to left follows from the fact that 5B(X n Y) c -BX. For the other inclusion, take b E 6BX

On B Y. Then, for

some x EX and y Y, we have: b < x, b < y. Thus, b < xAy. As X and Y are

open in A+, x A y E X n Y. It follows that b E B (X n Y). For (ii), suppose that there is an a E rBB X n A+ that does not belong to X. By

Fact 3.1 we obtain (a) Vu E B+: u < a -= 6Bu 0n BX 0. Since X is regular in

A+, we also get (b) 3w E A+: w < a & 6Aw n X = 0. Now, it is easy to see that from (a) it follows that for all u E B+, u < a there is a z E 6BX with u A z > 0. Now, clearly there exists an x E X with z < x, and thus u A x > 0, for some x E X. However, from (b) it follows that there is a w E A+ (hence, a w E B+) with w A x = 0, for all x E X. This is a contradiction.

To prove (iii), again suppose the contrary. Thus, we have an a E rAX such that a ' rB3BX. From the former, we get (1) Vu E A+, u < a: Au n X $ 0. From the latter, (2) 3b E B+,b b a:

Ob nb BX = 0. Now, from (1) above it follows that

VA({-a} U X) = 1. Indeed, suppose there

is an upper bound q of {-la} U X in A \ {1}. Then, in particular, q > -a, hence,

-q < a, and thus, 6A(-q) N X $ 0. This means, that for some xo E X, we



938 TOMASZ KOWALSKI

have 0 < xo < -?q, which forces 1 > -xo > q > x for every x E X. Hence, 1 > -axo > xo > 0, a contradiction.

On the other hand, from (2) we have: -b > -ia, and b A x = 0, for all x E X. Thus,

-,b > x, for all x E X. Therefore, --b is an upper bound of {--a} U X in B.

Since b E B+, -b < 1. Thus, at least one join existing in A is not preserved in B. This contradicts the assumption and proves our claim.

For a BA B and its subalgebra A, it seems natural-and will prove useful-to ask how their minimal completions are related to each other. The answer is provided in the next lemma. Let A C B be BAs.

LEMMA 4. If the embedding i : A -- B preserves alljoins that exist in A, then i extends to a complete embedding e: AC ' BC, defined by: e(S) - rBGBS, for a regular S C

At+.

PROOF. We have: e(0) - rBB B = BO r = O ; similarly, e(A+) =

rBBA+ -

rBB+ - B+. Thus, top and bottom are preserved. Take S, R regular in A+. We have: S n R C S, thus, 6B(S n R) C BS; further,

as both sides of this inclusion are open in B+, rB5B(S n R) C rB6BS. This proves rB6B (S n R) C rBGBSn rBB R, and e(SA R) < e(S) Ae(R) follows. For the other direction, as 6BS, 6BR are open in B+, we have, by Fact 3.1(iv), e(S) A e(R) rB BS n rB BR C rB(BSn rB BR) C rB (rB(BS n6BR)) = rB(6BS odBR). Since S and R are open in A+, Fact 3.3(i) yields 6BS n 8BR

= 6BS n R. Thus, rB (6BSnoBR) - rBBe(S n R) = e(SAR). This establishes e(S)Ae(R) < e(SAR). Thus, e preserves meets; hence e preserves the order.

Now, we will prove preservation of arbitrary joins. Put C = AC, D = Be. Let Z be any subset of C; thus all z E Z are regular subsets of A+. We will prove that e (Vc Z) = VD{e(z): z E Z}. Since e has already been shown to preserve order, we conclude that e (Vc Z) ? VD{e(z): z E Z} always holds. Thus, we only need to prove the reverse inequality. By definition of C, we have: VC Z

- rA (U z).

Calculate: e (VC Z) - e (rA (U z)) - rB5B (rA (U Z)), by definition ofe. Applying

downward closure in B+ to Fact 3.3(ii), we get: 6BrA (U z) C 6eBrB8BU Z. Since rB6BU Z is open in B+, its downward closure is equal to itself; thus, we obtain: PBrA (U z) C rB•B U Z. From this, we further get: rBeBrA (U z) C rB rBBU Z = rBaBU Z. As, by Fact 3.1(v), downward closure commutes with union, this further equals: rB (U{6Bz: z E Z}). Now, each 6Bz is open in B+, thus, 6Bz C rB6Bz. It follows that: rB (U{(Bz: z E Z}) C rB (U{rBaBz: z E Z}), and the latter, by definition of e, equals: rB(U{e(z): z E z}) - VD{e(z): z E Z}. This finally establishes: e(Vc Z) < VD{e(z): z E Z}, and finishes this part of the argument.

Preservation of complements follows from what we have just established. Namely, for a S E C, we have: e(S) A e(-S) - e(SA -iS) - e(0) - 0, and thus, e(-iS) < -e(S). Analogously, e(S) V e(-S) = e(S V -S) = e(1) = 1, and thus, e(-S) > -e(S).

It remains to show that e is one-one. Let S -

R, and assume that a E S \ R. Since S is regular in A+, by Fact 3.3(iii) we have: a E rB3BS. However, as a E A+, we get that a g rB•BR, by Fact 3.3(ii). Thus rB8BS rBSBR as required.




As a matter of curiosity let us add that without the preservation requirement on i: A - B, the map e defined as above will not only cease to be complete, but may fail to be an embedding altogether.

Let B(X), B(Y) be BAs freely generated by X and Y; moreover let Y c X and X = IYI - co. By normal form theorem, we can assume that all elements of

B(X) are in conjunctive normal form over X. To spell it out: for any a E B(X), a - Jo A ... A Jm-1, with elementary joins Ji (0 < i < m - 1); an elementary join being a term V Xo, where Xo is a finite subset of X =- X U {--x: x E X}. Since Y C X, this also represents all elements of B( Y) in conjunctive normal form over Y.

Whenever a E B(X)\ {1}, we write aIy for V(Xo n Y) A ... A V(Xk-I n Y:), where Xi - X n Ji, for i {0, ..., k - 1}. If a - 1, we put a I y a. Clearly, for any a E B(X), ayr E B(Y).

FACT 3.4. Let Z be a subset of B(Y) and let b E B(X) be an upper bound of Z. Then, b I y is also an upper bound of Z.

PROOF. We can assume that b -f 1, hence Z n { 1 } = 0. Then, each element of Z has the form Jo A ... A Jm-1, with all Ji (0 < i < m - 1) being elementary joins of elements from Y+ - Y U {-y: y Y}. Let {xo,..., xk-l} C X+. We claim that the following holds:

() Jo A ... A Jm1 <_

xo V ... V Xk-1

iff Jo A ... A Jm-1

< V({x0o.... Xk-1 )n Y).

Only the direction from left to right is not trivial. We proceed by induction on m. For m = 1, we have: Jo < xo V ... - Vxk- iffyoV0 . Vy_ n1 < xoV . . .

Vxk-_1, with Jo =

yo V . . V yn--1.

This is further equivalent to: (-'yo A. .. A -yn-1) V xo V - V Xk-1 (GYo V xo V . V Xk-1) A' ' ' A ('Ym-1 V x0 V . V xk- 1 ) 1, which can only happen if for all i c {0 ..., m - 1 }, yi V xo V ... V Xk-1 = 1. By independence of generators, this is the case iff Vi E {0,...,m - } 3j E {0,..., k- 1}: yi = xj. This says that all the elements yi occur in {xo ....,Xk -1 }, yielding: {xo, ... xk-1} n Y? 2 {yo,... IYm-1}. From this, yo V . -. V ym-1 ? V({X0o . . Xk-1} 0 +Y) follows immediately.

For the inductive step, consider: Jo A ... A Jm-1 A Jm < xo V ... V Xk-1. Writing Jm in full as yo V ... V yl-1, we obtain the equivalent statement: --J0 V ... V

-Jm-1 V -(yo

V . V YI-1) V xo V ... V Xk-1 = 1. By boolean identities, we get: Jo V . . V ---Jm1 V-I ((-yo V xo V . . . V Xk-1) A" ...A (Yl-1 V

Xo V"..

VXk-1)) = 1

This is the case iff Jo A . A Jm_ 1 <_ -'yi-1VxoV

. . V xkl, for all i E {0,... 1- 1}. As the inductive hypothesis works for any number of elements on the right-hand side, we obtain, for all i E {0,..., 1 - 1}:

Jo A ...A Jml1 < ~Yi-1 V xo V ... V Xk-1

iff Jo A . A Jm-1 < V({Yi, x0

.... .Xk-l}0 Y).

Taking the meet over all i {0 ...., 1 - 1 } and distributing suitably, we transform the right-hand side into:

Jo A . .

A AJ 1 < ( yo A A yll) V V({ xO ... )n r),



940 TOMASZ KOWALSKI

and further into:

Jo A ... A Jm-1A (Yo V . . . V yi-1) = Jo A ... A Jm-1 A Jm

< V({x0o...Xk-1} Y), as required by the claim.

Now, let b E B(X), be an upper bound of Z. Then, b = K0 A .. A Ki-1, with Ko,..., K,-1 being elementary joins. Thus, for any z E Z, we have: z < Ko A ... A K1-1. This happens iff z < Ki, for all i E {0,..., k - 1}. By (*) above, we obtain: for all z E Z, z < b y.

Before we proceed further, let's recall one more fact from boolean algebra theory. Let A c B be any BAs, and Z be a subset of A.

FACT 3.5. The join of Z is preserved and equals a iff the join of {z V -a : z E Z} is preserved and equals 1.

PROOF. It is folklore in BA theory, but to keep the paper reasonably self-contained, we present a proof. For the 'only if' part, assume VA Z = a = VB Z. Take any b E B with b > zV--a,for all z E Z. Then, we have: bAa > (zV-'a)Aa = zAa = z, for all z E Z. Since by assumption VB Z = a, we obtain: b A a > a, i.e., b > a. However, b is also greater than --a, thus b = 1, as needed. For the 'if' part, assume

VA{z A -a: z E Z} = 1 = VB{z A -a: z E Z}, but suppose that VA Z = a is not preserved in B. Thus, there is a b E B, with b < a and b > z, for all z E Z. Since b < a, we get that --,a V b < 1. But -a V b is an upper bound of {z v -'a: z E Z} in B. Hence, VB{z A-a: z E Z}

- 1, contradicting the assumption.

Now we come back to the main line of the argument. Let B(X) and B( Y) be as before.

LEMMA 5. The natural identity embedding of B(Y) into B(X) preserves all joins that exist in B( Y). Moreover, for any Z C B(Y), ifb c B(X) is an upper bound of Z, then there is a b' E B ( Y) such that b' < b and b' is an upper bound of Z.

PROOF. By Fact 3.5, it suffices to prove that for all Z C B( Y), if VB(Y)Z exists and equals 1, then VB(X) Z also exists and equals 1. Suppose the contrary, then there is an upper bound u < 1 of Z in B (X). We can assume that u is an elementary join, i.e., u = xo V .. V Xk-1, with Xi E X* for all i E {0,...,k - 1}. Then, putting w = V({xo,..., xk-1} n Y+), we immediately obtain that w > 1, and w is an upper bound of Z in B(Y). This contradiction ends the proof of the first part.

For the second part, put b' = bjy. It is clear that bIy < b and, thus, the conclusion follows by Fact 3.4.

Let (Bo, B1, B2) be a V-formation of BAs. We will now recall the notion of amalgamated free product of B1, B2 over B0. For our purpose here, it suffices to state the characterisation lemma (dual to [2, Corollary 11.22]) below.

LEMMA 6. An algebra B is the amalgamatedfreeproduct of B1, B2 over B0, with the embeddings el, e2 if el (B1) U e2(B2) generates B and for any two elements b1 E B1, b2 E B2 with el (bl) V e2(b2) = 1, there is a c E Bo such that bl

_ c and b2

_ --c. One immediate consequence of the above is that B is a superamalgam of (Bo, B1,

B2). Indeed, if we have an x E B1 and y E B2, with el(x) < e2(y), then -el(x) V




e2(y) = 1. Thus, there is a c E Bo such that -x >- c and y > c, which yields x < c < y as needed. Another is that the map el U e2 is one-one, and thus can be taken to be the identity map.

Despite that, amalgamated free products may fail to preserve infinite operations. However, if the algebras and embeddings in (Bo, B1, B2) are complete, i.e., preserve all joins, then the preservation is ensured in the product as well. Of course, the product itself, being freely generated, is hardly ever complete, except for finite or trivial cases.

LEMMA 7. If B0, B1, B2, and the identity embeddings are all complete, then the amalgamated free product B of (Bo, B1, B2) preserves all joins from Bi (i - 1, 2).

PROOF. Follows easily from [2, Lemma 12.16], but, again, we will recall the

proof. Let X C B1. We can assume VB' X - 1, and we only need to prove that

VB X = 1. Suppose the contrary, i.e., that there is an upper bound u of X in B strictly less than 1. By the normal form theorem, there are elements bl E B1 and b2 E B2, such that 1 > b1 V b2 > u; hence, bl v b2 > x, for all x E X. Thus, for any x E X, we have b2 V (bl V -x) = 1, and, by Lemma 6, for any x E X there is a c(x) E Bo with b2 > c(x), bl V --x > -c(x). The latter can be rewritten as bl V c(x) > x. Let c = VBO{c(x): x E X}. By completeness of Bo this is well-defined and belongs to Bo; by completeness of the identity embedding B0 -- B1, VBO{c(x): x E X} = VB' {c(x): x X}. Thus, from the inequalities above, we obtain: b2 > VB {c(x): x E X} = VBO{c(x): x c X} = c; and, in a similar fashion, bl V c > VBI X

- 1. The latter yields bl > - c. Therefore,

bl V b2 > -c V c = 1, a contradiction.

?4. Superamalgamation. As we have already seen (cf. Fact 1.1), in boolean reducts of dynamic algebras some but not all meets have to exist. On the other hand, if we deal with a complete BA, and if the basic programs are defined on a certain suitable subset of its universe, we can always "upgrade" such a structure to a fully-fledged dynamic algebra. Our construction relies on two facts: that without the star, the dynamic operators are modal operators amenable to a form of infinitary filtration, and that the whole power of the star may be encoded by an infinite meet (see Fact 1.1 again).

Let B be a complete BA, and A a subset of B such that for all a c A, and for all c E , 4a is defined, and the following two conditions hold:

(CO) 0 E A and ?0 = 0, for all 0 E E;

(Cl) if aEA and b A, thenaVbEA and ~aV b =(aVb),forall~ E.

Then, define (what will turn out to be) the dynamic operators on B, inductively.

DEFINITION 1. For any a, f E C, and any b, c E B:

(a) for ( E E, (-)c = A{(x : x c, x E A}, (b) (a U fp)c

= (a)c V (fl)c, (c) (a,6)c = (a)(f6)c. (d) (a*)c = A{b: c V (a)b < b}.

Let C stand for the algebra (B; (ir)(r E 11)).



942 TOMASZ KOWALSKI

FACT 4.1. The algebra C is a dynamic algebra.

PROOF. Induction on the complexity of dynamic operators. For 3(iii) and 3(iv), as no ? c E can be of the form af or a U fl, the base case holds emptily. The inductive step follows by the definitions of (af) and (au f).

For 3(i), if 7 - r , for 6 E ., then (?)0 =- 0 = 0, by (CO). If = afp, then (apf)O = (a) (fl)O0, by definition, and applying inductive hypothesis twice, we obtain that this equals 0, as needed. Similarly for 7r = a U Pf. If r -

a*, then (a*)0 A{b: 0 v (a)b < b}. By inductive hypothesis we get (a)0 0= , thus 0 e {b: 0 V (a)b < b}. Hence, (a*)O - 0.

For 3(ii), if 71 = , for E E, then (?)c V (?)b = A{•(y: y > c,y E A} V

A•{z: z > b, z E A}. This, by distributivity of any meets over finite joins (which holds in any BA, provided the required meets exist) equals A{(y V ?z: y > c, z > b, y,z E A}.

Further still, by (C1) this is equal to A{~(y V z): y > c, z > b, y, z E A}. Now, clearly, any member of {((y V z): y > c, z > b, y, z E A} belongs also to {(x: x > c v b, x E A}. Conversely, any ?x with x 2 c V b, x E A, can be rewritten as (x V x), which obviously belongs to {((y V z): y > c, z > b, y, z E A}. Thus, we obtain: {((y V z): y > c, z > b, y, z E A}- {x: x > c V b, x E A}, and, hence, finally: (W)c v (?)b A{~(Y V z): y > c, z

_ b, y, z E

- } A{~x: x >

c V b, x E A} = () (c V b), as desired. If C = aft, then (a#p)(c v b) = (a)(P)(c v b). Now, the inductive hypothesis

yields that it equals to (a)(fP)c V (a)(fP)b, which, by definition, is further equal to (af)c V (afl)b. The proof for r a U P proceeds analogously.

If 7 = a*, then (a*)(c V b) = A{a: c V b V (a)a < a}. On the other hand, (a*)c V (a*)b is equal to A{u: c V (a)u < u} V A{w: b V (a)w < w} = A{u V w : c V (a)u < u, b V (a)w < w}. Now, if a has cVbV (a)a < a, then a = a Va has also c v (a)a < a and b v (a)a K a. Conversely, if u, w have c V (a)u K u and b V (a)w < w, then c V b V (a)u V (a)w < u V w. By inductive hypothesis, we then get c V b V (a)(u v w) < u v w, as needed.

The last thing we have to show is that (a*)c E {b: c V (a)b < b}. By Fact 1.1, this will demonstrate that 3(v) is satisfied. Now, by definition, (a*) c < b, for any b with c V (a)b < b. Then, c V (a)(a*)c c V (a)b b; this step requires only monotonicity of (a), which has already been established. Thus, c v (a) (a*) c ?

A{b: c v (a)b < b} = (a*)c. Hence, (a*)c E {b: c V (a)b < b} as required.

Now we have in hand all the tools we need to prove superamalgamation for dynamic algebras. By Lemma 3, it suffices to show that the V-formation (F2 (X n Y), F~ (X), F ( Y)) can be amalgamated in 9, where F- (X n Y), Fc (X), F ( Y) are the free dynamic algebras finitely generated by X n Y, X, Y, respectively. As 9 has constants, the free algebras involved always exist. From now on we will drop the subscript ?q. Also, we will apply the preservation results from the previous section mostly in their dual form.

LEMMA 8. The identity embeddings in (F(X n Y), F(X), F(Y)) preserve all meets that exist in F(X n Y). Moreover, for any Z C F(X n Y), ifb E F(X) U F(Y) is an upper bound of Z, then there is a c E F (X n Y) such that c K b and c is an upper bound of Z.




PROOF. We can assume that F(X n Y), F(X), and F( Y) are all countable (in fact, they must be, since even the zero-generated free dynamic algebra is infinite, but if any of them were in fact finite, then the claim would follow trivially). Then, they are all atomless, and thus isomorphic as BAs to the countably generated free BA B(w). Fix any boolean isomorphisms bx: F(X) -) B(w) and by: F(Y) -- B(co). Then, the boolean reduct of F(X) is generated by bxj (o), and similarly for F( Y). Since F (X n Y) = F (X) n F(Y), we have that F(X n Y) is generated as a BA by

bx1 (o n F(X) n F(Y)) = by1(w n F(X) n F(Y)). Then, the claim follows by Lemma 5.

Let Gxny, Gx, and Gy with F(Z) C Gz, for Z E {X, YX n Y} such that Gz is the minimal completion of the boolean reduct of F(Z). By Lemma 8 above and Lemma 4, there are complete embeddings ex: Gxny --- Gx and ey: GxnY -Y Gr. Thus, the triple (Gxn y, Gx, G ry) is a V-formation of complete BAs.

Let E be the amalgamated free product of Gx and Gy over Gxny, and Ec its minimal boolean completion. By Lemma 7 and Fact 3.1, Ec preserves all meets from Gx and G y. Since the embeddings of Gxn Y into Gx and G y are both complete, Ec also preserves all meets from Gxny. This, together with Fact 3.2. and Lemma 4, yields also that all the infinite meets (and joins) that exist in F(X n Y), F(X) and F( Y) are preserved in Ec.

Let C C EC be the closure of Gx U Gy under join. Then, for all ( E E, and all c E C, define: cc - (A{(z E F(X): z > c}) V (A{(z E F(Y): z > c}). By the previous paragraph, we can assume that all the meets are taken in EC.

FACT 4.2. The structure C satisfies (CO) and (Cl).

PROOF. Since 0 E F(X) n F(Y), we get 0 = 0 by axioms of dynamic algebras, thus (CO) holds. For (Cl), take a, b e C. Then, (a V b) = A{(z: z c F(X), z > a Vb}VA{(z: z E F(Y),z a Vb}. On the other hand, ~a V~b = A{(u: u E F(X),u > a} V A{u: u E F(Y),u > a} V A{w: w E F(X),w > b} V

A{w: { w c F(Y), w > b} which further equals A{(u V w: u > a, w > b, u, w E F(X)} V A{(u V w : u > a, w > b, u, w E F(Y)}. Since ?u V ?w = S(u V w) in dynamic algebras, this in turn is equal to A{~(u V w): u V w > a V b, u,w E F(X)} V A{(u V w): u V w> a V b, u,w c F(Y)}. Now, clearly, {(z: z E F(X),z > a V b} = ({(u V w): u V w > a V b, u, w F(X)} and {(z: z EF(Y), z > aV b} = {((u V w): uVw > aV b, u, wE F(Y)}, by which the claim follows.

By Fact 4.1 and Fact 4.2, Ec can be expanded to a complete dynamic algebra D with the universe EC. The whole situation is pictured in the diagram below:

F(X) cx. Gx

F(X n Y) CXN-rGxny

E' cE

F(Y) -,

Gy CY

where ix, iy are the natural identity embeddings of free dynamic algebras; ex, ey their complete boolean extensions as defined in Lemma 4; cx, cx, CxnY, and cE



944 TOMASZ KOWALSKI

indicate the minimal completion embeddings; fx, f y stand for the amalgamated free product embeddings; finally (- denotes the expansion of the BA EC to the dynamic algebra D.

LEMMA 9. The algebra D is a superamalgam of the V-formation (F(X n Y), F(X), F(Y)).

PROOF. For the amalgamation part, we only have to show that D preserves dynamic operators. By symmetry, it is enough to prove that for F(X). The proof is by induction on the complexity of dynamic operators. Take an a E F(X), and a 7 E I. If a = r E, then it is straightforward from the definition that (?)a a- a.

If 2t = aft, then (afl)a = (a)(f)a = arpa, where the first equality holds by definition, and the second by inductive hypothesis. Similarly, if r =-

a U /, we have (aU fl)a = (a)a V (fl)a aa V fa = (aU fl)a; here, the last equality holds because F(X), being a dynamic algebra, satisfies (Cl).

If r = a*, then (a*)a = A{c E C: aV(a)c < c}. On the otherhand, computing a* in F(X) yields, by *-continuity: a*a = AF(x){b E F(X): a V ab b} =

VF(X){ana: n E w}. Now, by inductive hypothesis, we have ab = (a)b, and the preservation results ensure that the infinite meet and join involved may as well be taken in Ec. Thus, we obtain a*a - A{b E F(X): a V (a)b < b} - V{aea: n E w}. Clearly, {c e C: a V (a)c < c} contains {b E F(X): a V (a)b < b}, and thus (a*)a < a*a. To prove the reverse inequality, we show that a*a < c, for any c E C with a V (a)c < c. We will show by embedded induction that for alln E Co, ana < c. Forn =0wegeta < c byaV (a)c < c. Forni+ 1, we have a"a < c by the inductive hypothesis, thus, (a)ana ? (a)c. Since a is simpler than a* we can apply the hypothesis of the outer induction and get: (a)ana = aa"a - a"n+a < (a)c

_ c. This proves V{ana: n E o} = a*a < c,

as desired. Since D is a (boolean) superamalgam of (Gxn y, Gx, Gy) by definition, we now

have to make sure that D satisfies the superamalgamation condition also with respect to (F(X n Y), F(X), F( Y)). Let a E F(X) and b E F(Y), have: fx o cx(a) < fy o cy(b). By superamalgamation for (GxnY, Gx, Gy), we have an element Z E Gxnr with cx(a) < ex(Z) and ey(Z) < cy(b). Since Gxnv is the minimal completion of F(X n Y), Z is a regular subset of F(X n Y)+. We also have: cx(a) -= F(X)(a), cy(b) = rF(Y)(b). Moreover: ex(Z) = rF(X)aF(X)(Z), and ey(Z) = rF(Y)6F(Y) (Z). All this yields the following:

(1) aF(X)(a)C rF(X)3F(X)(Z),

(2) rF(Y)5F(Y)(Z) C F(Y)(b).

From (1) we conclude that any u E F(X) that majorises Z, majorises a as well. Indeed, take any u E F(X) with: u > z, for all z E Z. Then, 6F(X)(u) D Z, and thus, GF(X)(u) D F(X)(Z), from which it follows that: 8F(X)(u) - rF(X)8F(X)(u) D rF(X)5F(X)(Z) 2 3F(X)(a). Hence, u 2 a.

From (2) we obtain that Z c 6F(Y)(b). For, by Fact 3.3(ii), we have rF(XnY)Z C rF(Y) F(Y)(Z); and rF(XnY)Z = Z, since Z is regular in F(X n Y)+.

Clearly, Z C 6F(Y)(b) iff for all z E Z, b > z. Therefore, by Lemma 8, there is an element c E F (X U Y) with c 2 z, for all z E Z. Then, by the conclusion derived




from (1) above, we have: c > a. Altogether, a < c < b, precisely as required for superamalgamation.

THEOREM 1. Free algebras from 0 superamalgamate. Thus, 9 has SupAP. PROOF. By Lemmas 9, 3, and 2.

THEOREM 2. Full propositional dynamic logic has Craig interpolation property. PROOF. In [3] it is proved that full PDL has interpolation iff test-free PDL has

interpolation. Test-free PDL is precisely A(2). The rest follows by Theorem 1 and Lemma 2.

?5. Two corollariest. Two noteworthy logics closely related to full PDL are full PDL with converse (full PDL') and full deterministic PDL (full DPDL). Again, the reader is referred to [3] for more-including more references; here let us only remark that full PDL- is being applied in logical analysis of reasoning about actions, and there are certain uses for interpolation properties there.t

Full PDL' arises by augmenting full PDL with operations rn for all programs n7 (or, equivalently, for all basic programs), along with the axioms whose counterparts in our algebraic idiolect may be expressed as the identities:

(cl) x A nry ? <r(y A 7(rx), (c2) x A Tr y < 7C(y A 7tx).

Full DPDL is full PDL plus the identities: (d) ?x < •-,x, for all basic programs r E. It is proved (or left as an exercise) in [3] that Craig interpolation for full PDL

implies Craig interpolation for both full PDL' and full DPDL. Thus, we can note:

COROLLARY 1. Full PDL- and full DPDL both have Craig interpolation property. On a more algebraic note, we can also state:

COROLLARY 2. Epimorphisms are surjective in the variety of dynamic algebras. PROOF. One proof goes by way of noticing that Craig interpolation implies so-

called global interpolation (see, e.g., [3] for the definition and a proof), which in turn is equivalent to surjectivity of epimorphisms (for which see [1]). However, since there is a rather straightforward direct proof, we will present it here.

Recall that a homomorphism f : A --+ B, with A and B belonging to some class X, is an epimorphism in X if for all C E X and all homomorphisms h: B

-- C

and h': B --) C, we have: h o f = h' o f implies h = h'. Clearly, all surjections are epimorphisms, the converse is not true in general. Sup-

pose f: A -* B is a non-surjective epimorphism in the variety of dynamic algebras. Let C be f (A), thus, B \ C is nonempty. Take the V-formation (C, B0, B1, io, ii), where B0, B1 are isomorphic copies of B such that B0o n B1 C, and io = il is the identity map. By strong amalgamation for 9 and our conventions, there is an algebra D E

. and identity embeddings eo, el such that eo U el is one-one. In

particular, for any b e B \ C (by assumption there is one), we have eo(b) Z el (b). Consider the maps eo o io o f = eo of and el o i o f = el o f. Amalgamation

tThanks to the anonymous referee for pointing them out.

tThanks to Samir Chopra and Dongmo Zhang for telling me about it.



946 TOMASZ KOWALSKI

yields eo o f = el o f; the setup in turn forces eo e2. This contradicts f being an epimorphism.

REFERENCES

[1] E. HOOGLAND, Algebraic characterizations of various Beth definability properties, ILLC preprint, 1999.

[2] S. KOPPELBERG, Handbook ofboolean algebras (J. D. Monk and R. Bonnet, editors), vol. 1, North Holland, 1989.

[3] M. KRACHT, Tools and techniques in modal logic, Studies in logic and the foundations ofmathematics, vol. 142, Elsevier, 1999.

[4] J. X. MADARASZ, Interpolation and amalgamation; pushing the limits. Part I, Studia Logica, vol. 61 (1998), pp. 311-345, Part II, Studia Logica vol. 62 [1999], pages 1-19.

[5] L. L. MAKSIMOVA, Interpolation theorems in modal logics and amalgamable varieties of topological boolean algebras, Algebra and Logic, vol. 18 (1979), pp. 348-370.

[6] V. PRATr, Dynamic algebras: Examples, constructions, applications, Studia Logica, vol. 50 (1991), pp. 571-601.

JAPAN ADVANCED INSTITUTE OF SCIENCE AND TECHNOLOGY 1-1 ASAHIDAI, TATSUNOKUCHI

923-1292 ISHIKAWA, JAPAN On leave from: Department of Logic, Jagiellonian University, ul. Grodzka 52, 31-044 Krak6w,

Poland E-mail: [email protected]



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PDL Has Interpolation