Transfinite Step-indexing: Decoupling Concrete and Logical Steps (Technical Appendix)

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1 Trnsfinite Step-indexing: Decoupling Concrete nd Logicl Steps (Technicl Appendix) Ksper Svendsen, Filip Sieczkowski nd Lrs Birkedl October 30, 2015 Contents 1 Complete ordered fmilies of equivlences over ω Uniform predictes Solving Recursive Domin Equtions in U Syntx nd opertionl semntics 11 3 Logicl reltion Comptibility lemms Soundness Exmple Complete ordered fmilies of equivlences over ω 2 Definition 1 (o.f.e.). An ordered fmily of equivlence reltions (o.f.e.) over ω 2 is pir (X, ( =) ω 2), consisting of set X nd n ω 2 -indexed set of equivlence reltions =, stisfying x, y X. x 0,0 = y x, y X., b ω 2. b x b = y x = y x, y X. ( ω 2. x = y) x = y Definition 2. Let (X, ( = X ) ω 2) nd (Y, ( = Y ) ω 2) be o.f.e s. A function f : X Y is non-expnsive if nd is contrctive if x 1, x 2 X. ω 2. x 1 =X x 2 f(x 1 ) = Y f(x 2 ) x 1, x 2 X. ω 2. ( b <. x 1 b =X x 2 ) f(x 1 ) = Y f(x 2 ) Definition 3 (Coherent fmilies nd limits). Let (X, ( =) ω 2) be n o.f.e., I subset of ω 2 nd (x ) I n I-indexed sequence of elements in X. Then (x ) I is coherent fmily if, b I. b x = xb An element x X is limit of the sequence (x ) I if I. x = x 1

2 Definition 4. An o.f.e. (X, ( =) ω 2) hs chosen prtil limits iff for ny b ω 2 there exists function lim <b x tht mps b -indexed coherent fmily (x ) <b to limit, such tht for ny two b -indexed coherent fmilies (x ) <b nd (y ) <b, ( < b. x = y ) lim <b x = lim <b y Definition 5 (Completeness). An o.f.e. (X, ( =) ω 2) is complete iff ll ω 2 -indexed coherent fmilies hve limits nd it hs chosen prtil limits. Definition 6 (U). Let U denote the ctegory of complete ordered fmily of equivlence reltions nd nonexpnsive mps. 1.1 Uniform predictes Definition 7 (Uniform predictes, UPred(X)). A uniform predicte, UPred(X), over set X is defined s UPred(X) = def {p P(ω 2 X) (, x) p. b. (b, x) p} with the following A-indexed equivlence reltions: p = def q iff p = q where p = {(b, x) p b < } Lemm 1. UPred(X) is n o.f.e. Proof. The first o.f.e. condition follows from the fct tht p 0,0 = for ll p UPred(X). The second o.f.e. condition follows esily from the fct tht p b = p for ll p UPred(X) nd, b ω 2 such tht b. Lstly, the third o.f.e. condition follows esily from the fct tht for every ω 2 there exists n b ω 2 such tht < b. Lemm 2. UPred(X) hs chosen prtil limits. Proof. Let b ω 2 nd define lim <b s follows: lim p def = {(, x) ω 2 X + 1 < b (, x) p +1 } <b Let (p ) <b nd (q ) <b be two coherent fmilies such tht < b. p = q. To show tht lim <b p = +1 lim <b q ssume (, x) lim <c p. Then + 1 < c nd (, x) p +1. Hence, p +1 = q +1 nd thus (, x) q +1. It follows tht (, x) lim <c q s required. lim <b q lim <b p follows by symmetric rgument. We lso need to show tht lim <b p UPred(X) for coherent fmily (p ) <b. To tht end, let (c, x) lim <b p nd d c. By definition, c + 1 < b nd (c, x) p c+1. Since p c+1 is uniform predicte, it follows tht (d, x) p c+1. Furthermore, since d c it follows tht d + 1 c + 1 nd thus by the coherent d+1 property tht p d+1 = p c+1. Hence, (d, x) p d+1 nd thus (d, x) lim <b p s required. Lstly, it remins to show tht lim <b p is limit for coherent fmily (p ) <b. Let c < b. To show tht lim <b p c p c c, ssume (d, x) lim <b p c. Then d < c, d + 1 < b nd (d, x) p d+1. Since d+1 d < c it follows tht d + 1 c nd thus by the coherent property tht p d+1 = p c. Thus, (d, x) p c nd (d, x) p c c. To show tht p c c lim <b p c, ssume (d, x) p c c. Then d < c nd (d, x) p c. Thus d+1 d + 1 c nd by the coherence property tht p d+1 = p c. It thus follows tht (d, x) p d+1. Lstly, since d + 1 c < b we hve tht (d, x) lim <b p. Lemm 3. UPred(X) hs limits of ω 2 -indexed coherent fmilies. 2

3 Proof. Let (p ) ω 2 be coherent fmily in UPred(X). Let lim p = {(, x) ω 2 X (, x) p +1 }. To show tht p UPred(X), ssume (, x) p, b. Hence, by definition of p, (, x) p +1. Since p +1 UPred(X) it follows tht (b, y) p +1. Since b it follows tht b nd thus b+1 p b+1 = p +1. Hence, (b, y) p b+1 nd (b, y) lim p. It thus remins to show tht lim p is limit of (p ) ω 2. Let b ω 2. To show tht lim p b p b b ssume tht (c, x) lim p b. Then c < b nd (c, x) p c+1. Since c < b it follows tht c + 1 b nd c+1 thus p c+1 = p b. Hence, (c, x) p b nd (c, x) p b b, s required. To show tht p b b lim p b, ssume tht (c, x) p b b. Then, c < b nd (c, x) p b. Since c < b it follows tht c + 1 b nd by coherence, c+1 p c+1 = p b. Hence, (c, x) p c+1 nd (c, x) lim p, s required. Lemm 4. The chosen prtil limits inupred(x) respect the prtil order on UPred(X). In prticulr, for ny coherent fmilies (p ) <b nd (q ) <b, ( < b. p q ) lim <b p lim <b q Proof. Assume (c, x) lim <b p. Then c + 1 < b nd (c, x) p c+1 q c+1. Thus, (c, x) lim <b q, s required. Lemm 5. Limits in UPred(X) for ω 2 -indexed coherent fmilies respect the prtil order on UPred(X). In prticulr, for ny coherent fmilies (p ) ω 2 nd (q ) ω 2, ( ω 2. p q ) lim p lim q Lemm 6. (UPred(X), ) is complete Heyting lgebr with meets nd joins given by S def = S = {(, x) ω 2 X p S. (, x) p} S def = S = {(, x) ω 2 X p S. (, x) p} nd impliction given by Constructions p q = def {(, x) b. (b, x) p (b, x) q} Definition 8. Let X = (X, ( = X ) ω 2) nd Y = (Y, ( = Y ) ω 2) be o.f.e. Then the o.f.e. of non-expnsive functions X ne Y is defined s follows: X ne Y = def ({f : X Y f is non-expnsive}, ( = X ney ) ω 2) where f = X ney g = def x X. f(x) = Y g(x). Lemm 7. If X nd Y re o.f.e.s nd Y hs chosen prtil limits then X ne Y hs chosen prtil limits. Proof. Let b ω 2 nd define the chosen limits function for coherent fmily (f ) <b, lim f = λx X. lim f (x) <b <b By the definition of = X ney it follows tht (f (x)) <b is coherent fmily in Y for ny x X nd thus tht lim <b f is well-defined. 3

4 c To show tht lim <b f is non-expnsive nd thus n element of X ne Y, ssume x 1 = x2. If c < b then (lim <b f )(x 1 ) = lim <b f (x 1 ) c = f c (x 1 ) c = f c (x 2 ) c = lim <b f (x 2 ) = (lim <b f )(x 2 ) by non-expnsiveness of f c, s required. If b c then f (x 1 ) = f (x 2 ) for ll b nd thus, (lim <b f )(x 1 ) = lim <b f (x 1 ) = lim <b f (x 2 ) = (lim <b f )(x 2 ) To show tht lim <b f is limit, let ω 2 such tht < b nd x X. Then, s required. (lim <b f )(x) = lim <b f (x) = f (x) = (f )(x) Lstly, let (f ) <b nd (g ) <b be two coherent fmilies such tht < b. f = g To show tht lim <b f = lim <b g, let x X. Then, it follows tht f (x) = g (x) for ll < b nd thus (lim <b f )(x) = lim <b f (x) = lim <b g (x) = (lim <b g )(x) Lemm 8. If X nd Y re o.f.e.s nd Y hs limits for ω 2 -indexed coherent fmilies then so does X ne Y. Proof. Let (f ) ω 2 be coherent fmily in X ne Y. Define the limit s follows, lim f = λx X. lim f (x) This is well-defined s (f (x)) ω 2 defines coherent fmily in Y for ll x X. To show tht lim f is b non-expnsive, ssume x 1 = x2. Then, (lim f )(x 1 ) = lim f (x 1 ) b = f b (x 1 ) b = f b (x 2 ) b = lim f (x 2 ) = (lim f )(x 2 ) by non-expnsiveness of f b nd the limit property of Y. Lstly, to show tht lim f is limit let b ω 2 nd x X. Then by the limit property of Y. (lim f )(x) = lim f (x) b = f b (x) = (f b )(x) Definition 9. Let X = (X, ( = X ) ω 2) nd Y = (Y, ( = Y ) ω 2) be prtilly ordered o.f.e.s Then the o.f.e. of monotone non-expnsive functions X mon Y is defined s follows: X mon Y def = ({f : X mon Y f is non-expnsive}, ( = X ney ) ω 2) where f = X ney g = def x X. f(x) = Y g(x). Lemm 9. Let X nd Y be prtilly ordered o.f.e.s. If Y hs chosen prtil limits such tht for ny two coherent fmilies (x ) <b nd (y ) <b in Y such tht then X mon Y hs chosen prtil limits. ( < b. x Y y ) lim <b x Y lim <b y 4

5 Proof. Let b ω 2 nd define the chosen limits function s follows for coherent fmily (f ) <b, lim f = λx X. lim f (x) <b <b We need to show tht lim <b f is monotone. Thus, ssume x 1 X x 2. Then by monotonicity of f it follows tht f (x 1 ) Y f (x 2 ) for ll < b. Thus, by ssumption lim <b f (x 1 ) Y lim <b f (x 2 ), s required. The proof tht lim <b f is well-defined, limit nd sufficiently unique is the sme s in Lemm 8. Lemm 10. Let X nd Y be prtilly ordered o.f.e.s. If Y hs limits for ll ω 2 -indexed coherent fmilies such tht for ny two coherent fmilies (x ) ω 2 nd (y ) ω 2 in Y such tht (. x Y y ) lim x Y lim y then X mon Y hs limits for ll ω 2 -indexed coherent fmilies. Definition 10. Let X be set. Then (X) = (X, ( =) ω 2) where 0,0 = is the totl reltion nd n,m = is the identity for (n, m) (0, 0). Lemm 11. (X) is c.o.f.e. for ny non-empty set X. Proof. The proof tht (X) is n o.f.e. is trivil. To show tht (X) hs chosen prtil limits, let (x ) <b be coherent fmily in (X) nd pick n element x X. Now, define { lim x x if b = (0, 0) or b = (0, 1) = <b x (0,1) if b > (0, 1) If b = (0, 0) this is vcuously limit s b for ll. If b = (0, 1) this is lso vcuously limit s 0,0 = is the totl reltion. Lstly, if b > (0, 1) this is vcuously limit s x (0,1) = x for ll (0, 1) < b by the coherence property. Furthermore, for ny two coherent fmilies (x ) <b nd (y ) <b such tht x = y for ll < b we clerly hve tht lim <b x = lim <b y. Finlly, (X) obviously hs limits for ll ω 2 -indexed coherent fmilies, by tking the limit of (x ) ω 2 to be x (0,1). Definition 11 (Loclly non-expnsive nd loclly contrctive functor). A bi-functor F : U op U U is loclly non-expnsive iff nd loclly contrctive iff X, X, Y, Y obj(u). f, f : Hom U (X, X ). g, g : Hom U (Y, Y). ω 2. f = f g = g F (f, g) = F (f, g ) X, X, Y, Y obj(u). f, f : Hom U (X, X ). g, g : Hom U (Y, Y). ω 2. ( b ω 2. b < f b = f g b = g ) F (f, g) = F (f, g ) Definition 12 ( ). Let : U U denote the following functor, ( ) ( ) X, ( =) def ω 2) = X, ( ) ω 2 (f) = def f where 0,0 is the totl reltion on X, n,m+1 is n,m = nd n+1,0 is defined s follows x 1 n+1,0 x 2 iff m N. x 1 n,m = x2 5

6 Lemm 12. If F : U op U U is loclly non-expnsive then F is loclly contrctive. Proof. Let ω 2, f, f Hom U (X, X ) nd g, g Hom U (Y, Y) such tht By locl non-expnsiveness of F it follows tht b ω 2. b < f b = f g b = g b ω 2. b < F (f, g) b = F (X,Y ) nef (X,Y) F (f, g ) To show tht F (f, g) = F (X,Y ) ne F (X,Y) F (f, g ) tke n x F (X, Y ). By definition of to show tht F (f, g)(x) = F (X,Y) F (f, g )(x) it suffices to prove tht F (f, g)(x) b = F (X,Y) F (f, g )(x) for ll b <. This follows esily from the ssumption. 1.2 Solving Recursive Domin Equtions in U In this section we give the explicit construction of solution of recursive domin equtions in the ctegory of cofes indexed over ω 2. We proceed by building fixed-point of loclly contrctive bifunctor F : (U op U) U. The construction will proceed in two stges. First, we will construct prtil limits tht re indistinguishble for logicl steps (this construction is nlogous to the construction of the solution in ω-indexed spces), nd fterwrds we will use these to construct the overll fixed-point of the functor. Lemm 13 (prtil limits). For ny spce S U, nturl number n nd two functions p S : F (S, S) S n,0 nd e S : S F (S, S) such tht p S e S =id S nd e S p S = id F (S,S), there exists spce X U nd mps n,m p X : F (X, X) X nd e X : X F (X, X) such tht p X e X =id X nd e X p X = idf (X,X) for ny m. n,0 Additionlly, there re mps π X : X S nd ι X : S X such tht π X ι X =id S nd ι X π X = id X, nd tht π X p X = p S F (ι X, π X ). Proof. The proof follows the outline of the construction of the solution for the ω-indexed cofes. We begin by constructing F i, together with projections p i : F i+1 F i nd embeddings e i : F i F i+1 s: F 0 =S p 0 =p S e 0 =e S F k+1 =F (F k, F k ) p k+1 =F (e k, p k ) e k+1 =F (p k, e k ) First, we clim tht the properties of p S nd e S extend to the whole sequence: k. p k e k = id Fk (1) k. e k p k n,k = id Fk+1 (2) We prove these properties by induction. inductive step we hve: For (1), the bse cse holds by our ssumption, while for the p k+1 e k+1 = F (e k, p k ) F (p k, e k ) = F (p k e k, p k e k ) IH = F (id Fk, id Fk ) =id Fk+1. Similrly, for (2) the bse cse we lredy ssumed. For the inductive step, we proceed similrly: e k+1 p k+1 = F (p k, e k ) F (e k, p k ) = F (e k p k, e k p k ) n,k+1 = F (id Fk+1, id Fk+1 ) =id Fk+2, where the crucil (n, k + 1)-equlity holds by induction hypthesis (for (n, k) on the rguments) nd locl contrctiveness of F. 6

7 We cn define helpful, iterted versions of projections nd embeddings, written p l k : F k+l F k nd e l k : F k F k+l, s p 0 k =id Fk We immeditely get the following observtions: Now we re redy to define X. Let { e 0 k =id Fk p l+1 k = p l k p k+l e l+1 k = e k+l e l k. X = k, l. p l k e l k = id Fk (3) k, l. e l k p l k n,k = id Fk+l. (4) x k N F k k. xk = p k (x k+1 ) }, with equlity defined pointwise, i.e., x = p X x p iff k. x k =Fk x k, for p ω2. Clerly, X is nd object of U. We cn now extend projections nd embeddings to X: we define π k : X F k nd ι k : F k X s { e i k π k (x) = x k (ι k (s)) i = k (s) if i k p k i i (s) if i k Agin we cn use (3) to show tht both these mps re well-defined. Now we re finlly redy to define the mps p X nd e X. We tke (p X (z)) k = p k (F (ι k, π k )(z)) e X (x) = lim k N F (π k, ι k )(e k x k ) First, we need to check tht these mps re well-defined. For p X, this mounts to checking tht p X (z) X, since it is clerly non-expnsive. Thus, we hve p k ((p X (z)) k+1 ) = p k (p k+1 (F (ι k+1, π k+1 )(z))) = p k (F (e k, p k )F (ι k+1, π k+1 )(z)) = p k (F (ι k+1 e k, p k π k+1 )(z)) = p k (F (ι k, π k )(z)) = (p X (z)) k, where the identities ι k+1 e k = ι k nd p k π k+1 = p k re esy to check. For e X, we need to check tht the limit ctully exists. Since we re working in U, it is enough to show tht the chin is Cuchy: we proceed by showing tht F (π k, ι k )(e k x k ) n,k+1 = F (π k+1, ι k+1 )(e k+1 x k+1 ), with the Cuchy condition (up to n) following by simple induction. We hve: F (π k+1, ι k+1 )(e k+1 x k+1 ) n,k+1 = F (e k π k, ι k p k )(e k+1 x k+1 ) = F (π k, ι k )(F (e k, p k )(e k+1 x k+1 )) = F (π k, ι k )(p k+1 (e k+1 x k+1 )) = F (π k, ι k )(x k+1 ) n,k+1 = F (π k, ι k )(e k (p k (x k+1 ))) = F (π k, ι k )(e k x k ). n,k The (n, k + 1)-equlities follow ultimtely (esy check) from e k p k = id, (2) nd contrctiveness of F. Like with p X, it is n esy check tht e X is non-expnsive. n,m Finlly, we cn prove the two required properties: p X e X =id X, nd m. e X p X = idf (X,X). For the first one, we compute s follows: p X (e X (x)) k = p k (F (ι k, π k )( lim m N F (π m, ι m )(e m x m ))) = p k lim m N F (π m+k ι k, π k ι m+k )(e m+k x m+k ), where we only considered the til of the chin pst its k-th element in the second eqution, nd used the fct tht nonexpnsive mps preserve limits. Since, s it is esy to check, π k+m ι k = e m k, π k ιm + k = p m k nd F (e m k, pm k ) = pm k+1, we get: p k lim m N F (π m+k ι k, π k ι m+k )(e m+k x m+k ) = p k lim m N F (e m k, p m k )(e k+m x k+m ) = p k lim m N p m k+1(e k+m x k+m ) = p k lim m N x k+1 = p k (x k+1 ) = x k 7

8 For the other direction, we wnt to show tht for ny m nd z, e X (p X (z)) n,m = z. Unfolding the definitions, we get e X (p X (z)) = lim k N (F (π k, ι k ) e k p k F (ι k, π k ))(z) It suffices to show tht k-th element of this chin is (n, k)-equl to z: then by the similrity of chins their limits will be (n, m)-equl for ny m. Thus, we hve F (π k, ι k ) e k p k F (ι k, π k ) n,k = F (π k, ι k ) F (ι k, π k ) = F (ι k π k, ι k π k ) n,k = F (id X, id X ) =id X The first of the (n, k)-equl steps follows by (2), while the second follows from ι k π k n,k = id X, which in turn depends on (4). Finlly, we tke π X = π 0 nd ι X = ι 0 ; the properties follow trivilly from the definition nd (4), while the finl property is exctly the definition of p X t index 0. Theorem 1 (solutions of recursive domin equtions). For ny loclly contrctive functor F : (U U) U such tht F (1, 1) is inhbited (by F (1,1) ) there exists spce X : U such tht X = F (X, X). Proof. The ide behind the construction is to iterte the construction from Lemm 13 to construct tower of progressively closer pproximtions of the solution, nd then to construct limit of it. To this end, we write G(S, p S, e S ) to denote the construction of the lemm, i.e., quintuple (X, p X, e X, π X, ι X ) whose existence the lemm shows. Next, we proceed with the construction of sequence of spces (X i : U) i N together with mps π i : X i+1 X i, ι i : X i X i+1, p i : F (X i, X i ) X i nd e i : X i F (X i, X i ). First, let (X 0, p 0, e 0,, ) = G(1,! F (1,1), 1 F (1,1) ), where! F (1,1) : F (1, 1) 1 is the unique mp into the unit type. Next, we proceed by induction, tking: (X n+1, p n+1, e n+1, π n, ι n) = G(F (X n, X n ), F (e n, p n ), F (p n, e n )), nd setting π n = p n πn, ι n = ι n e n. In order for this definition to be vlid, we hve to check severl properties. Firstly, for the first ppliction 0,0 of G, we need! F (1,1) ( 1 F (1,1) ) =id 1, which holds trivilly, nd ( 1 F (1,1) )! F (1,1) = id F (1,1), which lso holds trivilly, since ny two objects re (0, 0)-equl. Next, we need to check tht these conditions lso hold for the inductive step. We hve F (e n, p n ) F (p n, e n ) = F (p n e n, p n e n ) = F (id Xn, id Xn ) =id F (Xn,X n), where we use the fct tht p n e n = id Xn, which comes from Lemm 13. Similrly, we hve F (p n, e n ) F (e n, p n ) = F (e n p n, e n p n ) n+1,0 = F (id F (Xn,X n), id F (Xn,X n)) =id F (F (Xn,X n),f (X n,x n)). Here, the crucil (n+1, 0)-equivlence comes from locl contrctiveness of F : it mens we only need to show n,m tht e n p n = idf (Xn,X n) for ny m, which is precisely wht Lemm 13 gives us. Finlly, we lso check tht π n : X n+1 X n nd ι n : X n X n+1. The construction gives us πn : X n+1 F (X n, X n ), so the composition p n πn indeed hs the right type; similrly for ι n. We clim tht n. π n ι n = id Xn (5) n. ι n π n n,0 = id Xn+1, (6) which is esily checked by unfolding the definitions nd using properties of p n, e n, π n nd ι n. Furthermore, we prove n dditionl clims, nd its two simple corrolries n. F (ι n, π n ) = π n p n+1 (7) n. π n p n+1 = p n F (ι n, π n ) (8) n. F (ι n, π n ) e n+1 = π n (9) 8

9 The first property, which is resttement of the finl property of Lemm 13 is proved s follows: F (ι n, π n ) = F (ι n e n, p n π n) = F (e n, p n ) F (ι n, π n) = π n p n+1, where the lst equlity is the direct ppliction of the property from the lemm with the definitions used in the inductive step of the construction. The two corollries follow using definitions nd, in the second cse, (5). As for the previous construction, we define iterted versions of π nd ι: π 0 n = id Xn ι 0 n = id Xn π k+1 n = π k n π n+k ι k+1 n = ι n+k ι k n, nd prove, by simple induction, the iterted versions of properties: n, k. π k n ι k n = id Xn (10) n, k. ι k n π k n n,0 = id Xn+k (11) Now we cn construct wht we clim to be the fixed-point of the recursive domin eqution. We tke X U s: { X = x } X n k. xk = π k (x k+1 ), n N gin, with equlity defined pointwise. Similrly to the previous construction, we extend the projections nd embeddings to X: πk X : X X k nd ι X k : X k X s { πk X (x) = x k (ι X ι i k k (s)) i = k (s) if i k π k i k (s) if i k Finlly, we cn define p X : F (X, X) X nd e X : X F (X, X). We tke (p X (z)) k = p k (F (ι X k, π X k )(z)) e X (x) = lim k N F (π X k, ι X k )(e k (x k )) we define First, we check tht these mps re well-defined. For p X, we only need to show tht p X (z) X. We compute π k ((p X (z)) k+1 ) = π k (p k+1 (F (ι X k+1, π X k+1)(z))) = p k (F (ι k, π k )(F (ι X k+1, π X k+1)(z))) = p k (F (ι X k+1 ι k, π k π X k+1)) = p k (F (ι X k, π X k )(z)) = (p X (z)) k. In the computtion bove, the identities ι X k = ιx k+1 ι k nd π k πk+1 X = πx k re esy to check, nd the remining one needed is (8). In order for e X to be well-defined, we need to check tht the limit exists. To this end, it is enough to show tht the sequence is Cuchy: we re going to show tht F (πk X, ιx k )(e k(x k )) k,0 = F (πk+1 X, ιx k+1 )(e k+1(x k+1 )). We hve: F (π X k+1, ι X k+1)(e k+1 (x k+1 )) k,0 = F (ι k π X k, ι X k π k )(e k+1 (x k+1 )) = F (π X k, ι X k )(F (ι k, π k )(e k+1 (x k+1 ))), where the (k, 0)-equlity follows from contrctiveness nd (6). By (9) we hve F (ι k, π k ) e k+1 = πk, nd so F (π X k, ι X k )(F (ι k, π k )(e k+1 (x k+1 ))) = F (π X k, ι X k )(π k(x k+1 )) k,0 = F (π X k, ι X k )((e k p k π k)(x k+1 )) = F (π X k, ι X k )(e k (x k )). We now turn to showing tht p X nd e X form n isomorphism. First, we clim tht p X (e X (x)) = x. To show this, we pick n index k nd compute: p X (e X (x)) k = p k (F (ι X k, π X k ) lim n N F (π X n, ι X n )(e n (x n ))) = lim n N (p k F (ι X k, π X k ) F (π X k+n, ι X k+n) e k+n )(x k+n ) = lim n N (p k F (π X k+n ι X k, π X k ι X k+n) e k+n )(x k+n ) = lim n N (p k F (ι n k, π n k ) e n+k )(x n+k ) = lim n N (πn k p n+k e n+k )(x n+k ) = lim πk n (x n+k ) = lim x k = x k n N 9

10 The identities πn+k X ιx k = ιn k nd πx k ιx n+k = πn k re esy to check nd nlogous to the cse in Lemm 13. The remining identity p k F (ι n k, πn k ) = πn k p k+n we prove by induction on n s follows. The bse cse holds trivilly, since ι 0 k = π0 k =id X k. For the inductive cse, we hve p k F (ι n+1 k, π n+1 k ) = p k F (ι n k ι k+n, π k+n π n k ) = p k F (ι n k, π n k ) F (ι k+n, π k+n ) = π n k p k+n F (ι k+n, π k+n ) = π n k π k+n p k+n+1 = π n+1 k p k+n+1, where the second-to-lst equlity follows by (8). We re left with the finl obligtion, showing tht e X (p X (z)) = z. To this end we show tht e X (p X (z)) = lim n N (F (π X k, ι X k ) e k p k F (ι X k, π X k ))(z) = lim n N z = z. We show tht the two chins under the limit pproximte ech other, t progressively greter, unbounded ordinls, nd so the limits re equl. Precisely, we show tht k-th elements of the chin re (k, 0) equl: F (π X k, ι X k ) e k p k F (ι X k, π X k ) k,0 = F (π X k, ι X k ) F (ι X k, π X k ) = F (ι X k π X k, ι X k π X k ) k,0 = F (id X, id X ) =id F (X,X). Similrly to the first construction, the first (k, 0)-equlity follows from (6), nd the second from (11). 10

11 2 Syntx nd opertionl semntics τ, σ ::= 1 τ σ τ σ τ ref α. τ α ::=, α ε Γ ::= Γ, x : τ ε v Vl ::= fix f(x). e l pck v (v 1, v 2 ) e Exp ::= v e 1 e 2 (e 1, e 2 ) fst e snd e! e e 1 := e 2 ref e pck e unpck e 1 s x in e 2 K ECtx ::= K e v K (K, e) (v, K) fst K snd K! K K := e v := K ref K pck K unpck K s x in e Typing rules ; Γ e : τ τ σ τ σ τ, α τ 1 τ σ τ σ τ ref α. τ, α α τ ; Γ, x : τ x : τ ; Γ : 1 ; Γ, f : τ σ, x : τ e : σ ; Γ fix f(x). e : τ σ ; Γ e 1 : τ σ ; Γ e 2 : τ ; Γ e 1 e 2 : σ ; Γ e : τ ref ; Γ e 1 : τ ref ; Γ e 2 : τ ; Γ e : τ ; Γ e : σ[τ/α] ; Γ! e : τ ; Γ e 1 := e 2 : 1 ; Γ ref e : τ ref ; Γ pck e : α. σ ; Γ e 1 : α. τ, α; Γ, x : τ e 2 : σ σ ; Γ unpck e 1 s x in e 2 : σ ; Γ e 1 : τ ; Γ e 2 : σ ; Γ (e 1, e 2 ) : τ σ ; Γ e : τ σ ; Γ fst e : τ ; Γ e : τ σ ; Γ snd e : σ Opertionl semntics e, h e, h EvlFix (fix f(x). e) v, h e[fix f(x). e/f, v/x], h EvlUnpck unpck (pck v) s x in e, h e[v/x], h EvlRed l dom(h)! l, h h(l), h EvlFst fst (v 1, v 2 ), h v 1, h EvlWrite l dom(h) l := v, h, h[l v] EvlSnd snd (v 1, v 2 ), h v 2, h EvlAlloc l dom(h) ref v, h l, h[l v] EvlCtx e, h e, h K[e], h K[e ], h nd e, h 0 e, h e, h e, h e, h n e, h e, h n+1 e, h 11

12 Lemm 14. If K[e], h i e, h then there exists i 1, i 2, e nd h such tht Proof. By induction on i. e, h i1 e, h K[e ], h i2 e, h i = i 1 + i 2 If i = 0 then tke i 1 = i 2 = 0, e = e nd h = h. Otherwise, K[e], h e, h nd e, h i 1 e, h. If K = then we simply tke i 1 = i, i 2 = 0, e = e nd h = h. Otherwise, we proceed by cse nlysis on the K[e], h e, h derivtion: Cse EvlCtx: then e = K[e ] nd e, h e, h. Thus, by the induction hypothesis, there exists i 1, i 2, e nd h such tht e, h i1 e, h, K[e ], h i2 e, h nd i 1 = i 1 + i 2. We thus hve tht, K[e], h i1+1 K[e ], h nd K[e ], h i2 e, h. Cse EvlFix, EvlRed, EvlWrite, EvlAlloc, EvlUnpck, EvlFst, EvlSnd: then e Vl. Thus, e, h 0 e, h nd K[e], h i e, h. 12

13 3 Logicl reltion Assume c.o.f.e. Inv over ω 2 nd isomorphism Define Type, World nd Inv s follows ξ : Inv = ((N fin Inv) mon UPred(Hep Hep))) World = def N fin Inv Type = def World mon UPred(Vl Vl) def Inv = World mon UPred(Hep Hep) Vlue reltion V [ τ ] : Type Type V [ 1] ρ (W ) = def {(n, m,, )} V [ τ σ ] ρ (W ) = def {(n, m, v I, v S ) v 1I, v 2I, v 1S, v 2S. v I = (v 1I, v 2I ) v S = (v 1S, v 2S ) (n, m, v 1I, v 1S ) V [ τ ] ρ (W ) (n, m, v 2I, v 2S ) V [ σ ] ρ (W )} V [ τ σ ] ρ (W ) = def {(n,, v I, v S ) n < n. W W. ( m. (n, m, u I, u S ) V [ τ ] ρ (W )) (n + 1, v I u I, v S u S ) E(V [ σ ] ρ )(W )} V [ τ ref] ρ (W ) = def {(n, m, l I, l S ) ι dom(w ). ξ(w (ι)) n,m = inv(v [ τ ] Înv ρ, l I, l S )} V [, α α] ρ (W ) = def ρ(α)(w ) V [ α. τ ] ρ (W ) = def {(n, m, pck v I, pck v S ) ν Type. (n, m, v I, v S ) V [, α τ ] ρ[α ν] (W )} Reference invrint inv : Type L L World mon UPred(Hep Hep) inv(ν, l I, l S ) = def λw. {(n, m, h I, h S ) l I dom(h I ) l S dom(h S ) (n, m, h I (l I ), h S (l S )) ν(w )} World stisfction W = def {(n, h I, h S ) n = 0 r I, r S : dom(w ) Hep. h I = Π ι r I (ι) h S = Π ι r S (ι) Expression closure ι dom(w ). m. (n 1, m, r I (ι), r S (ι)) ξ(w (ι))(w )} E(ν) = def λw. {(n, e I, e S ) n n. i < n. h I, h S Hep. W W. (n, h I, h S ) W e I, h I i e I, h I v S, h S. W W. e S, h S v S, h S e I Vl (n i, h I, h S) W m. (n i, m, e I, v S ) ν(w )} Context reltion V [ Γ] : Type World mon UPred(Vl Γ Vl Γ ) V [ Γ] ρ (W ) = def {(n, m, σ I, σ S ) (x : τ) Γ. (n, m, σ I (x), σ S (x)) V [ τ ] ρ (W )} 13

14 Logicl reltion ; Γ = e I log e S : τ ; Γ = e I log e S : τ = def n N. W World. σ I, σ S Vl Γ. ρ Type. ( m. (n, m, σ I, σ S ) V [ Γ] ρ (W )) (n, σ I (e I ), σ S (e S )) E(V [ τ ] ρ )(W ) Lemm 15. W 1, W 2 World. h I, h S Hep. n, m N. W 1 n,m = World W 2 (n, h I, h S ) W 1 (n, h I, h S ) W 2 Proof. Since the conclusion is trivil for n = 0, ssume n > 0. By ssumption, there exists r I, r S : dom(w 1 ) Hep such tht h I = Π ι dom(w1)r I (ι) h S = Π ι dom(w1)r S (ι) nd ι dom(w 1 ). m. (n 1, m, r I (ι), r S (ι)) ξ(w 1 (ι))(w 1 ). n,m Since W 1 = W2 it follows tht dom(w 1 ) = dom(w 2 ). Let ι dom(w 2 ) nd m N. Thus, by ssumption, (n 1, m n,m, r I (ι), r S (ι)) ξ(w 1 (ι))(w 1 ). Since W 1 = W2 it follows tht ξ(w 1 (ι)) n,m = Înv ξ(w 2(ι)) nd thus ξ(w 1 (ι))(w 1 ) n 1,m +1 = ξ(w 2 (ι))(w 2 ). Hence, (n 1, m + 1, r I (ι), r S (ι)) ξ(w 2 (ι))(w 2 ). Lemm 16. Proof. Assume W 1, W 2 World. ν Type. n, m N. e I, e S Exp. W 1 n,m = World W 2 (n, e I, e S ) E(ν)(W 1 ) (n, e I, e S ) E(ν)(W 2 ) i < k n W 2 W 2 (k, h I, h S ) W 2 e I, h I i e I, h I Then there exists W 1 such tht W 1 W 1 nd W 1 There thus exists v S, h S, nd W 1 W 1 such tht n,m = W 2. Hence, by Lemm 15, (k, h I, h S ) W 1. e I Vl e S, h S v S, h S (k i, h I, h S) W 1 m. (k i, m, e I, v S ) ν(w 1 ) Hence, there exists W 2 such tht W 2 W 2 nd W 1 n,m = W 2 from which it follows tht (k i, h I, h S) W 2 m. (k i, m, e I, v S ) ν(w 2 ) by Lemm 15 nd non-expnsiveness. Lemm 17. ν Type. l I, l S Loc. inv(ν, l I, l S ) World mon UPred(Hep Hep) Lemm 18. The vlue reltion is well-defined. In prticulr, V [ τ ] ρ is non-expnsive for ll ρ Type, 14

15 V [ τ ] ρ is monotone for ll ρ Type, V [ τ ] ρ (W ) is downwrds-closed for ll ρ Type nd W World. Proof. By induction on the τ derivtion. Cse τ = 1: trivil. Cse τ = σ ref: to show tht the vlue reltion is non-expnsive, ssume W 1 n,m = W2, (n, m ) < (n, m), nd (n, m, l I, l S ) V [ τ ] ρ (W 1 ). Hence, there n ι dom(w 1 ) such tht ξ(w 1 (ι)) n,m = Înv inv(v [ σ ] ρ, l I, l S ) Since ξ is non-expnsive, ξ(w 1 (ι)) n,m = Înv ξ(w 2(ι)). By trnsivity it follows tht ξ(w 2 (ι)) n,m = Înv inv(v [ σ ] ρ, l I, l S ) nd thus, (n, m, l I, l S ) V [ τ ] ρ (W 2 ). The vlue reltion is esily seen to be monotone in worlds nd downwrds-closed in the step-index. Cse τ = σ 1 σ 2 : to show tht the vlue reltion is non-expnsive, ssume W 1 n,m = W2, (n, m ) < (n, m), nd (n, m, v I, v S ) V [ τ ] ρ (W 1 ). Hence, k < n. W 1 W 1. ( m. (k, m, u I, u S ) V [ σ 1 ] ρ (W 1)) (k + 1, v I u I, v S u S ) E(V [ σ 2 ] ρ )(W 1) To show tht (n, m, v I, v S ) V [ τ ] ρ (W 2 ), let k < n, W 2 W 2 such tht m. (k, m, u I, u S ) V [ σ 1 ] ρ (W 2). Hence, there exists W 1 such tht W 1 W 1 nd W 1 n,m = W 2. Hence, m. (k, m, u I, u S ) V [ σ 1 ] ρ (W 1), since V [ σ 1 ] ρ is non-expnsive by the induction hypothesis. It thus follows tht (k + 1, v I u I, v S u S ) E(V [ σ 2 ] ρ )(W 1) nd, by Lemm 16, (k + 1, v I u I, v S u S ) E(V [ σ 2 ] ρ )(W 2). The vlue reltion is clerly monotone nd downwrds-closed in the step-index. Cse τ = α. σ: follows esily from the induction hypothesis. Cse τ = α: follows from the type of ρ. Lemm 19. Proof. Assume n N. v I, v S Vl. W World. ν Type. ( m. (n, m, v I, v S ) ν(w )) (n, v I, v S ) E(ν)(W ) 0 < n n i < n W W (n, h I, h S ) W v I, h I i e I, h I Since v I is vlue it follows tht i = 0, e I = v I, nd h I = h I. Thus, we pick W = W, v S = v S nd h S = h S. By downwrds-closure it follows tht s required. (n i, h I, h S ) W m. (n i, m, e I, v S ) ν(w ) Lemm 20. ; Γ, x : τ = x x : τ. 15

16 Proof. Assume m. (n, m, σ I, σ S ) V [ Γ, x : τ ] ρ (W ) Thus, m. (n, m, σ I (x), σ S (x)) V [ τ ] ρ (W ) nd by Lemm 19, (n, σ I (x), σ S (x)) E(V [ τ ] ρ )(W ) s required. 3.1 Comptibility lemms Lemm 21. If ; Γ, f : τ σ, x : τ = e I e S : σ then ; Γ = fix f(x). e I fix f(x). e S : τ σ. Proof. We prove by induction on n tht n N. W World. ρ Type. σ I, σ S Vl Γ. ( m. (n, m, σ I, σ S ) V [ Γ] ρ (W )) ( m. (n, m, σ I (fix f(x). e I ), σ S (fix f(x). e S )) V [ τ σ ] ρ (W )) from which the conclusion follows esily, by Lemm 19, s fix f(x). e I nd fix f(x). e S re vlues. The bse cse follows trivilly, s (n < 0) for ll n. For the inductive cse, ssume m. (n + 1, m, σ I, σ S ) V [ Γ] ρ (W ) To show tht m. (n + 1, m, σ I (fix f(x). e I ), σ S (fix f(x). e S )) V [ τ σ ] ρ (W ) let n < n + 1, W W such tht m. (n, m, u I, u S ) V [ τ ] ρ (W ). Then it remins to show tht (n + 1, σ I (fix f(x). e I ) u I, σ S (fix f(x). e S ) u S ) E(V [ σ ] ρ )(W ) Assume i < n n W W (n, h I, h S ) W σ I (fix f(x). e I ) u I, h I i e I, h I Hence, i > 0 nd σ I (fix f(x). e I ) u I, h I σ I (e I )[u I /x, σ I (fix f(x). e I )/f], h I i 1 e I, h I. Since n n, by downwrds closure nd monotonicity, it follows tht m. (n, m, σ I, σ S ) V [ Γ] ρ (W ) Thus, by the induction hypothesis, m. (n, m, σ I (fix f(x). e I ), σ S (fix f(x). e S )) V [ τ σ ] ρ (W ) We thus hve tht m. (n, m, σ I, σ S) V [ Γ, f : τ σ, x : τ ] ρ (W ) for σ I = σ I[x u I, f fix f(x). e I ] nd σ S = σ S[x u S, f fix f(x). e S ]. From the ; Γ, f : τ σ, x : τ = e I e S : σ ssumption it thus follows tht (n, σ I(e I ), σ S(e S )) E(V [ σ ] ρ )(W ) The rest is esy. 16

17 Lemm 22. If ; Γ = e 1I e 1S τ σ nd ; Γ = e 2I e 2S : τ, then ; Γ = e 1I e 2I e 1S e 2S : σ. Proof. Assume m. (n, m, σ I, σ S ) V [ Γ] ρ (W ) nd i < n n W W (n, h I, h S ) W σ I (e 1I e 2I ), h I i e I, h I Hence, by Lemm 14, there exists i 1, i 2, h 1I, nd e 1I such tht nd i = i 1 + i 2. σ I (e 1I ), h I i1 e 1I, h 1I e 1I σ I (e 2I ), h 1I i2 e I, h I From the ; Γ = e 1I e 1S τ σ ssumption there exists W W, v 1S nd h 1S such tht σ S (e 1S ), h S v 1S, h 1S (n i 1, h 1I, h 1S ) W m. (n i 1, m, e 1I, v 1S ) V [ τ σ ] ρ (W ) nd e 1I Vl. Thus, by Lemm 14, there exists i 3, i 4, h 2I nd e 2I such tht nd i 2 = i 3 + i 4. σ I (e 2I ), h 1I i3 e 2I, h 2I e 1I e 2I, h 2I i4 e I, h I From the ; Γ = e 2I e 2S : τ ssumption there exists W W, v 2S, nd h 2S such tht σ S (e 2S ), h 1S v 2S, h 2S (n i 1 i 3, h 2I, h 2S ) W m. (n i 1 i 3, m, e 2I, v 2S ) V [τ ](W ) nd e 2I Vl. By (n i 1, 0, e 1I, v 1S) V [ τ σ ] ρ (W ) it follows tht (n i 1 i 3, e 1I e 2I, v 1S v 2S ) E(V [ σ ] ρ )(W ) Hence, there exists W W, v S nd h S such tht v 1S v 2S, h 2S v S, h S (n i, h I, h S) W m. (n i, m, e I, v S ) V [ σ ] ρ (W ) nd e I Vl. Lstly, σ S(e 1S e 2S ), h S v S, h S. Lemm 23. l I, l S Loc. h I, h S Hep. ν Type. ι N. W World. n, m N. inv(ν, l I, l S ) n,m = Înv ξ(w (ι)) (n, h I, h S ) W n > 0 l I dom(h I ) l S dom(h S ) m N. (n 1, m, h I (l I ), h S (l S )) ν(w ) Proof. By definition of hep stisfiction, there exists r I, r S : dom(w ) Hep such tht h I = Π ι dom(w ) r I (ι) h S = Π ι dom(w ) r S (ι) m. (n 1, m, r I (ι), r S (ι)) ξ(w (ι))(w ) Since inv(ν, l I, l S ) n,m = ξ(w (ι)) it follows tht inv(ν, l Înv I, l S )(W ) n 1,1 = Înv ξ(w (ι))(w ) nd thus Hence, l I dom(r I (ι)), l S dom(r S (ι)). (n 1, 0, r I (ι), r S (ι)) inv(ν, l I, l S )(W ) To show tht m. (n 1, m, r I (ι)(l I ), r S (ι)(l S )) ν(w ), ssume m N. Since (n 1, m + 1) < (n, m) it follows tht inv(ν, l I, l S )(W ) n 1,m +1 = Înv ξ(w (ι))(w ) nd thus Thus, (n 1, m, r I (ι)(l I ), r S (ι)(l S )) ν(w ). (n 1, m, r I (ι), r S (ι)) inv(ν, l I, l S )(W ) 17

18 Lemm 24. If ; Γ = e I e S : τ ref then ; Γ =! e I! e S : τ. Proof. Assume m. (n, m, σ I, σ S ) V [ Γ] ρ (W ) nd i < n n W W (n, h I, h S ) W σ I (! e I ), h I i e I, h I Hence, by Lemm 14, there exists i 1, i 2, v 1I nd h 1I such tht nd i = i 1 + i 2. σ I (e I ), h I i1 e 1I, h 1I! e 1I, h 1I i2 e I, h I From the ; Γ = e I e S : τ ref ssumption there exists W W, v 1S nd h 1S such tht σ S (e 1S ), h S v 1S, h 1S (n i 1, h 1I, h 1S ) W m. (n i 1, m, e 1I, v 1S ) V [ τ ref] ρ (W ) nd e 1I Vl. By the vlue reltion there exists n ι dom(w ) such tht ξ(w (ι)) n i 1,0 = Înv inv(v [ τ ] ρ, e 1I, v 1S ) Since n i 1 > 0 it follows by Lemm 23 tht e 1I dom(h 1I ), v 1S dom(h 1S ) nd m. (n i 1 1, m, h 1I (e 1I ), h 1S (v 1S )) V [Γ τ ] ρ (W ) Lstly, since e 1I dom(h 1I ) it follows tht h I = h 1I, v I = h 1I (e 1I ) nd i 2 = 1. Lemm 25. l I, l S Loc. h I, h S Hep. ν Type. ι N. W World. n, m N. inv(ν, l I, l S ) n,m = Înv ξ(w (ι)) (n, h I, h S ) W m. (n 1, m, v I, v S ) ν(w ) (n, h I [l I v I ], h S [l S v S ]) W Proof. As the conclusion is trivil if n < 1, ssume n 1. By definition of hep stisfiction, there exists r I, r S : dom(w ) Hep such tht h I = Π ι dom(w ) r I (ι) h S = Π ι dom(w ) r S (ι) m. (n 1, m, r I (ι), r S (ι)) ξ(w (ι))(w ) Since ξ(w (ι))(w ) n 1,1 = inv(ν, l I, l S ) it follows tht nd thus l I dom(r I (ι)) nd l S dom(r S (ι)). (n 1, 0, r I (ι), r S (ι)) inv(ν, l I, l S ) Let r I = r I[ι r I (ι)[l I v I ]] nd r S = r S[ι r S (ι)[l S v S ]]. Then, h I [l I v I ] = Π ι dom(r I )r I(ι) h S [l S v S ] = Π ι dom(r S )r S(ι) It thus remins to show tht x dom(w ). m. (n 1, m, r I(x), r S(x)) ξ(w (x))(w ) This holds trivilly for ll x ι. For x = ι, let m N. Then by ssumption (n 1, m, v I, v S ) ν(w ) nd inv(ν, l I, l S )(W ) n 1,m +1 = ξ(w (ι))(w ) Thus, (n 1, m, r I (x), r S (x)) inv(ν, l I, l S )(W ) nd (n 1, m, r I (x), r S (x)) ξ(w (ι))(w ). 18

19 Lemm 26. If ; Γ = e 1I e 1S : τ ref nd ; Γ = e 2I e 2S : τ then ; Γ = e 1I := e 2I e 1S := e 2S : 1. Proof. Assume m. (n, m, σ I, σ S ) V [ Γ] ρ (W ) nd i < n n W W (n, h I, h S ) W σ I (e 1I := e 2I ), h I i e I, h I Hence, by Lemm 14, there exists i 1, i 2, e 1I, nd h 1I such tht nd i = i 1 + i 2. σ I (e 1I ), h I i1 e 1I, h 1I e 1I := e 2I, h 1I i2 e I, h I Since ; Γ = e 1I e 1S : τ ref, it follows tht there exists W W, v 1S, nd h 1S such tht σ S (e 1S ), h S v 1S, h 1S (n i 1, h 1I, h 1S ) W m. (n i 1, m, e 1I, v 1S ) V [ τ ref] ρ (W ) nd e 1I Vl. By Lemm 23 it thus follows tht e 1I dom(h 1I) nd v 1S dom(h 1S ). By Lemm 14 there exists i 3, i 4, e 2I, nd h 2I such tht nd i 2 = i 3 + i 4. σ I (e 2I ), h 1I i3 e 2I, h 2I e 1I := e 2I, h 2I i4 e I, h I Since ; Γ = e 2I e 2S : τ, it follows tht there exists W W, v 2S, nd h 2S such tht σ S (e 2S ), h 1S v 2S, h 2S (n i 1 i 3, h 2I, h 2S ) W m. (n i 1 i 3, m, e 2I, v 2S ) V [ τ ] ρ (W ) nd e 2I Vl. Thus e 1I dom(h 1I) dom(h 2I ) nd v 1S dom(h 1S ) dom(h 2S ). Hence, by the evlution of e 1I := e 2I, h I = h 2I[e 1I e 2I ], v I =, nd v 1S := v 2S, h 2S, h 2S [v 1S v 2S ] Lstly, by Lemm 25, we hve tht (n i 1 i 3, h 2I [e 1I e 2I ], h 2S[v 1S v 2S ]) W. Lemm 27. l I, l S Loc. h I, h S Hep. ν Type. ι N. W World. n N. (n, h I, h S ) W ι dom(w ) l I dom(h I ) l S dom(h S ) m. (n 1, m, v I, v S ) ν(w ) (n, h I [l I v I ], h S [l S v S ]) W [ι ξ 1 (inv(ν, l I, l S ))] Proof. Since the conclusion is trivil for n = 0, ssume n > 0. By definition of hep stisfiction, there exists r I, r S : dom(w ) Hep such tht h I = Π ι dom(w ) r I (ι) h S = Π ι dom(w ) r S (ι) m. (n 1, m, r I (ι), r S (ι)) ξ(w (ι))(w ) Let r I = r I[ι [l I v I ]] nd r S = r S[ι [l S v S ]]. Then, h I [l I v I ] = Π ι dom(r I )r I(ι) h S [l S v S ] = Π ι dom(r S )r S(ι) It thus remins to show tht x dom(w ). m. (n 1, m, r I(x), r S(x)) ξ(w (x))(w ) where W = W [ι ξ 1 (inv(ν, l I, l S )). This holds trivilly for ll x ι. For x = ι, the proof obligtion reduces to, m. (n 1, m, [l I v I ], [l S v S ]) inv(ν, l I, l S )(W ) This follows from the m. (n 1, m, v I, v S ) ν(w ) ssumption, by monotonicity. 19

20 Lemm 28. If ; Γ = e 1I e 1S : τ then ; Γ = ref e 1I ref e 1S : τ ref. Proof. Assume m. (n, m, σ I, σ S ) V [ Γ] ρ (W ) nd i < n n W W (n, h I, h S ) W σ I (ref e 1I ), h I i e I, h I By Lemm 14, there exists i 1, i 2, e 1I, nd h 1I such tht σ I (e 1I ), h I i1 e 1I, h 1I ref e 1I, h 1I i2 e I, h I nd i = i 1 + i 2. Since ; Γ = e 1I e 1S : τ it follows tht there exists W W, v 1S, nd h 1S such tht σ S (e 1S ), h S v 1S, h 1S (n i 1, h 1I, h 1S ) W m. (n i 1, m, e 1I, v 1S ) V [ τ ] ρ (W ) nd e 1I Vl. Thus, e I Loc, e I dom(h 1I), h I = h 1I[e I e 1I ] nd i 2 = 1. Pick v S Loc such tht v S dom(h 1S ) nd ι N such tht ι dom(w ). Let W = W [ι ξ 1 (inv(v [ τ ] ρ, e I, v S ))] By Lemm 27, (n i 1, h 1I [e I e 1I], h 1S [v S v 1S ]) W Lstly, m. (n i 1 1, m, e I, v S) V [ τ ref] ρ (W ) nd ref v 1S, h 1S v S, h 1S [v S v 1S ]. Lemm 29. V [, α τ ] ρ[α V [ σ ]ρ] = V [ τ[σ/α]] ρ Proof. By induction on the τ derivtion. Cse 1,, α α: trivil. Cse τ σ, τ σ, τ ref, α. τ: follows directly from the induction hypothesis. Lemm 30. If τ then ρ Type. ν Type. V [ τ ] ρ = V [, α τ ] ρ[α ν] Proof. By induction on the τ derivtion. Cse 1,, α α: trivil. Cse τ σ, τ σ, τ ref, α. τ: follows directly from the induction hypothesis. Lemm 31. If ; Γ e 1I e 1S : σ[τ/α] then ; Γ pck e 1I pck e 1S : α. σ. Proof. Assume m. (n, m, σ I, σ S ) V [ Γ] ρ (W ) nd i < n n W W (n, h I, h S ) W σ I (pck e 1I ), h I i e I, h I By Lemm 14, there exists i 1, i 2, e 1I, nd h 1I such tht σ I (e 1I ), h I i1 e 1I, h 1I pck e 1I, h 1I i2 e I, h I 20

21 nd i = i 1 + i 2. Since ; Γ = e 1I e 1S : σ[τ/α] it follows tht there exists W W, v 1S, nd h 1S such tht σ S (e 1S ), h S v 1S, h 1S (n i 1, h 1I, h 1S ) W m. (n i 1, m, e 1I, v 1S ) V [ σ[τ/α]] ρ (W ) nd e 1I Vl. Thus, i 2 = 0, e I = pck e 1I nd h I = h 1I. By Lemm 29, m. (n i 1, m, e 1I, v 1S) V [, α σ ] ρ[α V [ τ ]ρ](w ) nd thus m. (n i 1, m, pck e 1I, pck v 1S ) V [ α. σ ] ρ (W ). Lemm 32. If ; Γ e 1I e 1S : α. τ,, α; Γ, x : τ e 2I e 2S : σ nd σ then ; Γ unpck e 1I s x in e 2I unpck e 1S s x in e 2S : σ. Proof. Assume m. (n, m, σ I, σ S ) V [ Γ] ρ (W ) nd i < n n W W (n, h I, h S ) W σ I (unpck e 1I s x in e 2I ), h I i e I, h I By Lemm 14, there exists i 1, i 2, e 1I, nd h 1I such tht σ I (e 1I ), h I i1 e 1I, h 1I unpck e 1I s x in σ I (e 2I ), h 1I i2 e I, h I nd i = i 1 + i 2. Since ; Γ = e 1I e 1S : α. σ it follows tht there exists W W, v 1S, nd h 1S such tht σ S (e 1S ), h S v 1S, h 1S (n i 1, h 1I, h 1S ) W m. (n i 1, m, e 1I, v 1S ) V [ α. τ ] ρ (W ) nd e 1I Vl. Hence, e 1I = pck v 1I nd v 1S = pck v 1S nd there exists ν Type such tht (n i 1, 0, v 1I, v 1S) V [, α τ ] ρ[α ν] (W ) Furthermore, unpck e 1I s x in σ(e 2I ), h 1I σ I (e 2I )[v 1I /x], h 1I i2 1 e I, h I nd i 2 1. By downwrds-closure, monotonicity nd Lemm 30, it follows tht m. (n i 1 1, m, σ I [x v 1I ], σ S [x v 1S]) V [, α Γ, x : τ ] ρ[α ν] (W ) Since i 2 1 < n i 1 1 it follows from the, α; Γ, x : τ e 2I e 2S : σ ssumption tht there exists W W, v S, nd h S such tht σ S[x v 1S ](e 2S), h 1S v S, h S nd (n i 1 i 2, h I, h S) W m. (n i 1 i 2, m, e I, v S ) V [, α σ ] ρ[α ν] (W ) nd e I Vl. Lstly, since σ it follows tht m. (n i 1 i 2, m, e I, v S ) V [ σ ] ρ (W ) 21

22 Lemm 33. If ; Γ e 1I e 1S : τ nd ; Γ e 2I e 2S : σ, then ; Γ (e 1I, e 2I ) (e 1S, e 2S ) : τ σ. Proof. Assume m. (n, m, σ I, σ S ) V [ Γ] ρ (W ) nd i < n n W W (n, h I, h S ) W σ I (e 1I, e 2I ), h I i e I, h I By Lemm 14, there exists i 1, i 2, e 1I, nd h 1I such tht nd i = i 1 + i 2. σ I (e 1I ), h I i1 e 1I, h 1I (e 1I, σ I (e 2I )), h 1I i2 e I, h I Since ; Γ = e 1I e 1S : τ it follows tht there exists W W, v 1S, nd h 1S such tht σ S (e 1S ), h S v 1S, h 1S (n i 1, h 1I, h 1S ) W m. (n i 1, m, e 1I, v 1S ) V [ τ ] ρ (W ) nd e 1I Vl. By Lemm 14, there exists i 3, i 4, e 2I, nd h 2I such tht nd i 2 = i 3 + i 4. σ I (e 2I ), h 1I i3 e 2I, h 2I (e 1I, e 2I), h 2I i4 e I, h I Since ; Γ = e 2I e 2S : σ it follows tht there exists W W, v 2S, nd h 2S such tht σ S (e 2S ), h 1S v 2S, h 2S (n i 1 i 3, h 2I, h 2S ) W m. (n i 1 i 3, m, e 2I, v 2S ) V [ σ ] ρ (W ) nd e 2I Vl. Thus e I = (e 1I, e 2I ), h I = h 2I nd i 4 = 0. Lstly, by downwrds-closure nd monotonicity. m. (n i, m, (e 1I, e 2I), (v 1S, v 2S )) V [ τ σ ] ρ (W ) Lemm 34. If ; Γ e I e S : τ σ then ; Γ fst e I fst e S : τ. Proof. Assume m. (n, m, σ I, σ S ) V [ Γ] ρ (W ) nd i < n n W W (n, h I, h S ) W σ I (fst e I ), h I i e I, h I By Lemm 14, there exists i 1, i 2, e I, nd h I such tht nd i = i 1 + i 2. σ I (e I ), h I i1 e I, h I fst e I, h I i2 e I, h I Since ; Γ = e I e S : τ it follows tht there exists W W, v S, nd h S such tht σ S (e S ), h S v S, h S (n i 1, h I, h S) W m. (n i 1, m, e I, v S) V [ τ σ ] ρ (W ) nd e I Vl. nd Thus, (n i 1, 0, e I, v S ) V [ τ σ ] ρ (W ) nd there exists v 1I, v 2I nd v 1S, v 2S such tht e I = (v 1I, v 2I ) v S = (v 1S, v 2S ) (n i 1, 0, v 1I, v 1S ) V [ τ ] ρ (W ) (n i 1, 0, v 2I, v 2S ) V [ σ ] ρ (W ) Thus, by downwrds-closure, Lstly, e I = v 1I, h I = h I nd i 2 = 1. m. (n i 1 1, m, v 1I, v 1S ) V [ τ ] ρ (W ) 22

23 3.2 Soundness Theorem 2 (Fundmentl theorem of logicl reltions). If ; Γ e : τ then ; Γ = e e : τ. Theorem 3. If = e I e S : 1 nd e I, [] v I, h I then then there exists h S such tht e S, [], h S. Proof. Trivil. 3.3 Exmple Lemm 35. Let τ denote the type Then, where f is defined s follows Proof. Assume α. (1 α) (α N) ; x : τ x log f(x) : τ f = def λx.unpck x s y in pck (λz.ref (fst y z), λz.snd y (!z)) (n, 0, σ I, σ S ) V [ ; x : τ ] [] (W ) Thus, there exists ν Type nd v 1I, v 2I, v 1S nd v 2S such tht σ I (x) = pck (v 1I, v 2I ) σ S (x) = pck (v 1S, v 2S ) (n, 0, v 1I, v 1S ) V [α 1 α] [α ν] (W ) nd (n, 0, v 2I, v 2S ) V [α α N] [α ν] (W ). To show tht (n, σ I (x), σ S (f(x))) E(V [ τ ] [] )(W ) ssume i < n n W W (n, h I, h S ) W σ I (x), h I i e I, h I Since σ I (x) is vlue, h I = h I, e I = σ I(x) nd i = 0. Since σ S (f(x)), h S v S, h S, where v S = pck (λz.ref ((fst (v 1S, v 2S ))(z)), λz.snd (v 1S, v 2S )(!z)), it just remins to show tht (n, h I, h S ) W (n, σ I (x), v S ) V [ τ ] [] (W ) The first proof obligtion holds trivilly. For the second proof obligtion, let nd ρ = [α ν ], where ν = λw. {(n, m, v I, v S ) ι dom(w ). ξ(w (ι)) n,m = Înv S(v I, v S )} S(v I, l S ) = def λw. {(n, m, h I, h S ) l S dom(h S ) (n, m, v I, h S (l S )) ν(w )} Then it remins to show tht m. (n, m, v 1I, λz.ref ((fst (v 1S, v 2S ))(z))) V [α 1 α] ρ (W ) m. (n, m, v 2I, λz.snd (v 1S, v 2S )(!z)) V [α α N] ρ (W ) 23

24 First function: Assume n < n W W m. (n, m, u I, u S ) V [α 1] ρ (W ) To show tht (n + 1, v 1I u I, (λz.ref ((fst (v 1S, v 2S ))(z))) u S ) E(V [α α] ρ )(W ), ssume j < k n + 1 W 1 W (k, h 1I, h 1S ) W 1 v 1I u I, h 1I j e 1I, h 1I By downwrds-closure nd monotonicity, it follows tht (n +1, 0, v 1I, v 1S ) V [α 1 α] [α ν] (W 1 ). Thus, there exists v 1S, h 1S nd W 1 W 1 such tht e 1I Vl nd v 1S u S, h 1S v 1S, h 1S (k j, h 1I, h 1S) W 1 m. (k j, m, e 1I, v 1S ) V [α α] [α ν] (W 1) = ν(w 1) Pick fresh loction l S dom(h 1S ). Then (λz.ref ((fst (v 1S, v 2S ))(z))) u S, h 1S l S, h 1S[l S v 1S ] Pick fresh invrint nme ι dom(w 1) nd let W 1 = W 1[ι ξ 1 (S(e 1I, l S))]. Then it remins to show tht (k j, h 1I, h 1S[l S v 1S ]) W 1 m. (k j, m, e 1I, l S ) V [α α] ρ (W 1 ) The first obligtion reduces to proving tht m. (k j 1, m, [], [l S v 1S ]) S(e 1I, l S )(W 1 ) which further reduces to proving tht m. (k j 1, m, e 1I, v 1S) ν(w 1 ). This follows esily by downwrdsclosure nd monotonicity. The second proof obligtion reduces to, m. ι dom(w 1 ). ξ(w 1 (ι)) k j,m = Type S(e 1I, l S ) which is esily seen to hold, s k j,m = is n equivlence reltion. Second function: Assume n < n W W m. (n, m, u I, u S ) V [α α] ρ (W ) = ν (W ) To show tht (n + 1, v 2I u I, (λz.snd (v 1S, v 2S )(!z)) u S ) E(V [α N] ρ )(W ), ssume j < k n + 1 W 2 W (k, h 2I, h 2S ) W 2 v 2I u I, h 2I j e 2I, h 2I To show tht m. (k 1, m, u I, h 2S (u S)) ν(w 2 ), ssume m N. By ssumption, (n, m + 1, u I, u S ) ν (W ) nd thus, there exists n ι dom(w ) such tht ξ(w (ι)) n,m+1 = Type S(u I, u S ). Furthermore, by the extension ordering on worlds, W 2 (ι) = W (ι) nd by world stisfction, there exists h 2I nd h 2S such tht h 2I h 2I h 2S h 2S m. (k 1, m, h 2I, h 2S) ξ(w (ι))(w ) Since (k 1, m) < (n, m + 1) it follows tht (k 1, m, h 2I, h 2S ) S(u I, u S )(W 2 ) nd thus, (k 1, m, u I, h 2S(u S )) ν(w 2 ) 24