Operational and Denotational Semantics. Solution.

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1 Operational and Denotational Semantics Please turn in by April 5, 2011 (-1 point penalty per day late) Solution. We consider an enriched, typed λ-calculus with constants for arithmetic expressions and recursion. Note that the essential question is question 9. This is the one that takes the longest to answer, and the one that will really tell me whether you understand (variants of) techniques we have seen in theλ-calculus. The types are σ,τ ::= Nat σ τ. All the terms in the language come with explicit types. In particular, we assume an infinite, countable set of variables of each type τ, and write x τ for a variable of type τ. The terms t of typeτ (in short, t : τ) are defined by induction on their size by : every variable x τ is of typeτ ; ifn : τ, then λx σ N is a term of type σ τ ; ifm : σ τ andn : σ, then MN is a term of type τ ; for each n N, there is a distinct constant n : Nat ; there are distinct constants pred : Nat Nat (subtract one), succ : Nat (add one), ifz τ : Nat τ τ τ (test if first argument equals 0), and Y τ : (τ τ) τ (fixpoint, recursion). As in the λ-calculus, the terms are understood up to α-renaming. This takes the special form thatx σ can be replaced by any fresh variable y σ of the same typeσ inλx σ t. We drop type indices whenever they are irrelevant or can be reconstructed from context. We define its semantics not through reduction, but by building a machine directly. The contexts are defined by the grammar E := EN succe prede ifz E N P, where N, P denote terms.. We see contexts as particular terms, with a unique occurrence of a specific variable that does not occur in any term, called the hole. A context E is of type σ τ when E is of type τ, assuming the hole of type σ. E[M] denotes the replacement of the hole by M (of typeσ). Our machine is a transition system whose configurations are pairs E M. Intuitively, in such a configuration, the machine is in the process of evaluating E[M], and the focus is currently on the subterm M. 1

2 The machine rules come into two groups. The first form the redex discovery rules. E.g., in (DApp), the machine tries to evaluate MN, and proceeds by pushing the argument N into the context and focusing on the function part M. (DApp) E MN E[ N] M (DIf) E[ MNP] ifz E[ifz N P] M (DPred) E[ N] pred E[pred ] N (DSucc) E[ N] succ E[succ ] N The second group of rules computes : (β) E[ N] λx P E P[x := N] (Y) E[ N] Y E N(YN) (pred) E[pred ] n+1 E n (succ) E[succ ] n E n+1 (ifz0) E[ifz N P] 0 E N (ifz1) E[ifz N P] n+1 E P We do not take the closure of under contexts, whatever this may mean. The relation is entirely specified by the rules above. We also consider a denotational semantics of the above terms. First, we define the cpo τ of all values of typeτ : Nat isn, the set of all natural numbers plus an added, so-called bottom element. These are ordered by m n iff m = or m = n. (Think as all element of N being incomparable and above.) σ τ is the dcpo [ σ τ ] of all continuous maps from σ to τ. They are ordered byf g iff f(x) g(x) in τ for every x σ. 1. Show that every term t has exactly one type. This is by induction on the size oft, using the fact that every term has at least one type by definition, and conversely that the type is determined from a unique typing rule. E.g., if our term is MN, then M has a unique type, N has a unique type, and since MN has a type at all, it must be obtained by the rule if M : σ τ and N : σ then MN is a term of type τ, which determines τ uniquely. If our term is λx σ N, then N has a unique type τ, and therefore the unique type of λx σ N is σ τ. This is the only interesting case, and justifies why we explicitly decorate the variable x σ with its type. Finally, the types of variables and constants are uniquely determined. 2. Show that τ has a least element τ, for every type τ. By abuse of language, we shall write instead of τ, and call it bottom. It is useful to think of as non-termination. By induction on τ. N has as its least element, and if τ is the least element of τ, then the constant map with value τ is the bottom element of σ τ. Then we define the semantics t ρ of terms t : τ as values in τ in environments ρ mapping each variable x σ to an element of σ, by : x σ ρ = ρ(x σ ) ; MN ρ = M ρ( N ρ) ; λx σ N ρ is the function that maps each v σ to N (ρ[x σ := v]) ; n ρ = n, for every n N ; 2

3 pred ρ is the map that sendsn+1 to n N, and0and to ; succ ρ is the map that sendsn N to n+1, and to ; ifz τ ρ is the map that sendsu,v,w to ifu =, to v if u = 0, and to w ifu = 0, ; Y τ ρ is the map that sendsf [ τ τ ] to sup n N f n ( τ ). 3. Given any continuous map f [ τ τ ], show that lfp(f) = sup n N f n ( τ ) exists in τ, and is the least fixpoint of f. A fixpoint of f is an element v such that f(v) = v. It is least iff v w for every other fixpoint w. (This is meant to explain the definition of Y τ ρ.) We first check that (f n ( τ )) n N is a chain, i.e., f 0 ( τ ) f 1 ( τ )... f n ( τ )... It is enough to check that f n ( τ ) f n+1 ( τ ) for every n N, and this is done by induction on n. If n = 0, this is because f 0 ( τ ) = τ is least. Otherwise, f n ( τ ) = f(f n 1 ( τ )) f(f n ( τ )) = f n+1 ( τ ) by induction hypothesis and the fact thatf is monotonic. Now lfp(f) is a fixpoint : f(lfp(f)) = f(sup n N f n ( τ ) = sup n N f n+1 ( τ ), because f is continuous and (f n ( τ )) n N is a chain ; we conclude because sup n N f n ( τ ) = sup(f 0 ( τ ),sup n N f n+1 ( τ )) = sup( τ,sup n N f n+1 ( τ )) = sup n N f n+1 ( τ ). We claim that lfp(f) is the least one. Assume x is another fixpoint. By induction on n, f n ( τ ) x : this is clear if n = 0, since τ is least, otherwise f n+1 ( τ ) = f(f n ( τ )) f(x) (by induction hypothesis, and sincef is monotonic)= x (sincexis a fixpoint). Taking sups on each side, lfp(f) x. 4. Establish soundness : if E M n, then E[M] ρ = n, for every n N, and every environment ρ. We first show that : (1) ife M E M, then E[M] ρ = E [M ] ρ. By induction on the number of steps, it will follow that if E M n, then E[M] ρ = [n] ρ = n. Claim (1) is obvious in the cases of the rules(dapp),(dif),(dpred) and(dsucc), since in these cases E[M] = E [M ]. For the other rules, we note that for any configuration E M, E M ρ = E (ρ[ M ρ]). This is a form of the substitution lemma we have seen in the lectures, and is proved by induction on the size ofe. It remains to show that : (β) (λx P)N ρ = P[x := N] ρ : the left-hand side is (v P (ρ[x := v]))( N ρ) = P (ρ[x := N ρ]), which is equal to the right-hand side by the substitution lemma. (Y) YN ρ = N(YN) ρ : the left-hand side is lfp(f), wheref = N ρ, while the right-hand side is f(lfp(f)). These two are equal because lfp(f) is a fixpoint of f, as we have seen. (pred) predn+1 ρ = n ρ : both sides equaln. (succ) succn ρ = n+1 ρ : both sides equalsn+1. 3

4 (ifz0) ifz 0 N P ρ = N ρ : obvious. (ifz1) ifz n+1 N P ρ = P ρ : obvious. The rest of the questions aim at establishing the converse of soundness, a property known as computational adequacy. This is harder : notice in particular that computational adequacy entails that if E[M] ρ = n for every n N, and every environment ρ, then E M must terminate (on the configuration n). We first show that we can only hope to prove computational adequacy in a limited setting. First, we only consider ground configurations : call a term ground iff it has no free variable, and a configuration E M ground iff E[M] is ground. Then E[M] ρ does not depend on ρ, and we shall write it E[M]. 5. One might think of proving computational adequacy at every type τ, i.e., that if E M is ground, and E[M] is a value v τ (the same for every environment ρ), then E M M for some canonical ground term M : τ with M = v. By canonical, we mean that there is a unique canonical term M such that M = v. Show that this is hopeless : canonicality fails, at least for some type τ other than Nat. Take E =, M 1 = λx σ succ1 and M 2 = λx σ 2. These two terms have the same value, namely the constant2function. Canonicality would imply that M 1 and M 2 would rewrite to the same M. But M 1 and M 2 do not rewrite at all. Define σ on ground terms by M σ N iff for every context E of typeσ Nat, if E M n then E N n. Extend σ to non-ground terms by M σ N iff Mθ σ Nθ for every well-typed substitution θ of ground terms for all variables in M and N. A substitution θ is well-typed iff x σ θ is a term of typeσ for every variable x σ in its domain. 6. Show that : M[x σ := N] τ (λx σ M)N wheneverm : τ, N : σ ; M(YM) τ YM provided M : τ τ ; ifn Nat M then n+1 Nat succm ; ifn+1 Nat M then n Nat predm ; if0 Nat M then N τ ifz M N P (where N : τ,p : τ), ifn+1 Nat M then P τ ifz M N P (where N : τ, P : τ). We show the claims assuming each side of the inequalities ground. The general case follows by applying some well-typed substitutionθ throughout. M[x σ := N] τ (λx σ M)N : ife M[x σ := N] n, thene (λx σ M)N E[ N] λx σ M (by(dapp)) E M[x σ := N] (by(β)) n (by assumption). M(YM) τ YM : similarly,e YM E[ M]Y E M(YM) by(dapp) and (Y). Ifn+1 Nat M thenn Nat predm. This is slightly different. Assume thate n m for some m N. So E[pred ] n+1 E n (by (pred)) m. Since n+1 Nat M, and using the context E[pred ], E[pred ] M m. But then E predm E[ M] pred (by(dapp)) E[pred ] M (by(dpred)) m. 4

5 If n+1 Nat M then n Nat predm. Similar, using(succ) and(dsucc) instead. If 0 Nat M then N τ ifz M N P. Assume E N m, and use the context E[ifz N P]. Since0 Nat M ande[ifz N P] 0 E N (by(ifz0)) m, we must havee[ifz N P] M m, soe ifz M N P E[ P] ifzmn E[ NP] ifzm E[ MNP] ifz E[ifz N P] M E N m, using(dapp) three times, then (DIf). If n+1 Nat M then P τ ifz M N P. Similar, using(ifz1) instead. The main tool in the proof of computational adequacy is the use of a so-called logical relation. This is a notion similar to the reducibility sets RED τ defined in the lectures, except there are now binary relations, i.e., sets of pairs. Here, a logical relation is a family of binary relations R τ, indexed by types τ, between values in τ and ground terms M : τ. These are defined by induction on types as follows. For short, we say for all N R σ v instead of for every term N : σ, every value v σ, if N R σ v then. M R Nat u iff u =, oruis a natural number n andn Nat M. M R σ τ f iff for all N R σ v, MN R τ f(v). 7. Given M : τ, let MR τ be the set {u τ M R τ u}. Show that MR τ is Scott-closed, i.e., both downward closed (ifu v andm R τ v thenm R τ u) and stable under directed sups (if (u i ) i I is a directed family in τ, andm R τ u i for every i I, thenm R τ sup i I u i ). Show also thatmr τ is non-empty, i.e., contains. By induction on types. MR Nat is { } {n N n Nat M, so it contains and is downward closed (the only elements below n N are n and ). It is also closed under directed sups because every non-empty subsetaof N is : either A = { } and this is clear, oracontains somen N, then everym A is below some element above both n andmby directedness, which implies m n since n is maximal in N. MR σ τ contains σ τ, since for all N R σ v, MN R τ σ τ (v) = τ, by induction hypothesis onτ. If f MR σ τ and g f, then for all N R σ v, MN R τ f(v) by definition hence MN R τ g(v) since MNR τ is downward closed by induction hypothesis on τ. So g MR σ τ. If(f i ) i I is a directed family inmr τ, then for allnr σ v,mnr τ f i (v). SinceMNR τ is closed under directed sups by induction hypothesis on τ, MN R τ sup i I (f i (v)) = (sup i I f i )(v). SoM R σ τ sup i I f i. 8. Show that if M τ N and M R τ u, then N R τ u. In other words, the set R τ u = {M : τ M R τ u} is upward closed in τ. By induction on types again. If M Nat N and M R Nat u, then either u = and N R Nat u is by definition, or u = n N, n Nat M. Now, Nat is transitive. Precisely, for every E of the right type, and every well-typed substitution θ of ground terms for variables, for everym N, the latter states that ife n m thene M m, andm Nat N then implies thate N m. Son Nat N. Sinceu = n,nr Nat u. 5

6 On function types σ τ, the key argument is that whenever M σ τ N, then for every P : σ, MP τ NP, i.e., that for every E of the right type, and every well-typed substitutionθ of ground terms for variables, for everym N, ife MP m then E NP m. The only reduction E MP m must use (DApp) as the first rule, hence be of the forme MP E[ P] M m. SinceM σ τ N one must havee[ P] N m, hencee NP m, using(dapp). So assume that M σ τ N and M R σ τ f. For all P R σ v, MP R τ f(v). We have seen thatmp τ NP, so by induction hypothesis onτ, NP R τ f(v). SoN R σ τ f. 9. Given a well-typed substitution θ, and an environment ρ, write θ R ρ iff x σ θ R σ ρ(x σ ) for every variable x σ in the domain ofθ. Prove the Basic Lemma : ifθr ρ, and M is a term of type τ, then MθR τ M ρ. By induction on the size ofm this time. If M is a variablex σ, then this is the induction hypothesis. IfM is an applicationnp, withp : σ, then by induction hypothesisnθr σ τ N ρ and Pθ R σ P ρ, so Mθ = (Nθ)(Pθ) R τ N ρ( P ρ) = M ρ by definition of R σ τ. If M is an abstraction λx σ N, and τ = σ σ, then we must show that for all P R σ v ((λx σ N)θ)P R σ N (ρ[x σ := v]). Using α-renaming, we may assume x σ fresh, and therefore θ [x σ := P] to be a meaningful well-typed substitution θ such that Nθ = Nθ[x σ := P]. Note that for every variable y in the domain of θ, yθ is in logical relation to ρ (y), where ρ = ρ[x σ := v]. This is by assumption if y is in the domain of θ, and follows from P R σ v if y = x σ. So by induction hypothesis Nθ R σ N ρ. We conclude since Nθ = Nθ[x σ := P] σ ((λx σ N)θ)P (Question 6, first item) and using Question 8. If M is a constantn,n N, then we must show thatn Nat n. This is obvious. If M is the constant pred, then we must show that for all P R Nat v, predp R Nat pred ρ(v). If v = or if v = 0, then pred ρ(v) =, so this results from the fact that predpr Nat contains (Question 7, last item). Otherwise, v = n+1 for somen N, and we must show thatpredp R Nat n under the assumption thatp R Nat n+1. Equivalently, we must shown Nat predp under the assumptionn+1 Nat P. This is Question 6, item 4. The casem = succ similarly follows from Question 6, item 3. The case M = ifz similarly follows from Question 6, item 5 (if v = 0) or item 6 (if v = n+1 for somen N). Finally, we deal with the casem = Y τ. We must show that for allp R τ τ f,yp R τ sup n N f n ( ). Since YPR τ is Scott-closed (Question 7), it is enough to show that YP R τ f n ( ) for every n N. We show this by induction on n N. If n = 0, this is obvious since YPR τ contains (Question 7, last item). Otherwise, by induction hypothesis YP R τ f n 1 ( ). Since P R τ τ f, by definition of R τ τ, P(YP) R τ 6

7 f n ( ). ButP(YP) τ YP by Question 6, item 2. SinceR τ f n ( ) is upward-closed (Question 8),YP R τ f n ( ), and we conclude. 10. Conclude that computational adequacy holds : given a ground configuration E M such that E[M] : Nat, E[M] = n N if and only ife M n. One direction is Question 4. Conversely, if E[M] = n N, then E[M] R Nat n by Question 9, using the identity substitution (since E M is ground). By definition, n Nat E[M]. Using an empty context, E[M] n. Now the first machine rules in the latter reduction must be redex discovery rules, until we reach the configuration E M. Formally, we show that if E E[M] n, then E [E] M n, by induction on the size of E. If E =, this is clear. If E = E 1 N, then the only applicable rule is (DApp), yieldinge E[M] E [ N] E 1 [M] n. IfE = succe 1, then the first two rules must be(dapp) and(dpred), yieldinge E[M] E [ E 1 [M]] succ E [succ ] E 1 [M] n; ife = prede 1, then these are(dapp) and(dsucc) ; if E = ifz E 1 N P, then these are(dapp) (three times) and(dif). 7