Alonzo Church ( ) Lambda Calculus. λ-calculus : syntax. Grammar for terms : Inductive denition for λ-terms

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1 Alonzo Church ( ) Lambda Calculus 2 λ-calculus : syntax Grammar for terms : t, u ::= x (variable) t u (application) λx.t (abstraction) Notation : Application is left-associative so that t 1 t 2... t n means (... (t 1 t 2 )... t n ). Inductive denition for λ-terms t is a λ-term t u is a λ-term x is a variable (Var) x is a λ-term u is a λ-term (App) t is a λ-term (Lamb) λx.t is a λ-term λx 1... x n.t means λx λx n.t. 3 4

2 Free and bound variables λx.(z x y (λz.z y)) The set of free and bound variables are dened as follows fv(x) = {x} bv(x) = fv(t u) = fv(t) fv(u) bv(t u) = bv(t) bv(u) fv(λx.t) = fv(t) \ {x} bv(λx.t) = bv(t) {x} But for t = x (λx.x) Alpha-conversion The relation = α, called alpha-conversion, is the congruence generated by the axiom α. Formally, λx.t α λy.t[x /y] for y fresh t= α t λx.t= α λx.t λx.t = α λy.t[x /y] t= α t u= α u t u= α t u fv(x (λx.x)) = {x} = bv(x (λx.x)) 5 6 Dening the notion of fresh replacement The operation t[x /y] means the replacement of all the free occurrences of x in t by a fresh variable y. Examples x[x /y] = y z[x /y] = z (t u)[x /y] = t[x /y] u[x /y] (λx.t)[x /y] = λx.t (λz.t)[x /y] = λz.t[x /y] λx.(λx.x z) α λy.(λx.x z) = α λy.(λy.y z) x (λx.x) = α x (λz.z) 7 8

3 Barendregt variable convention From now on we assume the following variable convention : 1. No variable is both free and bound. 2. Bound variables have all dierent names. Example : x (λz.z) is OK, x (λx.x) is not OK, λx.λy.x z is OK but λx.λx.x z is not OK. Theorem : For every λ-term t there is λ-term u verifying the Barendregt convention such that t = α u. Indeed, x (λx.x) = α x (λz.z) and λx.λx.x z = α λy.λx.x z. Operational semantics of λ-calculus A one-step β-reduction is given inductively by t t (λx.t) u t{x/u} λx.t λx.t t t u u t u t u t u t u What is exactly _{_/_}? 9 10 Towards a notion of substitution : warning! A simple notion of higher-order substitution t{x/u} means replace all the free occurrences of x in t by u. (λx.(λy.x)) y (λy.x){x/y} = λy.y This operation is dened modulo α-conversion as follows : Incorrect (λx.(λy.x)) y = α (λx.(λz.x)) y Correct (λz.x){x/y} = λz.y x{x/u} = u y{x/u} = y (λy.v){x/u} = λy.v{x/u} if x y and y / fv(u) (no capture holds) (t v){x/u} = (t{x/u} v{x/u}) 11 12

4 A terminating β-reduction sequences A non terminating β-reduction sequences (λx.λf.x f y) (λz.z) (λw.w w) (λf.x f y) {x/λz.z} (λw.w w) = (λf.(λz.z) f y) (λw.w w) (λf.f y) (λw.w w) (f y){f/λw.w w} = (λw.w w) y (w w) {w/y} = (y y) Let Π = λx.f (x x). Π Π (f (x x)){x/π} = f (Π Π) f (f (x x)){x/π} = f (f (Π Π)) Curry-Howard Isomorphism Properties of λ-calculus [Conuence] The reduction relation is conuent. [Free variables decrease] If t t, then fv(t) fv(t ). Logical system Language Propositions Types Proofs Programs Proof normalisation Program Evaluation 15 16

5 Adding (simply) types to λ-calculus Curry'58, Howard'68 Grammar for types : A, B ::= τ (base types) A B (functional types) Notation : is right-associative so that for example A 1 A 2 A 3 means A 1 (A 2 A 3 ) Typing Environment A typing environment Γ is a nite function from variables to types, usually written x 1 : A 1,..., x n : A n. Thus for example, x : A, y : B and y : B, x : A are two dierent notations for the same typing environment. The domain of Γ = x 1 : A 1,..., x n : A n, written dom(γ), is the set {x 1,..., x n }. We write Γ, x : A for the typing environment extending Γ with the pair x : A. It is only dened i x / dom(γ). Natural deduction as typed λ-calculus (ax) Γ, x : A x : A Γ, x : A t : B Γ t : A B Γ u : A ( i) Γ λx.t : A B Γ t u : B We denote by Γ t : A the derivability/typing relation. We say that t is typable i there is Γ and A such that Γ t : A. Remark : Γ t : A denotes also a sequent! ( e) 19 20

6 Examples Example of a typable term : (λx.x)(λy.y). y : σ σ y : σ σ x : σ x : σ λy.y : (σ σ) (σ σ) λx.x : σ σ (λy.y)(λx.x) : σ σ Example of non-typable term : λx.xx. Typed Properties [Subject Reduction] If Γ t : A and t t, then Γ t : A. [Strong Normalization] Every typed term is normalising : if Γ t : A, then t SN β Some General Remarks When using typed terms, the notation t{x/u} means that x and u have the same type. t SN β i there is no innite β-reduction sequence starting at t i every β-reduction sequence starting at t is nite t SN β can also be dened as follows If t is a β-normal form, then t SN β If t [(t t ) implies t SN β ], then t SN β If t SN β, then every subterm of t is also SN β. SN β is not stable by substitution. Example : x x SN β, λy.y y SN β, but (x x){x/λy.y y} = / SN β. u SN β i λy.u SN β. u 1,..., u n SN β i x u 1... u n SN β. Given t SN β we dene µ β (t) as the maximal lenght of a reduction sequence starting at t. We observe that t t implies µ β (t ) < µ β (t). Every type A can be written as A 1... A n τ, where A 1,..., A n (n 0) are arbitrary types and τ is a base type. The standard order between types is given by A < A B and B < A B. Thus base types are minimal with respect this order

7 First Proof of the SN property Dénition : Let M be of type A = A 1... A n τ. Then M SC i R i : A i SC M R 1... R n SN β. The denition implies 1. SC SN. 2. SC is closed under β. 3. x SC for every variable x (using 1). Lemma : [Basic Lemma] If t, R 1,..., R n (n 1) SN β and t{x/r 1 }R 2... R n SN β, then (λx.t)r 1 R 2... R n SN β. Proof. It is sucient to show that all the reducts of (λx.t)r 1... R n are in SN β. We reason by induction on µ(t) + Σ i µ(r i ). The reducts are (λx.t )R 1... R n, where t t. Then µ(t ) < µ(t), we conclude by the i.h. (λx.t)r 1... R i... R n, where R i R i. Then µ(r i ) < µ(r i), we conclude by the i.h. t{x/r 1 }R 2... R n. We conclude by the hypothesis Lemma : Let M be a typed term. Let σ be a type preserving substitution mapping all the free variables of M to terms in SC. Then Mσ SC. Proof. We proceed by induction on the typed term M. If M = x, then xσ = σ(x) SC by hypothesis. If M = NL, then let R n SC. We have Nσ and Lσ in SC by i.h. Then (N L)σ R n = Nσ Lσ R n SN β by denition. If M = λx.n, then (λx.n)σ = λx.nσ. Since σ {x/x} veries the hypothesis of the lemma, then by the i.h. N(σ {x/x}) = Nσ SC SN β. To show λx.nσ SC we show (λx.nσ)p 1... P n SN β for P 1... P n SC SN β. This follows from the Basic Lemma. Lemma : Every typed term is SC. Proof. Using the previous lemma with the identity substitution (which veries the hypothesis of the lemma). Theorem : Every typed term is in SN β. Proof. Using the previous lemma and the fact the SC SN β

8 Second proof of the SN property 1. Dene SN inductively : M 1,..., M n SN implies x M 1... M n SN. M SN implies λx.m SN. M{x/N} P n SN and N SN implies (λx.m) N P n SN. 2. Dene Λ A inductively : If x is a variable of type A, then x Λ A. If t Λ C and x is a variable of type B, then λx.t Λ B C. If t Λ B A and u Λ B, then t u Λ A. 3. Dene SN A = SN Λ A. 5. Show SN = SN β (in fact SN SN β suces). 6. Show Λ A B = Λ A Λ B 7. Show SN A SN B SN A B (easy). 8. If N SN A1 SN A2... SN Am with A m a base type and P SN B, then P {x/n} SN B (induction on SN using 7). 9. Show SN A B SN A SN B (using 8). 10. Show that Λ A SN A (by induction using 9). 4. Dene X Y = {M N X (MN) Y } Third Proof of the SN property Lemma : If t and u are typed and belong to SN β, then t{x/u} SN β. Proof. By induction on type(u), µ(t), size(t). The base case base type, 0, 1 is trivial. Case t = λy.v is straightforward (size(t) strictly decreases). Case t = y c n with x y is straightforward (µ(t) decreases and size(t) strictly decreases.). Case t = x. We have x{x/u} = u SN β by hypothesis. Case t = x b c n. By i.h. B = b{x/u} and C i = c i {x/u} are in SN β. We want to show that u B C n SN β. It is sucient to show that all its reducts are in SN β. We reason by induction on µ(u) + µ(b) + Σ i µ(c i ). The reducts are u B C n, where u u. Apply the i.h. u B Cn, where B B. Apply the i.h. u B C 1... C i... C n, where C i C i. Apply the i.h. u 1 {y/b} C n, where u = λy.u 1. But u 1 {y/b} C n = (zc n ){z/u 1 {y/b}} and type(u 1 {y/b}) < type(u). We thus conclude by the i.h. since zc n and u 1 {y/b} are typed and in SN β by the i.h.. Case t = (λz.b) c d. By i.h. B = b{x/u} and C = c{x/u} and D i = d i {x/u} are in SN β. Suppose t{x/u} = (λz.b) C D n / SN β. Then B{z/C} D n / SN β. But B{z/C} D n = (b{z/c} d n ){x/u} and µ(b{z/c} d n ) < µ(t). Thus B{z/C} D n SN β by the i.h. Contradiction. Thus t{x/u} = (λz.b) C D n SN β

9 Theorem : If t is typable, then t SN β. Proof. By induction on the typing derivation of t. Case t = x is trivial. Case t = λy.u holds by the i.h. For the case t = u v use the fact that t = (z v){z/u} and apply previous lemma (verication of the hypothesis is easy). Fourth proof of the SN property See for example Gandy's proof by Alexandre Miquel. A combinatorial proof of strong normalisation for the simply typed lambda-calculus. snlam.pdf 33 34