Komponenten- und Service-orientierte Softwarekonstruktion

Komponenten- und Service-orientierte Softwarekonstruktion

Lecture 2: Types and combinators

Jakob Rehof

LS XIV – Software Engineering

TU Dortmund Sommersemester 2017

Simple types

Definition 1 (Simple types)

LetVdenote a denumerable set oftype variables, ranged over by metavariables α, β, γ, . . ., and letbrange over a setBoftype constants. The setTofsimple types, ranged over byτ, σ, ρ, . . ., is defined inductively by:

α∈V⇒α∈T b∈B⇒b∈T

τ ∈T, σ∈T⇒τ →σ∈T

Type variables and type constants are referred to asatoms. We writeτ→σ→ρfor τ →(σ→ρ).

Atype environmentis a finite setΓoftype assumptionsof the form Γ ={(x1:τ1), . . . ,(xn:τn)}withxi6≡xj fori6=j. We let

Dm(Γ) ={x∈ V | ∃τ∈T.(x:τ)∈Γ}. We writeΓ, x:τ forΓ∪ {(x:τ)}.

Simple typed λ-calculus, λ

Γ, x : τ ` x : τ (var)

Γ, x : τ ` M : σ Γ ` λx.M : τ → σ (→I)

Γ ` M : τ → σ Γ ` N : τ

Γ ` M N : σ (→E)

Explicit typing

Γ, x:τ`^∗x:τ(var)

Γ, x:τ`<sup>∗</sup>M:σ Γ`^∗λx:τ.M :τ →σ(→I)

Γ`<sup>∗</sup>M:τ →σ Γ`^∗N :τ Γ`^∗M N:σ (→E)

Exercise 1

Show by induction onM: ifΓ`<sup>∗</sup>M :σandΓ`^∗M :τ, thenσ≡τ.

Basis lemma

Let Γ ↓

= {(x : τ ) ∈ Γ | x ∈ V }.

Lemma 2

If Γ ⊆ Γ

then Γ ` M : τ implies Γ

` M : τ , If Γ ` M : τ , then FV(M ) ⊆ Dm(Γ),

If Γ ` M : τ , then Γ ↓

` M : τ .

Exercise 2

Prove Lemma 2.

Inversion

Lemma 3 (Generation lemma)

Suppose thatΓ`M :τ.

1 IfM≡x, thenΓ = Γ⁰, x:τ,

2 IfM≡λx.N, thenτ≡ρ→σandΓ, x:ρ`N :σ,

3 IfM≡P Q, thenΓ`P:σ→τ andΓ`Q:σfor someσ.

Proof.

Immediate by inspection of the last rule used in the derivation ofΓ`M:τ.

Exercise 3

Call a termM typable, if Γ`M :τ for someΓandτ.

Show that every subterm of a typable term is typable.

Substitutivity

A type substitution is a map S : V → T, and it is lifted homomorphically to a map S : T → T . Let S(Γ) = {(x : S(τ )) | (x : τ ) ∈ Γ}.

Lemma 4 (Substitution)

1

If Γ, x : τ ` M : σ and Γ ` N : τ , then Γ ` M [x := N ] : σ.

2

If Γ ` M : τ , then S(Γ) ` M : S(τ ), for any type substitution S.

Exercise 4

Prove Lemma 4 by induction on derivations.

Subject reduction

Theorem 5 (Subject Reduction)

If Γ ` M : τ and M

N , then Γ ` N : τ .

Proof.

(Sketch) First prove the statement for the case when M is a redex and N its reduct, using the substitution lemma. Then prove the statement when M →

N by reduction of a redex R in M, by induction on C where M ≡ C[R]. Then prove the statement by induction in the length of the reduction M

N .

Exercise 5

Complete the proof sketch of Theorem 5.

Strong normalization

Definition 6

Aλ-termM is calledstrongly normalizing, if everyβ-reduction sequence starting from M is finite, andweakly normalizing, if there exists a finite reduction sequence starting fromM.

Theorem 7

Every typable term inλ^→ is strongly normalizing.

Proof.

See lecture notes.

Combinators

LetCbe a set ofcombinator symbols, ranged over byX, Y, Z. The setΞC of combinatory expressions, ranged over byF, G, H are defined inductively by:

X∈ C ⇒X∈ΞC, x∈ V ⇒x∈ΞC, F, G∈ΞC⇒(F G)∈ΞC

Consider the set SKI={S,K,I}of combinator symbols (referred to as a combinatory base) and define the notion ofweak reductionwonΞSKIby setting, for all

X, Y, Z∈ΞSKI:

IF w F

KF G w F

SF GH w (F H)(GH)

Let→w and^w be the reduction relations onΞSKIinduced byw, by closingw

under contextsC::= []|(CF)|(F C), (F∈ΞSKI).

Combinatory bases

By choosing different sets B ⊆ C of combinators (such as

B = SKI = {S, K, I}) we can study different combinatory calculi, since in each case we can consider a B-calculus generated from the combinators in B. In such cases, we refer to the set B as a combinatory base.

Exercise 6

Show that the combinator I can be coded in terms of S and K. Hint:

Notice that (KF)(KF )

F .

In other words, the base SKI = {S, K, I} is redundant. Or, in yet other words, the base SK = {S, K} is complete with respect to SKI-calculus.

For this reason, one also talks about SK-calculus.

Combinatory bases

Sch¨ onfinkel used, in addition to S and K, the combinators B and C with the definitions

BF GH

F GH CF GH

F HG But they are not strictly needed, for we can take

B ≡ S(KS)K

C ≡ S(S(K(S(KS)K))S)(KK)

Combinatory bases

Exercise 7 (One point basis)

Define the combinator X by the rule

(XF )

((F S)K)

Show that (XX)

SK(KK).

Show that X(X(XX))

K.

Show that X(X(X(XX)))

S.

Conclude that {X} is complete with respect to SKI-calculus.

Ξ

7→ Λ

Define the map( )Λ: ΞSKI→Λby induction on expressions inΞSKI: (x)Λ ≡ x, forx∈ V

(I)Λ ≡ λx.x (K)Λ ≡ λxy.x (S)Λ ≡ λxyz.(xz)(yz) (F G)Λ ≡ (F)Λ(G)Λ

Proposition 1

IfF ^wG, then(F)Λβ (G)Λ.

Exercise 8

Prove Proposition 1 by induction on the length ofF wG.

Λ 7→ Ξ

Define, for eachx∈ V, the “bracket abstraction” map [x]: ΞSKI→ΞSKIby induction on expressions inΞSKI:

[x]x ≡ I

[x]F ≡ KF, ifx6∈FV(F) [x](F G) ≡ S([x]F)([x]G), otherwise

Proposition 2 (Combinatory completeness)

1 ∀F∈ΞSKI.∀G∈ΞSKI.([x]F)GwF[x:=G]

2 ∀F∈ΞSKI.([x]F)Λβ λx.(F)Λ

3 ∀x∈ V.∀F ∈ΞSKI.∃H∈ΞSKI.∀G∈ΞSKI. HG^wF[x:=G]

Exercise 9

Λ 7→ Ξ

Define the map( )Ξ: Λ→ΞSKIby induction onλ-terms:

(x)Ξ ≡ x, forx∈ V (M N)Ξ ≡ (M)Ξ(N)Ξ

(λx.M)Ξ ≡ [x](M)Ξ

Proposition 3

For allM∈Λ, one has((M)Ξ)ΛβM.

Exercise 10

Prove Proposition 3.

Combinatory logic SKI

Γ, x:τ `skix:τ(var)

Γ`_skiI:τ →τ(I)

Γ`_skiK:τ →σ→τ(K)

Γ`<sub>ski</sub>S: (τ→σ→ρ)→(τ →σ)→τ →ρ(S) Γ`_skiF:τ→σ Γ`_skiG:τ

Γ`_ski(F G) :σ (→E)

Notice that variablesxhave fixed,monomorphic types, whereas combinatorsS,K,I have infinitely many types (their types areschematicorpolymorphic). We shall return to this important point in Lecture5.

Combinatory logic SKI

Lemma 8 (Deduction theorem for SKI)

IfΓ, x:σ`<sub>ski</sub>F :τ, thenΓ`_ski[x]F :σ→τ.

Proposition 4

1 IfΓ`<sub>ski</sub>F:τ, thenΓ`Λ(F)Λ:τ.

2 IfΓ`ΛM:τ, thenΓ`_ski(M)Ξ:τ.

Exercise 11

Prove Proposition 4. Hint: The first statement is proven by induction on the derivation ofΓ`skiF :τ. The second statement is proven by induction on the derivation of Γ`ΛM :τ using Lemma 8.

Decision problems

Type checking(Γ`M :τ?). GivenΓ,M andτ, doesΓ`M :τ hold?

Typability(?`M :?). GivenM, are thereΓandτ such thatΓ`M:τ? Inhabitation(Γ`? :τ). GivenΓandτ, does there exist a term (“inhabitant”)M such thatΓ`M:τ?

Type checking and typability forλ^→are linear time solvable.

Inhabitation forλ^→ ispspace-complete. See Lecture3.

Inhabitation can be seen as a program synthesis problem:

ConstructprogramM satisfyingspecificationτ.