Nameless Representation of Terms

Transcription

1 Nameless Representation of Terms CIS500: Software Foundations Nameless Representation of Terms p.1/29

2 First, some review... A Proof on λ-terms Nameless Representation of Terms p.2/29

3 Proof (1) We want to prove that if z FV ([x v]u) then In other words, z (FV (u) \ {x}) FV (v) FV ([x v]u) (FV (u) \ {x}) FV (v) Proof by induction on the structure of u. Nameless Representation of Terms p.3/29

4 Proof (2) Case u = x: Then [x v]u = v, and FV (v) FV (u) \ {x} FV (v) Case u = y, where y x: Then [x v]u = y, and FV (u) = FV (y) = {y} ({y} \ {x}) FV (v) = (FV (u) \ {x}) FV (v) Nameless Representation of Terms p.4/29

5 Proof (3) Case u = λy. t, where y x: Then [x v]u = λy. [x v]t By the IH, FV ([x v]t) (FV (t) \ {x}) FV (v). So FV ([x v]u) = FV (λy. [x v] t) = FV ([x v] t) \ {y} ((FV (t) \ {x}) FV (v)) \ {y} (FV (t) \ {x} \ {y}) FV (v) = (FV (t) \ {y} \ {x}) FV (v) = (FV (u) \ {x}) FV (v) Nameless Representation of Terms p.5/29

6 Proof (4) Case u = t 1 t 2 : Exercise. Nameless Representation of Terms p.6/29

7 Now on to the main topic... Nameless Representation of Terms Nameless Representation of Terms p.7/29

8 Representing Terms t ::= x λx. t t 1 t 2 Choosing a concrete way to represent terms is necessary when using computers to work with λ-terms. Implementing programming language evaluators. Writing machine-checkable definitions and proofs of theorems. Nameless Representation of Terms p.8/29

9 Variable Capture [x λy. z](λz. x) λz. λy. z How can we be sure that our implementation doesn t make this mistake? Nameless Representation of Terms p.9/29

10 Idea: Rename During Substitution Rename z to z before applying substitution. [x λy. z](λz. x) = λz. λy. z Nameless Representation of Terms p.10/29

11 Idea: Barendregt Convention We can make sure our terms never use the same variable name twice. So we must always start with [x λy. z](λz. x) Nameless Representation of Terms p.11/29

12 Idea: Barendregt Convention We can make sure our terms never use the same variable name twice. So we must always start with [x λy. z](λz. x) But then what happens here? [x λy. z](λz. x x) Nameless Representation of Terms p.11/29

13 More Extreme Proposals Explicit Substitutions: Make substitutions part of the syntax and encode renaming into the evaluation rules. Combinators: Find a language with applications but no variables or binding, and translate terms to this langauge. Nameless Representation of Terms p.12/29

14 Devise Canonical Representation Maybe we can think of a unique representation for α-equivalent terms. Nameless Representation of Terms p.13/29

15 Devise Canonical Representation Maybe we can think of a unique representation for α-equivalent terms. For λx. λy. x (y x) we could write λ. λ. 1 (0 1) Is this representation unique? Nameless Representation of Terms p.13/29

16 Devise Canonical Representation Maybe we can think of a unique representation for α-equivalent terms. For λx. λy. x (y x) we could write λ. λ. 1 (0 1) Is this representation unique? What about free variables? Nameless Representation of Terms p.13/29

17 Formal Definition of de Bruijn Terms We will define a family of sets T n so that the set T i can represent terms with at most i free variables. 0 k < n k T n t T n n > 0 λ.t T n 1 t 1 T n t 2 T n (t 1 t 2 ) T n Nameless Representation of Terms p.14/29

18 Free Variables What do we do with y? λx. y x Nameless Representation of Terms p.15/29

19 Free Variables What do we do with y? λx. y x We need some sort of context of definitions, for example Γ = x 4,y 3,z 2,a 1,b 0 Then we should be able to define a function db Γ, such that db Γ (x (y z)) = 4 (3 2) db Γ (λx. y x) = λ. 4 0 Nameless Representation of Terms p.15/29

20 Naming Contexts Let s simplify Γ to be a sequence of variable names. Γ = x n 1,...,x 1,x 0 Then we ll define dom(γ) = {x n 1,...,x 1,x 0 } And Γ(x) = rightmost index of x in Γ Nameless Representation of Terms p.16/29

21 Converting to Nameless Representation db Γ (x) = Γ(x) db Γ (λx. t) = λ.db Γ,x (t) db Γ (t 1 t 2 ) = db Γ (t 1 ) db Γ (t 2 ) Nameless Representation of Terms p.17/29

22 Converting to Nameless Representation db Γ (x) = Γ(x) db Γ (λx. t) = λ.db Γ,x (t) db Γ (t 1 t 2 ) = db Γ (t 1 ) db Γ (t 2 ) What is the type of db Γ? Nameless Representation of Terms p.17/29

23 Converting to Nameless Representation db Γ (x) = Γ(x) db Γ (λx. t) = λ.db Γ,x (t) db Γ (t 1 t 2 ) = db Γ (t 1 ) db Γ (t 2 ) What is the type of db Γ? db Γ : T λ T len(γ) Nameless Representation of Terms p.17/29

24 Conversion Example We will work with Γ = x,y,z and will convert the term λx. y x Then we have db x,y,z (λx. y x) = λ. db x,y,z,x (y x) = λ. db x,y,z,x (y) db x,y,z,x (x) = λ. 2 0 Nameless Representation of Terms p.18/29

25 Defining Substitution Nameless Representation of Terms p.19/29

26 Substitution on Nameless Terms We must define for terms in T n. But how? [k s]t Nameless Representation of Terms p.20/29

27 Substitution on Nameless Terms We must define [k s]t for terms in T n. But how? We want to guarantee db Γ ([x s]t) = [Γ(x) db Γ (s)]db Γ (t) for all Γ such that FV (s) FV (t) {x} dom(γ) Nameless Representation of Terms p.20/29

28 First Attempt { s if k = j [j s]k = k otherwise [j s](λ.t) = λ. [j s]t [j s](t 1 t 2 ) = ([j s]t 1 ) ([j s]t 2 ) Nameless Representation of Terms p.21/29

29 Counter-Example [x λz. z](x (λy. y)) Let Γ = x. db Γ ([x λz. z](x (λy. y))) = db Γ ((λz. z) (λy. y)) = (λ. 0) (λ. 0) but [Γ(x) db Γ (λz. z)]db Γ (x (λy. y)) = [0 λ. 0](0 (λ.0)) = (λ. 0) (λ. [0 λ. 0]0) = (λ. 0) (λ. λ. 0) Nameless Representation of Terms p.22/29

30 Second Attempt { s if k = j [j s]k = k otherwise [j s](λ.t) = λ. [j + 1 s]t [j s](t 1 t 2 ) = ([j s]t 1 ) ([j s]t 2 ) Nameless Representation of Terms p.23/29

31 Counter-Example [x λy. w]λz. x Let Γ = x,w. but db Γ ([x λy. w]λz. x) = db Γ (λz. λy. w) = λ. db Γ,z (λy. w) = λ. λ. db Γ,z,y (w) = λ. λ. 2 [Γ(x) db Γ (λy. w)]db Γ (λz. x) = [1 λ. 1]λ. 2 = λ. [2 λ. 1]2 = λ. λ. 1 Nameless Representation of Terms p.24/29

32 Third Attempt (Shifting) (k) = k + 1 (λ. t) = λ. (t) (t 1 t 2 ) = (t 1 ) (t 2 ) { s if k = j [j s]k = k otherwise [j s](λ.t) = λ. [j + 1 (s)]t [j s](t 1 t 2 ) = ([j s]t 1 ) ([j s]t 2 ) Nameless Representation of Terms p.25/29

33 Counter-Example [x λy. w y]λz. x Let Γ = x,w. but db Γ ([x λy. w y]λz. x) = db Γ (λz. λy. w y) = λ. λ. db Γ,z,y (w y) = λ. λ. 2 0 [Γ(x) db Γ (λy. w y)]db Γ (λz. x) = [1 λ. 1 0]λ. 2 = λ. [2 (λ. 1 0)]2 = λ. [2 λ. 2 1]2 = λ. λ. 2 1 Nameless Representation of Terms p.26/29

34 Third Attempt (Shifting with Cut-Off) { k if k < c c (k) = k + 1 if k c c (λ. t) = λ. c+1 (t) c (t 1 t 2 ) = c (t 1 ) c (t 2 ) { s if k = j [j s]k = k otherwise [j s](λ.t) = λ. [j (s)]t [j s](t 1 t 2 ) = ([j s]t 1 ) ([j s]t 2 ) Nameless Representation of Terms p.27/29

35 Generalized Shifting { k if k < c d c (k) = k + d if k c d c (λ. t) = λ. d c+1 (t) d c (t 1 t 2 ) = d c (t 1 ) d c (t 2 ) Nameless Representation of Terms p.28/29

36 Evaluation of de Bruijn Terms The evaluation rule we want is (λ. t 12 ) v ([0 1 0 (v 2 )]t 12 ) E-APPABS Nameless Representation of Terms p.29/29

37 Evaluation of de Bruijn Terms The evaluation rule we want is (λ. t 12 ) v ([0 1 0 (v 2 )]t 12 ) E-APPABS Consider this example. Let s say our context is Γ = z,y,x. db Γ ((λw. w x y) x y z) = (λ ) ( 1 0 ([0 1 0 (0)](0 1 2))) 1 2 = ( 1 0 ([0 1](0 1 2))) 1 2 = ( 1 0 (1 1 2)) 1 2 = = db Γ (x x y y z) Nameless Representation of Terms p.29/29