Nominal Syntax and Semantics

Transcription

1 Nominal Syntax and Semantics Andrew Pitts University of Cambridge Computer Laboratory APPSEM 2005, 1 - p. 1

2 How best to reconcile Mathematics of syntax syntactical issues to do with name-binding and α-conversion with a structural approach to semantics? Specifically: improved forms of structural recursion and structural induction for syntactical structures. APPSEM 2005, 1 - p. 2

3 How best to reconcile Mathematics of syntax syntactical issues to do with name-binding and α-conversion with a structural approach to semantics? Specifically: improved forms of structural recursion and structural induction for syntactical structures. Lectures provide a taster of the nominal sets model of names and name-binding. (Simplified version of [Gabbay-Pitts, 2002].) APPSEM 2005, 1 - p. 2

4 Lecture 1 Introduction: from structural to α-structural recursion. Nominal sets first look. Lecture 2 Nominal sets, continued. α-structural recursion proof sketch. Nominal signatures. Lecture 3 Extended example: NBE. Mechanization [extra]. Lecture materials available at: APPSEM 2005, 1 - p. 3

5 Lecture 1 APPSEM 2005, 1 - p. 4

6 Structural recursion and induction APPSEM 2005, 1 - p. 5

7 Structural recursion and induction position APPSEM 2005, 1 - p. 5

8 Structural recursion and induction positionality APPSEM 2005, 1 - p. 5

9 Structural recursion and induction Compositionality APPSEM 2005, 1 - p. 5

10 Structural recursion and induction Compositionality is crucial in [programming language] semantics it s preferable to give meaning to program constructions rather than just to whole programs. APPSEM 2005, 1 - p. 5

11 Structural recursion and induction In particular, as far as semantics is concerned, concrete syntax letrec f x = if x > 100 then x 10 else f ( f ( x + 11 ) ) in f ( x ) is unimportant compared to abstract syntax (ASTs): letrec f x if > x 100 x f x 101 f + x 11 APPSEM 2005, 1 - p. 5

12 Structural recursion and induction ASTs enable two fundamental (and inter-linked) tools in programming language semantics: Definition of functions on syntax by recursion on its structure. Proof of properties of syntax by induction on its structure. APPSEM 2005, 1 - p. 5

13 Concrete syntax: ASTs: Running example t ::= a t t λa.t letrec a a = t in t Λ µs.(v + (S S) + (V S) + (V V S S)) where V is some fixed, countably infinite set (of names a of variables). APPSEM 2005, 1 - p. 6

14 letrec f a = if a > 100 then a 10 else f(f(a + 11)) in f(a + 101) letrec f a if > a 100 a f a 101 f + a 11 APPSEM 2005, 1 - p. 7

15 Structural recursion for Λ [LN p 10] Given a set S and functions f V : V S f A : S S S f L : V S S f F : V V S S S, there is a unique function ˆf : Λ S satisfying ˆf a 1 = f V a 1 ˆf(t 1 t 2 ) = f A ( ˆf t 1, ˆf t 2 ) ˆf(λa 1.t 1 ) = f L (a 1, ˆf t 1 ) ˆf(letrec a 1 a 2 = t 1 in t 2 ) = f F (a 1, a 2, ˆf t 1, ˆf t 2 ) for all a 1, a 2 V and t 1, t 2 Λ. APPSEM 2005, 1 - p. 8

16 Structural recursion for Λ [LN p 10] Given a set S and functions f V : V S f A : S ĝ a 1 S= g V as 1 ĝ(t 1 t 2 ) = g A (t 1, t 2, ĝ t 1, ĝ t 2 ) f L : V S S ĝ(λa 1.t 1 ) = g L (a 1, t 1, ĝ t 1 ) f F : V V S S S, A more complicated version ( primitive recursion instead of iteration ) is derivable: ĝ(letrec a 1 a 2 = t 1 in t 2 ) = g F (a 1, a 2, t 1, t 2, ĝ t 1, ĝ t 2 ) there is a unique function ˆf : Λ S satisfying ˆf a 1 = f V a 1 ˆf(t 1 t 2 ) = f A ( ˆf t 1, ˆf t 2 ) ˆf(λa 1.t 1 ) = f L (a 1, ˆf t 1 ) ˆf(letrec a 1 a 2 = t 1 in t 2 ) = f F (a 1, a 2, ˆf t 1, ˆf t 2 ) for all a 1, a 2 V and t 1, t 2 Λ. APPSEM 2005, 1 - p. 8

17 Finite set of free variables fv t of an AST t: fv a 1 {a 1 } fv(t 1 t 2 ) (fv t 1 ) (fv t 2 ) fv(λa 1.t 1 ) (fv t 1 ) {a 1 } fv(letrec a 1 a 2 = t 1 in t 2 ) (fv t 1 ) {a 1, a 2 } (fv t 2 ) {a 1 } APPSEM 2005, 1 - p. 9

18 Finite set of free variables fv t of an AST t: fv a 1 {a 1 } fv(t 1 t 2 ) (fv t 1 ) (fv t 2 ) fv(λa 1.t 1 ) (fv t 1 ) {a 1 } fv(letrec a 1 a 2 = t 1 in t 2 ) (fv t 1 ) {a 1, a 2 } (fv t 2 ) {a 1 } defined by structural recursion using S P fin (V) (finite sets of variables), f V a 1 {a 1 }, f A (A 1, A 2 ) A 1 A 2, f L (a 1, A 1 ) A 1 {a 1 }, f F (a 1, a 2, A 1, A 2 ) (A 1 {a 1, a 2 }) (A 2 {a 1 }). APPSEM 2005, 1 - p. 9

19 Finite set of all variables var t of an AST t: var a 1 {a 1 } var(t 1 t 2 ) (var t 1 ) (var t 2 ) var(λa 1.t 1 ) (var t 1 ) {a 1 } var(letrec a 1 a 2 = t 1 in t 2 ) {a 1, a 2 } (var t 1 ) (var t 2 ) APPSEM 2005, 1 - p. 10

20 t{b/a} replace all occurrences of a with b in an AST t: a 1 {b/a} if a 1 = a then b else a 1 (t 1 t 2 ){b/a} (t 1 {b/a}) (t 2 {b/a}) (λa 1. t 1 ){b/a} λa 1 {b/a}. t 1 {b/a} (letrec a 1 a 2 = t 1 in t 2 letrec (a 1 {b/a})(a 2 {b/a}) = t 1 {b/a} in t 2 {b/a} APPSEM 2005, 1 - p. 11

21 Given a set S and functions Structural recursion for Λ f V : V S f A : S S S f L : V S S f F : V V S S S, there is a unique function ˆf : Λ S satisfying ˆf a 1 = f V a 1 ˆf(t 1 t 2 ) = f A ( ˆf t 1, ˆf t 2 ) ˆf(λa 1.t 1 ) = f L (a 1, ˆf t 1 ) ˆf(letrec a 1 a 2 = t 1 in t 2 ) = f F (a 1, a 2, ˆf t 1, ˆf t 2 ) for all a 1, a 2 V and t 1, t 2 Λ. APPSEM 2005, 1 - p. 12

22 Given a set S and functions Structural recursion for Λ f V : V S f A : S S S f L : V S S f F : V V S S S, there is a unique function ˆf : Λ S satisfying ˆf a 1 = f V a 1 ˆf(t 1 t 2 ) = f A ( ˆf t 1, ˆf t 2 ) ˆf(λa 1.t 1 ) = f L (a 1, ˆf t 1 ) ˆf(letrec a 1 a 2 = t 1 in t 2 ) = f F (a 1, a 2, ˆf t 1, ˆf t 2 ) for all a 1, a 2 V and t 1, t 2 Λ. Doesn t take binding into account! APPSEM 2005, 1 - p. 12

23 letrec f a = if a > 100 then a 10 else f(f(a + 11)) in f(a + 101) letrec f a if > a 100 a f a 101 f + a 11 APPSEM 2005, 1 - p. 13

24 letrec f a = if a > 100 then a 10 else f(f(a + 11)) in f(a + 101) letrec f a if > a 100 a f a 101 f + a 11 APPSEM 2005, 1 - p. 13

25 letrec f a = if a > 100 then a 10 else f(f(a + 11)) in f(a + 101) letrec f a if > a 100 a 10 f a 101 f + a 11 APPSEM 2005, 1 - p. 13

26 α-equivalence [LN p 11] Smallest binary relation = α on Λ closed under the rules: t 1 = α t 1 t 2 = α t 2 a V t 1 t 2 = α t 1 a = α a t 2 t 1 {a 1 /a 1} = α t 1 {a 1 /a 1 } a 1 / var(a 1, t 1, a 1, t 1 ) λa 1. t 1 = α λa 1. t 1 t 1 {a 1, a 2 /a 1, a 2 } = α t 1 {a 1, a 2 /a 1, a 2 } t 2 {a 1 /a 1} = α t 2 {a 1 /a 1 } a 1 a 2 a 1, a 2 / var(a 1, a 2, t 1, t 2, a 1, a 2, t 1, t 2 ) letrec a 1 a 2 = t 1 in t 2 = α letrec a 1 a 2 = t 1 in t 2 APPSEM 2005, 1 - p. 14

27 Exercise: prove that α-equivalence = α is transitive (and reflexive and symmetric). Smallest binary relation = α on Λ closed under the rules: t 1 = α t 1 t 2 = α t 2 a V t 1 t 2 = α t 1 a = α a t 2 t 1 {a 1 /a 1} = α t 1 {a 1 /a 1 } a 1 / var(a 1, t 1, a 1, t 1 ) λa 1. t 1 = α λa 1. t 1 t 1 {a 1, a 2 /a 1, a 2 } = α t 1 {a 1, a 2 /a 1, a 2 } t 2 {a 1 /a 1} = α t 2 {a 1 /a 1 } a 1 a 2 a 1, a 2 / var(a 1, a 2, t 1, t 2, a 1, a 2, t 1, t 2 ) letrec a 1 a 2 = t 1 in t 2 = α letrec a 1 a 2 = t 1 in t 2 [LN p 11] APPSEM 2005, 1 - p. 14

28 Abstract syntax / α Dealing with issues to do with binders and α-conversion is irritating (want to get on with more interesting aspects of semantics!) pervasive (very many languages involve binding operations; cf. POPLMark Challenge [TPHOLs 05]) difficult to formalise/mechanise without losing sight of common informal practice: APPSEM 2005, 1 - p. 15

29 Abstract syntax / α Dealing with issues to do with binders and α-conversion is irritating (want to get on with more interesting aspects of semantics!) pervasive (very many languages involve binding operations; cf. POPLMark Challenge [TPHOLs 05]) difficult to formalise/mechanise without losing sight of common informal practice: We identify expressions up to α-equivalence... APPSEM 2005, 1 - p. 15

30 Abstract syntax / α Dealing with issues to do with binders and α-conversion is irritating (want to get on with more interesting aspects of semantics!) pervasive (very many languages involve binding operations; cf. POPLMark Challenge [TPHOLs 05]) difficult to formalise/mechanise without losing sight of common informal practice: We identify expressions up to α-equivalence and then forget about it, referring to α-equivalence classes e = [t] α only via representatives, t. For example... APPSEM 2005, 1 - p. 15

31 E.g. capture-avoiding substitution (a := e)e 1 = substitute e Λ/= α for all free occurrences of a in e 1 Λ/= α, avoiding capture of free variables in e by binders in e 1. APPSEM 2005, 1 - p. 16

32 E.g. capture-avoiding substitution (a := e)a 1 if a 1 = a then e else a 1 (a := e)(e 1 e 2 ) ((a := e)e 1 )((a := e)e 2 ) (a := e)(λa 1.e 1 ) if a 1 / fv(a, e) then λa 1.(a := e)e 1 else don t care! (a := e)(letrec a 1 a 2 = e 1 in e 2 )? APPSEM 2005, 1 - p. 16

33 E.g. capture-avoiding substitution (a := e)a 1 if a 1 = a then e else a 1 (a := e)(e 1 e 2 ) ((a := e)e 1 )((a := e)e 2 ) (a := e)(λa 1.e 1 ) if a 1 / fv(a, e) then λa 1.(a := e)e 1 else don t care! (a := e)(letrec a 1 a 2 = e 1 in e 2 ) if a 1, a 2 / fv(a, e) & a 2 / fv(a 1, e 2 ) then letrec a 1 a 2 = (a := e)e 1 in (a := e)e 2 else don t care! APPSEM 2005, 1 - p. 16

34 E.g. capture-avoiding substitution (a := e)a 1 if a 1 = a then e else a 1 (a := e)(e 1 e 2 ) ((a := e)e 1 )((a := e)e 2 ) (a := e)(λa 1.e 1 ) if a 1 / fv(a, e) then λa 1.(a := e)e 1 else don t care! (a := e)(letrec a 1 a 2 = e 1 in e 2 ) if a 1, a 2 / fv(a, e) & a 2 / fv(a 1, e 2 ) then letrec a 1 a 2 = (a := e)e 1 in (a := e)e 2 else don t care! Does uniquely specify a well-defined function on α-equivalence classes, (a := e)( ) : Λ/α Λ/α, but not via an obvious, structurally recursive definition of a function ˆf : Λ Λ respecting α-equivalence. APPSEM 2005, 1 - p. 16

35 E.g. denotational semantics of Λ/α in some suitable domain D: a 1 ρ ρ(a 1 ) e 1 e 2 ρ app( e 1 ρ, e 2 ρ) λa 1.e 1 ρ fun(λd D. e 1 (ρ[a 1 d])) letrec a 1 a 2 = e 1 in e 2 ρ where - ρ ranges over environments mapping variables to elements of D - D comes equipped with continuous functions app : D D D and fun : (D D) D. APPSEM 2005, 1 - p. 17

36 E.g. denotational semantics of Λ/α in some suitable domain D: a 1 ρ ρ(a 1 ) e 1 e 2 ρ app( e 1 ρ, e 2 ρ) λa 1.e 1 ρ fun(λd D. e 1 (ρ[a 1 d])) letrec a 1 a 2 = e 1 in e 2 ρ Why is this (very standard) definition independent of the choice of bound variable a 1? APPSEM 2005, 1 - p. 17

37 E.g. denotational semantics of Λ/α in some suitable domain D: a 1 ρ ρ(a 1 ) e 1 e 2 ρ app( e 1 ρ, e 2 ρ) λa 1.e 1 ρ fun(λd D. e 1 (ρ[a 1 d])) letrec a 1 a 2 = e 1 in e 2 ρ In this case we can use ordinary structural recursion to first define denotations of ASTs and then prove that they respect α-equivalence. But is there a quicker way, working directly with ASTs/α? APPSEM 2005, 1 - p. 17

38 α-structural recursion Is there a recursion principle for Λ/α that legitimises these definitions of (a := e)( ) : Λ/α Λ/α and : Λ/α D (and many other e.g.s)? APPSEM 2005, 1 - p. 18

39 α-structural recursion Is there a recursion principle for Λ/α that legitimises these definitions of (a := e)( ) : Λ/α Λ/α and : Λ/α D (and many other e.g.s)? Yes! α-structural recursion (and induction too see lecture notes). APPSEM 2005, 1 - p. 18

40 α-structural recursion Is there a recursion principle for Λ/α that legitimises these definitions of (a := e)( ) : Λ/α Λ/α and : Λ/α D (and many other e.g.s)? Yes! α-structural recursion (and induction too see lecture notes). What about other languages with binders? APPSEM 2005, 1 - p. 18

41 α-structural recursion Is there a recursion principle for Λ/α that legitimises these definitions of (a := e)( ) : Λ/α Λ/α and : Λ/α D (and many other e.g.s)? Yes! α-structural recursion (and induction too see lecture notes). What about other languages with binders? Yes! available for any nominal signature. APPSEM 2005, 1 - p. 18

42 α-structural recursion Is there a recursion principle for Λ/α that legitimises these definitions of (a := e)( ) : Λ/α Λ/α and : Λ/α D (and many other e.g.s)? Yes! α-structural recursion (and induction too see lecture notes). What about other languages with binders? Yes! available for any nominal signature. Great. What s the catch? APPSEM 2005, 1 - p. 18

43 α-structural recursion Is there a recursion principle for Λ/α that legitimises these definitions of (a := e)( ) : Λ/α Λ/α and : Λ/α D (and many other e.g.s)? Yes! α-structural recursion (and induction too see lecture notes). What about other languages with binders? Yes! available for any nominal signature. Great. What s the catch? Need to learn a bit of possibly unfamiliar math, to do with permutations and support. APPSEM 2005, 1 - p. 18

44 Pause APPSEM 2005, 1 - p. 19

45 Concrete syntax: ASTs: Running example (reminder) t ::= a t t λa.t letrec a a = t in t Λ µs.(v + (S S) + (V S) + (V V S S)) where V is some fixed, countably infinite set (of names a of variables). APPSEM 2005, 1 - p. 20

46 Structural recursion for Λ [LN p 10] Given a set S and functions f V : V S f A : S S S f L : V S S f F : V V S S S, there is a unique function ˆf : Λ S satisfying ˆf a 1 = f V a 1 ˆf(t 1 t 2 ) = f A ( ˆf t 1, ˆf t 2 ) ˆf(λa 1.t 1 ) = f L (a 1, ˆf t 1 ) ˆf(letrec a 1 a 2 = t 1 in t 2 ) = f F (a 1, a 2, ˆf t 1, ˆf t 2 ) for all a 1, a 2 V and t 1, t 2 Λ. APPSEM 2005, 1 - p. 21

47 α-structural recursion for Λ/α [LN p 31] Given a nominal set X f V : V X f A : X X X and functions f L : V X X f F : V V X X X, all supported by a finite subset A V, there is a unique function ˆf : Λ/α X such that... APPSEM 2005, 1 - p. 22

48 α-structural recursion for Λ/α [LN p 31]...! function ˆf : Λ/α X such that: ˆf a 1 = f V a 1 ˆf(e 1 e 2 ) = f A ( ˆf e 1, ˆf e 2 ) a 1 / A ˆf(λa 1.e 1 ) = f L (a 1, ˆf e 1 ) a 1, a 2 / A & a 1 a 2 & a 2 / fv(e 2 ) ˆf(letrec a 1 a 2 = e 1 in e 2 ) = f F (a 1, a 2, ˆf e 1, ˆf e 2 ) for all a 1, a 2 V & e 1, e 2 Λ/α, APPSEM 2005, 1 - p. 22

49 α-structural recursion for Λ/α [LN p 31]...! function ˆf : Λ/α X such that: ˆf a 1 = f V a 1 ˆf(e 1 e 2 ) = f A ( ˆf e 1, ˆf e 2 ) a 1 / A ˆf(λa 1.e 1 ) = f L (a 1, ˆf e 1 ) a 1, a 2 / A & a 1 a 2 & a 2 / fv(e 2 ) ˆf(letrec a 1 a 2 = e 1 in e 2 ) = f F (a 1, a 2, ˆf e 1, ˆf e 2 ) provided freshness condition for binders (FCB) holds for f L : ( a 1 / A)( x X) a 1 # f L (a 1, x) for f F : ( a 1, a 2 / A) a 1 a 2 & ( x 1, x 2 X) a 2 # x 1 a 1, a 2 # f F (a 1, a 2, x 1, x 2 ) APPSEM 2005, 1 - p. 22

50 α-structural recursion for Λ/α [LN p 31]. The..! freshness function relation ˆf : Λ/α ( ) X# such ( ) between that: names and elements of nominal ˆf asets 1 = generalises f V a 1 the ( ) / fv( ) relation ˆf(ebetween 1 e 2 ) = variables f A ( ˆf e 1, ˆf and e 2 ) ASTs. E.g. for a 1 the / Acapture-avoiding ˆf(λa 1.e 1 ) = substitution f L (a 1, ˆf e 1 ) example, fa L 1 (a, a 12, e) / A λa & a 1.e 1 and a 2 (FCB) & a 2 / holds fv(e 2 trivially ) because aˆf(letrec a a 2 = e 1 in e 2 ) = f F (a 1, a 2, ˆf e 1, ˆf 1 / fv(λa 1.e) (and similarly for f F ). e 2 ) provided freshness condition for binders (FCB) holds for f L : ( a 1 / A)( x X) a 1 # f L (a 1, x) for f F : ( a 1, a 2 / A) a 1 a 2 & ( x 1, x 2 X) a 2 # x 1 a 1, a 2 # f F (a 1, a 2, x 1, x 2 ) APPSEM 2005, 1 - p. 22

51 To be explained: Nominal sets, support and the freshness relation, ( ) # ( ). How is α-structural recursion proved? How to generalise α-structural recursion from the example language Λ to general languages with binders? What s involved with applying α-structural recursion in any particular case? Example: normalisation by evaluation. Machine-assisted support? APPSEM 2005, 1 - p. 23

52 Atoms [LN p 7] From now on assume bindable names in ASTs are drawn from a fixed, countably infinite set A (elements called atoms) Need different flavours of names (variables, references, channels, nonces,... ), so assume A is partitioned into countably infinite number of sorts. Write sort(a) for the sort of a A. There are infinitely many atoms of each sort. APPSEM 2005, 1 - p. 24

53 Atoms [LN p 7] From now on assume bindable names in ASTs are drawn from a fixed, countably infinite set A (elements called atoms) The mathematical model of bindable names we use is very abstract: in the world of nominal sets, the only attributes of an atom are identity and sort. Probably interesting & pragmatically useful to consider more structured atoms (!), e.g. linearly ordered ones, but we don t do that here. APPSEM 2005, 1 - p. 24

54 Permutations [LN p 12] Set Perm of atom-permutations consists of all bijections π : A A such that {a A π(a) a} is finite sort(π(a)) = sort(a) (all a A). Perm is a group: multiplication π π = function composition π π : π π (a) π(π (a)) identity element ι = identity function on A inverse π 1 of π Perm is inverse qua function. APPSEM 2005, 1 - p. 25

55 Permutations [LN p 12] Set Perm of atom-permutations consists of all bijections π : A A such that {a A π(a) a} is finite sort(π(a)) = sort(a) (all a A). Given a 1, a 2 A with sort(a 1 ) = sort(a 2 ), transposition (a 1 a 2 ) is the π Perm given by a 2 if a = a 1 π(a) a 1 if a = a 2 a otherwise APPSEM 2005, 1 - p. 25

56 Permutations [LN p 12] Set Perm of atom-permutations consists of all bijections π : A A such that {a A π(a) a} is finite sort(π(a)) = sort(a) (all a A). Given a 1, a 2 A with sort(a 1 ) = sort(a 2 ), transposition (a 1 a 2 ) is the π Perm given by Exercise: prove that every a 2 πif a Perm = a 1 can be expressed as aπ(a) composition a 1 of if a (finitely = a 2 many) transpositions. a otherwise APPSEM 2005, 1 - p. 25

57 Actions of permutations An action of Perm on a set S is a function Perm S S written (π, s) π s satisfying ι s = s and π (π s) = (ππ ) s A Perm-set is a set S equipped with an action of Perm on S. APPSEM 2005, 1 - p. 26

58 Actions of permutations An action of Perm on a set S is a function Perm S S written (π, s) π s satisfying ι s = s and π (π s) = (ππ ) s Three simple examples of Perm-sets: Natural numbers N with trivial action: π n = n. A with action: π a = π(a). Perm itself with conjugation action: π π = π π π 1. More examples in a mo. APPSEM 2005, 1 - p. 26

59 Finite support [LN p 12] Definition. A finite subset A A supports an element s S of a Perm-set S if (a a ) s = s holds for all a, a A (of same sort) not in A A nominal set is a Perm-set all of whose elements have a finite support. APPSEM 2005, 1 - p. 27

60 Lemma. If s S has a finite support, then it has a smallest one, written supp(s). APPSEM 2005, 1 - p. 28

61 Lemma. If s S has a finite support, then it has a smallest one, written supp(s). Proof Suffices to show that if the finite sets A 1 and A 2 support s, so does A 1 A 2. APPSEM 2005, 1 - p. 28

62 Lemma. If s S has a finite support, then it has a smallest one, written supp(s). Proof Suffices to show that if the finite sets A 1 and A 2 support s, so does A 1 A 2. So given a, a A (A 1 A 2 ), have to show (a a ) s = s. W.l.o.g. a a. APPSEM 2005, 1 - p. 28

63 Lemma. If s S has a finite support, then it has a smallest one, written supp(s). Proof Suffices to show that if the finite sets A 1 and A 2 support s, so does A 1 A 2. So given a, a A (A 1 A 2 ), have to show (a a ) s = s. W.l.o.g. a a. Pick any a (of same sort) in the infinite set A (A 1 A 2 {a, a }). Then (a a ) = (a a ) (a a ) (a a ) is a composition of transpositions each of which fixes s. APPSEM 2005, 1 - p. 28

64 Freshness relation [LN P 18] Given nominal sets X and Y and elements x X and y Y, write x # y to mean supp(x) supp(y) =. APPSEM 2005, 1 - p. 29

65 Freshness relation [LN P 18] Given nominal sets X and Y and elements x X and y Y, write x # y to mean supp(x) supp(y) =. So if a A, then a # x means a / supp(x). APPSEM 2005, 1 - p. 29

66 Freshness relation [LN P 18] Given nominal sets X and Y and elements x X and y Y, write x # y to mean supp(x) supp(y) =. So if a A, then a # x means a / supp(x). Hence Key fact for atoms a and a of the same sort: (a, a ) # x (a a ) x = x APPSEM 2005, 1 - p. 29

67 Languages/α form nominal sets [LN p 13] For example, there s a Perm-action on Λ/α satisfying: π a = π(a) π (e 1 e 2 ) = (π e 1 )(π e 2 ) π (λa.e) = λπ(a).(π e) π (letrec a 1 a 2 = e 1 in e 2 ) = letrec π(a 1 ) π(a 2 ) = π e 1 in π e 2 APPSEM 2005, 1 - p. 30

68 Languages/α form nominal sets [LN p 13] For example, there s a Perm-action on Λ/α satisfying: π a = π(a) π (e 1 e 2 ) = (π e 1 )(π e 2 ) π (λa.e) = λπ(a).(π e) π (letrec a 1 a 2 = e 1 in e 2 ) = letrec π(a 1 ) π(a 2 ) = π e 1 in π e 2 N.B. binding and non-binding constructs are treated just the same APPSEM 2005, 1 - p. 30

69 Languages/α form nominal sets [LN p 13] For example, there s a Perm-action on Λ/α satisfying: π a = π(a) π (e 1 e 2 ) = (π e 1 )(π e 2 ) π (λa.e) = λπ(a).(π e) π (letrec a 1 a 2 = e 1 in e 2 ) = letrec π(a 1 ) π(a 2 ) = π e 1 in π e 2 Proof (exercise) First define π ( ) : Λ Λ by structural recursion, and then prove that t = α t ( π Perm) π t = α π t. APPSEM 2005, 1 - p. 30

70 Languages/α form nominal sets [LN p 13] For example, there s a Perm-action on Λ/α satisfying: π a = π(a) π (e 1 e 2 ) = (π e 1 )(π e 2 ) π (λa.e) = λπ(a).(π e) π (letrec a 1 a 2 = e 1 in e 2 ) = letrec π(a 1 ) π(a 2 ) = π e 1 in π e 2 For this action, it is not hard to see (exercise) that e Λ/α is supported by any finite set of variables containing all those occurring free in e and hence a # e iff a / fv(e). APPSEM 2005, 1 - p. 30

71 End of lecture 1 APPSEM 2005, 1 - p. 31