1. 1 Semantics Q1 2007 S EMANTICS (Q1,’07) Week 4 Jacob Andersen PhD student andersen@daimi.au.dk

2. 2 Semantics Q1 2007 News…

3. 3 Semantics Q1 2007 Outline (Structural) Induction Finite vs. Infinite Stores Functional Languages (Let) Compound Definitions (Let++) Type-Checking Definitions (Let B ++)

4. 4 Semantics Q1 2007 (S TRUCTURAL ) I NDUCTION

5. 5 Semantics Q1 2007 Principle of Mathematical Induction Let P be a predicate (i.e. a boolean function): then we have that: Intuitive: ? P: N  { true, false }  n  N : P(n) P(0)  induction stepbase case Principle of mathematical induction: P(n)  P(n+1)  P(3) P(0)P(0) => P(1)P(1) => P(2)P(2) => P(3) 

6. 6 Semantics Q1 2007 Mathematical Induction example Prove the following theorem: All non-empty subsets of ℕ have a smallest element. or by contraposition: Live exercise… [Think 3 mins; then interactively on the whiteboard] ∀ A ⊆ ℕ : A ≠ ∅  (∃ l ∈ A : ∀ a ∈ A : l ≤ a ) ∀ A ⊆ ℕ : (∀ l ∈ A : ∃ a ∈ A : l>a )  A = ∅ Let A ⊆ ℕ be given. Assume: ∀ l ∈ A : ∃ a ∈ A : l>a Prove: A = ∅

7. 7 Semantics Q1 2007 Mathematical Induction example What about: Will work ??? Instead this will work: ∀ A ⊆ ℕ : (∀ l ∈ A : ∃ a ∈ A : l>a )  A = ∅ Let A ⊆ ℕ be given. Assume: ∀ l ∈ A : ∃ a ∈ A : l>a Prove: A = ∅ P(n) ≝ n ∉ A P(n)  P(n+1) NO ! P(n) ≝ ∀ m ∈ ℕ : m ≤ n  m ∉ A  n  ℕ : P(n) P(0) P(n)  P(n+1)    ∀ n, m ∈ ℕ : m ≤ n  m ∉ A or:  ∀ m ∈ ℕ : m ∉ A  A ∩ ℕ = ∅ (Since A ⊆ ℕ )   A = ∅ 

8. 8 Semantics Q1 2007 Structural Induction (for Exp) Given: Arithmetic Expressions ( e  Exp) – e ::= n | v | e 0 +e 1  e  Exp : P(e) P(n)  composite (inductive) case base cases Principle of structural induction: P(e 0 )  P(e 1 )  P(e 0 +e 1 )  P(v) and

9. 9 Semantics Q1 2007 Intuition: Induction vs. Str’ Induction Induction: Holds for ? Structural Induction: Holds for ? P(0)P(0) => P(1)P(1) => P(2)P(2) => P(3)  P(3) P(7+(x+y)) P(7) P(x) P(y) P(x+y) P(7+(x+y))     

10. 10 Semantics Q1 2007 Structural Induction Examples Given: Arithmetic Expressions ( e  Exp) – Determinism of SOS* for L-expressions : “One-step evaluation of expressions is deterministic” e ::= n | v | e 0 +e 1 P(e)   e’,e’’:[(  | _ e  e’   | _ e  e’’) => e’= e’’]  e: P(e) */ using the (side-effectless) small-step operational semantics Let  be given: Prove: (i.e., “prove property, P, holds for all expressions”)

11. 11 Semantics Q1 2007 Proof Structure (according to SI) Base cases: –Case [ n ]; show : –Case [ v ]; show : Inductive case: –Case [ e 0 +e 1 ]; show : –assume induction hypothesis [lhs]: P(e 0 )  P(e 1 ) –show inductive step [rhs]: P(e 0 +e 1 ) P(n)   e’,e’’:(  | _ n  e’   | _ n  e’’) => e’=e’’ P(v)   e’,e’’:(  | _ v  e’   | _ v  e’’) => e’=e’’ P(e 0 )   e’,e’’:(  | _ e 0  e’   | _ e 0  e’’) => e’=e’’ P(e 1 )   e’,e’’:(  | _ e 1  e’   | _ e 1  e’’) => e’=e’’ P(e 0 +e 1 )   e’,e’’:(  | _ e 0 +e 1  e’   | _ e 0 +e 1  e’’) => e’=e’’ P(n) P(v) P(e 0 )  P(e 1 ) => P(e 0 +e 1 )

12. 12 Semantics Q1 2007 Proof: Base Case [n] –Case [ n ]; show : –Show implication “ => ” (assume [lhs], show [rhs]): However, since [lhs] is trivially false (no rules for constants), the whole thing is trivially true (aka. “vacuously true”) »Recall that: P(n)   e’,e’’:(  | _ n  e’   | _ n  e’’ ) => e’ = e’’ P(n)   b : (false => b)  true P  Q   P  Q PQ P  Q false true falsetrue false true

13. 13 Semantics Q1 2007 Proof: Base Case [v] –Case [ v ]; show : –Show implication “ => ” (assume [lhs], show [rhs]): assume [lhs]: and for some e’,e’’ show [rhs]: – could only be because ; thus, e’ =  (v) …and similarly: – could only be because ; thus, e’’ =  (v) –Thus, we have that: e’ =  (v) = e’’ (as required)  P(v)   e’,e’’:(  | _ v  e’   | _ v  e’’) => e’=e’’ P(v)  | _ v  e’ e’ = e’’  | _ v  e’’  | _ v  e’  | _ v   (v)  | _ v  e’’  | _ v   (v)

14. 14 Semantics Q1 2007 Proof: Inductive Case [e 0 +e 1 ] –Case [ e 0 +e 1 ]; show : –assume induction hypothesis: P(e 0 )  P(e 1 ) –show inductive step: P(e 0 +e 1 ) –Show implication “ => ” (assume [lhs], show [rhs]): Assume [lhs]: and for some e’,e’’ Show [rhs]: When could we have that: (?) –Case Analysis (on e 0 +e 1 ) »Three possibilities ( [SUM 1 ], [SUM 2 ], and [SUM 3 ] )… P(e 0 )   e’,e’’:(  | _ e 0  e’   | _ e 0  e’’) => e’=e’’ P(e 1 )   e’,e’’:(  | _ e 1  e’   | _ e 1  e’’) => e’=e’’ P(e 0 +e 1 )   e’,e’’:(  | _ e 0 +e 1  e’   | _ e 0 +e 1  e’’) => e’=e’’ P(e 0 )  P(e 1 ) => P(e 0 +e 1 )  | _ e 0 +e 1  e’  | _ e 0 +e 1  e’’ e’ = e’’  | _ e 0 +e 1  e

15. 15 Semantics Q1 2007 Proof: Case Analysis (on e 0 +e 1 ) –Case Analysis (on e 0 +e 1 ): Case [SUM 1 (e 0  Z)]: – could only be because –But from our induction hypothesis, P(e 0 ) ; i.e. … : … we get that: » and implies that: e 0 ’ = e 0 ” »Thus we have that: e’ = e 0 ’+e 1 = e 0 ”+e 1 = e’’ (as required)  | _ e 0 +e 1  e’  |_ e0  e0’ |_ e0  e0’  | _ e 0 +e 1  e 0 ’+e 1 = e’  |_ e0  e0” |_ e0  e0”  | _ e 0 +e 1  e 0 ”+e 1 = e’’  | _ e 0 +e 1  e’’   |_ e0  e0’ |_ e0  e0’  |_ e0  e0” |_ e0  e0” P(e 0 )   e’,e’’:(  | _ e 0  e’   | _ e 0  e’’) => e’=e’’

16. 16 Semantics Q1 2007 Proof: Case Analysis (on e 0 +e 1 ) Case [SUM 2 (e 0  Z, e 1  Z)]: – (analoguous) Case [SUM 3 (e 0  Z, e 1  Z)]: – could only be because –Thus, we have that: e’ = m = n 0 +n 1 = m’ = e’’ (as required) Q.E.D.  | _ n 0 +n 1  e’  | _ n 0 +n 1  e’’  | _ n 0 +n 1  m = e’  | _ n 0 +n 1  m’ = e’’ m = n 0 +n 1 m’ = n 0 +n 1   

17. 17 Semantics Q1 2007 What did we do: Proof Structure Base cases: –Case [ n ]; showed : –Case [ v ]; showed : Inductive case: –Case [ e 0 +e 1 ]; showed : –assumed induction hypothesis [lhs]: P(e 0 )  P(e 1 ) –showed inductive step [rhs] via case analysis: P(e 0 +e 1 ) P(n)   e’,e’’:(  | _ n  e’   | _ n  e’’) => e’=e’’ P(v)   e’,e’’:(  | _ v  e’   | _ v  e’’) => e’=e’’ P(e 0 )   e’,e’’:(  | _ e 0  e’   | _ e 0  e’’) => e’=e’’ P(e 1 )   e’,e’’:(  | _ e 1  e’   | _ e 1  e’’) => e’=e’’ P(e 0 +e 1 )   e’,e’’:(  | _ e 0 +e 1  e’   | _ e 0 +e 1  e’’) => e’=e’’ P(n) P(v) P(e 0 )  P(e 1 ) => P(e 0 +e 1 )

18. 18 Semantics Q1 2007 I NFINITE vs. F INITE S TORES

19. 19 Semantics Q1 2007 Problem: Infinite Stores At this point Plotkin starts using stores with no upper bound. Problem: (Var infinite) –Solution: (for V  FIN Var) –Now define finite configurations: We shall not worry about this (like TM vs. DFA)   Store = Var  Z  v  Store V = V  Z  V,E := { | Var E ( e )  V    Store V } appropriate store all vars in “V”

20. 20 Semantics Q1 2007 D EFINITIONS (FUNCTIONAL LANGs)

21. 21 Semantics Q1 2007 Introducing Binding Constructs Consider the language ”Let”: Expressions ( e  Exp): –Example Let program: –Note: the actual identifier names are irrellevant (only the use-def links are significant) e ::= n | x | e  e’ | let x = e in e’   { +, -, * } …where: x + let x = 5 in x + let x = 1 + x in x * x Definition (def) Usage (use): Free use Bound use use-def link Terminology definition scope for ' x '

22. 22 Semantics Q1 2007 Free Variables Determine the free variables? FV: Exp  P(Var) –FV( n ) = Ø –FV( x ) = { x } –FV( e  e’ ) = FV( e )  FV( e’ ) –FV( let x=e in e’ ) = FV( e )  ( FV( e’ ) \ { x } ) e ::= n | x | e  e’ | let x = e in e'

23. 23 Semantics Q1 2007 Dynamic Semantics (Trans. Sys.) Define environments: recall terminology (no side-effects) Configurations: i.e., numbers Transition relation: ; writing for  = { } T = { | z  Z }   Env = Var  Z Note: this is done differently in the note [SOS]: through “relative transition systems” that are (implicitly) relative to an environment.           |- e  e (, )  ‘  ’

24. 24 Semantics Q1 2007 Dynamic Semantics [n] | [x] | [e  e] Structural Operational Semantics: [ BOP 1 ] [ BOP 2 ] [ BOP 3 ] [ VAR ]  |- e 0  e 1  e 0 ’  e 1  |- e 0  e 0 ’  |- z 0  z 1  z z = z 0  z 1  |- z 0  e 1  z 0  e 1 ’  |- e 1  e 1 ’  |- x  z z =  (x)

25. 25 Semantics Q1 2007 Dynamic Semantics [ let-in ] Structural Operational Semantics: »Note: the premises work in the extended environment Live exercise: eval in env: [ LET 2 ] [ LET 3 ] [ LET 1 ]  |- let x = e 0 in e 1  let x = e 0 ’ in e 1  |- e 0  e 0 ’  |- let x = z in n  n  |- let x = z in e 1  let x = z in e 1 ’  [x  z] |- e 1  e 1 ’ let x=y in x [y7][y7]

26. 26 Semantics Q1 2007 C OMPOUND D EFINITIONS

27. 27 Semantics Q1 2007 Compound Definitions Consider language Let++: –Expressions ( e  Exp): –Definitions (d  Def): e ::= n | x | e  e’ | let d in e d ::= nil | x = e | d ; d’ | d and d’ | d in d’ x + 3y = x+7 Graphically:  Z Z  Env [x1][x1] [x1][x1][y8][y8] 4   { +, -, * } …where:

28. 28 Semantics Q1 2007 Graphically: [x = e] | [d ; d’] Definition block: –E.g.: Sequential definition: –E.g.: let d in e d 0 ; d 1 let x=a ; y=b in b * c + y a b c x y b c >

29. 29 Semantics Q1 2007 Graphically: [d and d’] | [d in d’] Simultaneous definition: –E.g.: Private definition: –E.g.: d 0 and d 1 d 0 in d 1 >

30. 30 Semantics Q1 2007 C OMPOUND D EFINITIONS Semantics of Let++

31. 31 Semantics Q1 2007 Static Semantics: Let++ Problem: I.e. non-disjoint simultaneous definitions! Solution: static semantics def / exp “well-formedness relations”: ‘|- wfd ’ / ‘|- wfe ’: let x = 1 and x = 2 in x | _ wfe e | _ wfe e’ | _ wfe e  e’ Exp well-formedness: | _ wfe n | _ wfe x | _ wfd d | _ wfe e | _ wfe let d in e  Definition well-formedness: | _ wfd nil | _ wfe e | _ wfd x = e | _ wfd d | _ wfd d’ | _ wfd d ; d’ | _ wfd d | _ wfd d’ | _ wfd d and d’ | _ wfd d | _ wfd d’ | _ wfd d in d’ DV( d )  DV( d’ ) = Ø

32. 32 Semantics Q1 2007 Recall: Free Variables for Let Let: –FV( n ) = Ø –FV( x ) = { x } –FV( e  e’ ) = FV( e )  FV( e’ ) –FV( let x=e in e’ ) = FV( e )  ( FV( e’ ) \ { x } ) Let++: –FV( let d in e’ ) = FV( d )  ( FV( e’ ) \ DV( d ) ) –So we need FV( d ) and DV( d ) …

33. 33 Semantics Q1 2007 Defining Variables and Free Variables DV: Def  P(Var) –DV( nil ) = Ø –DV( x = e )= { x } –DV( d ; d’ ) = DV( d )  DV( d’ ) –DV( d and d’ )= DV( d )  DV( d’ ) –DV( d in d’ )= DV( d’ ) FV: Def  P(Var) –FV( nil ) = Ø –FV( x = e )= FV( e ) –FV( d ; d’ ) = FV( d )  ( FV( d’ ) \ DV( d ) ) –FV( d and d’ )= FV( d )  FV( d’ ) –FV( d in d’ )= FV( d )  ( FV( d’ ) \ DV( d ) ) + should be disjoint union d ::= nil | x = e | d ; d’ | d and d’ | d in d’

34. 34 Semantics Q1 2007 Dynamic Semantics Expression Configurations: –Problem: what should definitions evaluate to? Environments (like expressions evaluate to numbers) –Solution: “environments as results” (add intermediate syntax): Definition Configurations:  E = { | |- wfe e } T E = { | z  Z } only well-formed exps d ::= nil | x = e | d ; d’ | d and d’ | d in d’ |  0  D = { | |- wfd d } T D = { } only well-formed defs Question: Why not (just) define:  D = { } ∪ Env T D = Env ??? Note: This is a static property!

35. 35 Semantics Q1 2007 Dynamic Semantics: Exp Structural Operational Semantics: »Note: the premisis works in the  0 - extended environment [ LET 2 ] E [ LET 3 ] E [ LET 1 ] E  |- let d in e  let d’ in e  |- d  d’  |- let  0 in n  n  |- let  0 in e  let  0 in e’  [  0 ] |- e  e’

36. 36 Semantics Q1 2007 Structural Operational Semantics: [ NIL ] D  |- nil  [] [ DEF 1 ] D  |- x = e  x = e’  |- e  e’ [ DEF 2 ] D  |- x = n  [x  n] Dynamic Sem.: Def [nil] | [x=e]

37. 37 Semantics Q1 2007 Dynamic Sem.: Def [d;d’] Structural Operational Semantics: »Note: the premises works in the  0 - extended environment »Note: the result is the combined environment,  0 [  1 ] [ SEQ 2 ] D [ SEQ 3 ] D [ SEQ 1 ] D  |- d 0 ; d 1  d 0 ’ ; d 1  |- d 0  d 0 ’  |-  0 ;  1   0 [  1 ]  |-  0 ; d 1   0 ; d 1 ’  [  0 ] |- d 1  d 1 ’ Live exercise: evaluate in environment: x = 2 ; x = x+y [ y = 1]

38. 38 Semantics Q1 2007 Dynamic Sem.: Def [d and d’] Structural Operational Semantics: »Notice the independence of the two operands ( d 0 & d 1 ) »Note: the result is the disj. combined environment,  0  1 [ AND 2 ] D [ AND 3 ] D [ AND 1 ] D  |- d 0 and d 1  d 0 ’ and d 1  |- d 0  d 0 ’  |-  0 and  1   0  1  |-  0 and d 1   0 and d 1 ’  |- d 1  d 1 ’ Q: what happens if operands do not contain disjoint definitions (e.g. "x=1 and x=2") A: trans. sys. was only def'd for well-formed exps/defs! Hence, no such runtime errors!

39. 39 Semantics Q1 2007 Dynamic Sem.: Def [d in d’] Structural Operational Semantics: »Note: the premises works in the  0 - extended environment »Note: the result is only the last environment,  1 [ IN 2 ] D [ IN 3 ] D [ IN 1 ] D  |- d 0 in d 1  d 0 ’ in d 1  |- d 0  d 0 ’  |-  0 in  1   1  |-  0 in d 1   0 in d 1 ’  [  0 ] |- d 1  d 1 ’

40. 40 Semantics Q1 2007 T YPE C HECKING D EFINITIONS

41. 41 Semantics Q1 2007 Adding Boolean Variables: Let B ++ Consider language Let B ++: –Mixed Expressions ( e  Exp): –Definitions (d  Def): –Note: The type of a var is now context dependent I.e. a context-free (grammar) approach will not suffice –Examples: ? ? … e ::= n | t | x | ~ e | e  e’ | e ? e’ : e” | let d in e d ::= nil |  x = e | d ; d’ | d and d’ | d in d’   { +, -, *, =, or } …where: x or tt x * x type definition (annotation)   { bool, int } …where:

42. 42 Semantics Q1 2007 Static Semantics: Let B ++ FV E, DV D and FV D as before …adding: We now need type environments: Type = { bool, int } Define static semantics (type checking rel’s): “ e has type  (given type environment  )” “ d yields type env.  (given type env.  )” FV E ( e ? e’: e” ) = FV E ( e )  FV E ( e’ )  FV E ( e” ) TEnv V = Var  Type  |- e :   |- d : 

43. 43 Semantics Q1 2007 Static Semantics (expressions) Expressions: [ NUM ] E  |- n : int  |- t : bool [ TVL ] E [ VAR ] E  |- x :  (x) [ VAR ] E  |- ~ e : bool  |- e : bool [ LET ] E  |- let d in e :   |- d :   [  ] |- e :  [ IFE ] E  |- e 0 ? e 1 : e 2 :   |- e 0 : bool  |- e 1 :  1  |- e 2 :  2  = 1 =2 = 1 =2 [ BOP ] E  |- e 0  e 1 : type  (  0,  1 )  |- e 0 :  0  |- e 1 :  1

44. 44 Semantics Q1 2007 Static Semantics (definitions) Definitions: [ NIL ] D  |- nil : [] [ DEF ] D  |-  x = e : [x  ]  |- e :  ’  = ’ = ’ [ SEQ ] D  |- d 0 ; d 1 :  0 [  1 ]  |- d 0 :  0  [  0 ] |- d 1 :  1 [ AND ] D  |- d 0 and d 1 :  0  1  |- d 0 :  0  |- d 1 :  1 [ IN ] D  |- d 0 in d 1 :  1  |- d 0 :  0  [  0 ] |- d 1 :  1 DV( d 0 )  DV( d 1 ) = Ø Note: combined environment Note: only last environment Note: disjoint environment Note: type check

45. 45 Semantics Q1 2007 Dynamic Semantics: Let B ++ Type Environment (Compile-time): Type = { bool, int } Environment (Runtime): Val = B  Z, B = { tt, ff } –Define type correspondence relation ‘~’: –Note:   TEnv = Var  Type   Env = Var  Val  ~    x  V:  ( x ) = bool   ( x )  B  ( x ) = int   ( x )  Z (  0 ~  0   1 ~  1 )   0 [  1 ] ~  0 [  1 ]

46. 46 Semantics Q1 2007 Exp. Transition System for Let B ++ Let++ Expression Configurations: Let B ++ Expression Configurations: –   Var  Type: Env  := »i.e., “only type corresponding runtime environments”  = { | |- wfe e } T = { | z  Z } only well-formed exps   = { |   Env    :  |- e :  } T  = { |   Env   r  B  Z } {   ( Var  B  Z ) |  ~  }

47. 47 Semantics Q1 2007 Def. Transition System for Let B ++ Let++ Definition Configurations: Let B ++ Definition Configurations: –   Var  Type: Env  := »i.e., “only type corresponding runtime environments”  V = { | |- wfd d } only well-formed defs T V = { }   = { |   Env    :  |- V d :  } T  = { |   Env  } {   ( Var  B  Z ) |  ~  }

48. 48 Semantics Q1 2007 Exp. Dynamic Semantics: Let B ++ SOS of Expressions for Let B ++: –E.g. [LET]: [ LET 2 ] [ LET 3 ] [ LET 1 ]   Env   |- let d in e   let d’ in e  |- d   d’  |- let  0 in r   r  |- let  0 in e   let  0 in e’  [  0 ] |- e   [  0] e’ 0 ~ 00 ~ 0

49. 49 Semantics Q1 2007 Def. Dynamic Semantics: Let B ++ SOS of Definitions for Let B ++: –E.g [SEQ]: [ SEQ 2 ] [ SEQ 3 ] [ SEQ 1 ]  |- d 0 ; d 1   d 0 ’ ; d 1  |- d 0   d 0 ’  |-  0 ;  1    0 [  1 ]  |-  0 ; d 1    0 ; d 1 ’  [  0 ] |- d 1   [  0] d 1 ’   Env  0 ~ 00 ~ 0

50. 50 Semantics Q1 2007 "Three minutes paper" Please spend three minutes writing down the most important things that you have learned today (now). After 1 day After 1 week After 3 weeks After 2 weeks Right away

51. 51 Semantics Q1 2007 Next week (CCS): Concurrency and Communication Any Questions?