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CIS Software Foundations Fall 2004 6 October
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CIS 500, 6 October
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Midterm 1 is next Wednesday
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Todays lecture will not be covered by the midterm. Next Monday, review class. Old exams and review questions on webpage. No recitation sections next week. New oce hours next week, watch newsgroup for details.
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CIS 500, 6 October
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Where weve been: Inductive denitions abstract syntax inference rules Proofs by structural induction Operational semantics The lambda-calculus
Plans
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Typing rules and type soundness
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CIS 500, 6 October
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Where weve been: Inductive denitions abstract syntax inference rules Proofs by structural induction Operational semantics The lambda-calculus
Plans
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Typing rules and type soundness Where were going: Simple types for the lambda-calculus Formalizing more features of real-world languages (records, datatypes, references, exceptions, etc.) Subtyping
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Objects
3-a
CIS 500, 6 October
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The Simply Typed Lambda-Calculus
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CIS 500, 6 October
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t ::=
Lambda-calculus with booleans
terms x x.t t t true false if t then t else t variable abstraction application constant true constant false conditional values x.t true false abstraction value true value false value
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v
::=
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CIS 500, 6 October
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T ::= Bool TT
Simple Types
types type of booleans types of functions
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CIS 500, 6 October
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Typing rules
true : Bool false : Bool t1 : Bool t2 : T t3 : T (T-True) (T-False) (T-If)
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if t1 then t2 else t3 : T
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CIS 500, 6 October
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Typing rules
true : Bool false : Bool t1 : Bool t2 : T t3 : T (T-True) (T-False) (T-If)
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if t1 then t2 else t3 : T (T-Var) x:T
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CIS 500, 6 October
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Typing rules
true : Bool false : Bool t1 : Bool t2 : T t3 : T (T-True) (T-False) (T-If)
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if t1 then t2 else t3 : T x:T x:T (T-Var)
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CIS 500, 6 October
7-b
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Typing rules
t1 : Bool true : Bool false : Bool t2 : T t3 : T (T-True) (T-False) (T-If)
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if t1 then t2 else t3 : T x:T x:T (T-Var)
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CIS 500, 6 October
7-c
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Typing rules
t1 : Bool true : Bool false : Bool t2 : T t3 : T (T-True) (T-False) (T-If)
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if t1 then t2 else t3 : T x:T , x:T1 x:T t2 : T2 (T-Abs) (T-Var)
x:T1 .t2 : T1 T2
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CIS 500, 6 October
7-d
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Typing rules
t1 : Bool true : Bool false : Bool t2 : T t3 : T (T-True) (T-False) (T-If)
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if t1 then t2 else t3 : T x:T , x:T1 x:T t2 : T2 (T-Abs) t2 : T11 (T-Var)
x:T1 .t2 : T1 T2 (T-App)
t1 : T11 T12
t1 t2 : T12
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CIS 500, 6 October
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Typing Derivations
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What derivations justify the following typing statements? (x:Bool.x) true : Bool f:BoolBool f:BoolBool f (if false then true else false) : Bool x:Bool. f (if x then false else x) : BoolBool
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CIS 500, 6 October
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Properties of
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As before, the fundamental property of the type system we have just dened is soundness with respect to the operational semantics.
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CIS 500, 6 October
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Properties of
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As before, the fundamental property of the type system we have just dened is soundness with respect to the operational semantics. 1. Progress: A closed, well-typed term is not stuck If t : T, then either t is a value or else t t for some t .
2. Preservation: Types are preserved by one-step evaluation If t : T and t t , then t : T.
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CIS 500, 6 October
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Same steps as before...
Proving progress
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CIS 500, 6 October
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Same steps as before...
Proving progress
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inversion lemma for typing relation canonical forms lemma progress theorem
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CIS 500, 6 October
10-a
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Typing rules again (for reference)
t1 : Bool true : Bool false : Bool t2 : T t3 : T (T-True) (T-False) (T-If)
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if t1 then t2 else t3 : T x:T , x:T1 x:T t2 : T2 (T-Abs) t2 : T11 (T-Var)
x:T1 .t2 : T1 T2 (T-App)
t1 : T11 T12
t1 t2 : T12
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CIS 500, 6 October
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Lemma: 1. If 2. If
Inversion
true : R, then R = Bool. false : R, then R = Bool. t1 : Bool and
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3. If if t1 then t2 else t3 : R, then t2 , t3 : R.
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CIS 500, 6 October
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Lemma: 1. If 2. If
Inversion
true : R, then R = Bool. false : R, then R = Bool. t1 : Bool and
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3. If if t1 then t2 else t3 : R, then t2 , t3 : R. 4. If x : R, then
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CIS 500, 6 October
12-a
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Lemma: 1. If 2. If
Inversion
true : R, then R = Bool. false : R, then R = Bool. t1 : Bool and
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3. If if t1 then t2 else t3 : R, then t2 , t3 : R. 4. If x : R, then x:R .
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CIS 500, 6 October
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Lemma: 1. If 2. If
Inversion
true : R, then R = Bool. false : R, then R = Bool. t1 : Bool and
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3. If if t1 then t2 else t3 : R, then t2 , t3 : R. 4. If 5. If x : R, then x:R . x:T1 .t2 : R, then
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CIS 500, 6 October
12-c
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Lemma: 1. If 2. If
Inversion
true : R, then R = Bool. false : R, then R = Bool. t1 : Bool and
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3. If if t1 then t2 else t3 : R, then t2 , t3 : R. 4. If 5. If x : R, then x:R .
x:T1 .t2 : R, then R = T1 R2 for some R2 with , x:T1
t 2 : R2 .
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CIS 500, 6 October
12-d
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Lemma: 1. If 2. If
Inversion
true : R, then R = Bool. false : R, then R = Bool. t1 : Bool and
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3. If if t1 then t2 else t3 : R, then t2 , t3 : R. 4. If 5. If 6. If x : R, then x:R .
x:T1 .t2 : R, then R = T1 R2 for some R2 with , x:T1 t1 t2 : R, then
t 2 : R2 .
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CIS 500, 6 October
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Lemma: 1. If 2. If
Inversion
true : R, then R = Bool. false : R, then R = Bool. t1 : Bool and
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3. If if t1 then t2 else t3 : R, then t2 , t3 : R. 4. If 5. If x : R, then x:R .
x:T1 .t2 : R, then R = T1 R2 for some R2 with , x:T1
t 2 : R2 .
6. If t1 t2 : R, then there is some type T11 such that and t2 : T11 .
t1 : T11 R
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Lemma:
Canonical Forms
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Lemma:
Canonical Forms
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1. If v is a value of type Bool, then
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CIS 500, 6 October
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Lemma:
Canonical Forms
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1. If v is a value of type Bool, then v is either true or false.
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Lemma:
Canonical Forms
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1. If v is a value of type Bool, then v is either true or false. 2. If v is a value of type T1 T2 , then
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CIS 500, 6 October
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Lemma:
Canonical Forms
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1. If v is a value of type Bool, then v is either true or false. 2. If v is a value of type T1 T2 , then v has the form x:T1 .t2 .
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CIS 500, 6 October
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Progress
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Theorem: Suppose t is a closed, well-typed term (that is, t : T for some T). Then either t is a value or else there is some t with t t . Proof: By induction
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CIS 500, 6 October
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Progress
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Theorem: Suppose t is a closed, well-typed term (that is, t : T for some T). Then either t is a value or else there is some t with t t . Proof: By induction on typing derivations.
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CIS 500, 6 October
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Progress
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Theorem: Suppose t is a closed, well-typed term (that is, t : T for some T). Then either t is a value or else there is some t with t t . Proof: By induction on typing derivations. The cases for boolean constants and conditions are the same as before. The variable case is trivial (because t is closed). The abstraction case is immediate, since abstractions are values.
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CIS 500, 6 October
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Progress
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Theorem: Suppose t is a closed, well-typed term (that is, t : T for some T). Then either t is a value or else there is some t with t t . Proof: By induction on typing derivations. The cases for boolean constants and conditions are the same as before. The variable case is trivial (because t is closed). The abstraction case is immediate, since abstractions are values. Consider the case for application, where t = t1 t2 with t2 : T11 . t1 : T11 T12 and
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CIS 500, 6 October
14-c
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Progress
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Theorem: Suppose t is a closed, well-typed term (that is, t : T for some T). Then either t is a value or else there is some t with t t . Proof: By induction on typing derivations. The cases for boolean constants and conditions are the same as before. The variable case is trivial (because t is closed). The abstraction case is immediate, since abstractions are values. Consider the case for application, where t = t1 t2 with t1 : T11 T12 and t2 T11 : . By the induction hypothesis, either t1 is a value or else it can make a step of evaluation, and likewise t2 .
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CIS 500, 6 October
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Progress
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Theorem: Suppose t is a closed, well-typed term (that is, t : T for some T). Then either t is a value or else there is some t with t t . Proof: By induction on typing derivations. The cases for boolean constants and conditions are the same as before. The variable case is trivial (because t is closed). The abstraction case is immediate, since abstractions are values. Consider the case for application, where t = t1 t2 with t1 : T11 T12 and t2 : T11 . By the induction hypothesis, either t1 is a value or else it can make a step of evaluation, and likewise t2 . If t1 can take a step, then rule E-App1 applies to t. If t1 is a value and t2 can take a step, then rule E-App2 applies. Finally, if both t1 and t2 are values, then the canonical forms lemma tells us that t1 has the form x:T11 .t12 , and so rule E-AppAbs applies to t.
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CIS 500, 6 October
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Theorem: If
Proving Preservation
t : T and t t , then t : T.
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Proof: By induction
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CIS 500, 6 October
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Theorem: If
Proving Preservation
t : T and t t , then t : T.
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Proof: By induction on typing derivations. [Which case is the hard one?]
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CIS 500, 6 October
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Theorem: If
Proving Preservation
t : T and t t , then t : T.
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Proof: By induction on typing derivations. [Which case is the hard one?] Case T-App: Given t = t 1 t2 t1 : T11 T12 t2 : T11 t : T12
T = T12 Show
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CIS 500, 6 October
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Theorem: If
Proving Preservation
t : T and t t , then t : T.
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Proof: By induction on typing derivations. [Which case is the hard one?] Case T-App: Given t = t 1 t2 t1 : T11 T12 t2 : T11
T = T12 Show t : T12 By the inversion lemma for evaluation, there are three subcases...
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CIS 500, 6 October
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Theorem: If
Proving Preservation
t : T and t t , then t : T.
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Proof: By induction on typing derivations. [Which case is the hard one?] Case T-App: Given t = t 1 t2 t1 : T11 T12 t2 : T11
T = T12 Show t : T12 By the inversion lemma for evaluation, there are three subcases... Subcase: t1 = x:T11 . t12 t2 a value v2 t = [x v2 ]t12
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CIS 500, 6 October
15-d
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Theorem: If
Proving Preservation
t : T and t t , then t : T.
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Proof: By induction on typing derivations. [Which case is the hard one?] Case T-App: Given t = t 1 t2 t1 : T11 T12 t2 : T11
T = T12 Show t : T12 By the inversion lemma for evaluation, there are three subcases... Subcase: t1 = x:T11 . t12 t2 a value v2 t = [x v2 ]t12
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Uh oh.
CIS 500, 6 October
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The Substitution Lemma
t : T and s : S, then [x s]t : T.
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Lemma: Types are preserved under substitition. If , x:S
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CIS 500, 6 October
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The Substitution Lemma
t : T and s : S, then [x s]t : T.
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Lemma: Types are preserved under substitition. If , x:S Proof: ...
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CIS 500, 6 October
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On to real programming languages...
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CIS 500, 6 October
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t ::= ... unit v ::= ... unit T ::= ... Unit
The Unit type
terms constant unit values constant unit types unit type
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New typing rules unit : Unit
t:T
(T-Unit)
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CIS 500, 6 October
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t ::= ... t1 ;t2
Sequencing
terms
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t ::= ... t1 ;t2
Sequencing
terms
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t1 t1 t1 ;t2 t1 ;t2 unit;t2 t2 t1 : Unit t 2 : T2
(E-Seq)
(E-SeqNext)
(T-Seq)
t1 ;t2 : T2
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Syntatic sugar
Derived forms
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Internal language vs. external (surface) language
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CIS 500, 6 October
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Sequencing as a derived form
t1 ;t2
def
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=
(x:Unit.t 2 ) t1 where x FV(t2 ) /
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CIS 500, 6 October
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[board]
Equivalence of the two denitions
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CIS 500, 6 October
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New syntactic forms t ::= ... t as T
Ascription
terms ascription t t v1 as T v1 t1 t1 t1 as T t1 as T (E-Ascribe) (E-Ascribe1) t1 : T t1 as T : T t:T
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New evaluation rules
New typing rules
(T-Ascribe)
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CIS 500, 6 October
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Ascription as a derived form
t as T = (x:T. x) t
def
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CIS 500, 6 October
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New syntactic forms t ::= ... let x=t in t New evaluation rules
Let-bindings
terms let binding t t let x=v1 in t2 [x v1 ]t2 t1 t1 let x=t1 in t2 let x=t1 in t2 (E-LetV)
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(E-Let) t:T (T-Let)
New typing rules t 1 : T1 , x:T1 t2 : T2
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let x=t1 in t2 : T2
CIS 500, 6 October
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t ::= ... {t,t} t.1 t.2 v ::= ... {v,v} T ::= ... T1 T2
Pairs
terms pair rst projection second projection values pair value types product type
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Evaluation rules for pairs
{v1 ,v2 }.1 v1 {v1 ,v2 }.2 v2 t1 t1 t1 .1 t1 .1 t1 t1 t1 .2 t1 .2 t1 t1 {t1 ,t2 } {t1 ,t2 } t2 t2 {v1 ,t2 } {v1 ,t2 } (E-Pair2) (E-Pair1) (E-Proj2) (E-PairBeta1) (E-PairBeta2) (E-Proj1)
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Typing rules for pai...