Administration Homework 1 due September 11 CS 611 Advanced - PDF document

Administration • Homework 1 due September 11 CS 611 Advanced Programming Languages Andrew Myers Cornell University Lecture 6: Inductive definitions 6 Sep 00 CS 611 Lecture 6 – Andrew Myers, Cornell University 2 Proofs Well-founded induction • Goal: Prove property P( e ) holds for all • In PL, want to prove various things about elements e of a set … inductively defined sets 4 • Idea: generalize predecessor relation � – expression termination, equivalence of 3 expressions – abstract syntax –natural numbers: n � n + 1 2 • Aexp, C[ ] –inductive step: show P( n ) & n � n � � P( n � ) 1 – equivalence of semantics – legal executions � c , σ � � • Well-founded relation � is any relation σ � , � c , σ � → ∗ � skip , σ �� with no infinite descending chains • What are inductively defined sets, exactly? • What is the basis for inductive proofs? –must be irreflexive, no cycles – Winskel: well-founded induction ∀ e . ( ∀ e �� e . P( e � )) � P( e ) (well-founded – Alternative: induction on proof height induction) ∀ e . P( e ) … CS 611 Lecture 6 – Andrew Myers, Cornell University 3 CS 611 Lecture 6 – Andrew Myers, Cornell University 4 Structural induction Induction on derivation • Well-founded relation � : • Inductive hypothesis for well-founded structural induction: P( e � ) for all sub- def e � e � = e is a sub-expression of e � expressions e � 1+2 if x < 2 then skip else skip • Last time, inductive hypothesis slightly x skip 1 2 stronger: P( e � ) for all e with shorter AST • To prove ( ∀ e �� e . P( e � )) � P( e ) –based on course-of-values induction –for expressions e with no predecessors –only alluded to in Winskel Ch. 4 (atoms), prove P( e ) –caveat: rarely a difference in practice –for expressions e with ≥ 1 predecessors e � , • How does this work? prove P( e ) assuming P( e � ) CS 611 Lecture 6 – Andrew Myers, Cornell University 5 CS 611 Lecture 6 – Andrew Myers, Cornell University 6 1

Inductive definitions Rule operator • Set defined inductively by set of rules (proof • Given a set of elements A assumed to be members of the set being defined, define R ( A ) system) to be elements derived by applying all rule • By consistent substitution in agreement with instances to A side conditions, rules generate rule instances x ... x with form x 1 … x n R ( A ) { x | 1 n is a rule inst. & x A } = ∀ ∈ i 1 .. n i ∈ x x • x , x i are elements of set (no meta-variables) • R ( ∅ ) = ? • Meaning: if the elements x 1 … x n are all • R ( R ( ∅ )) = ? members of the set, so is x • R ( A 1 ∪ A 2 ) = ? CS 611 Lecture 6 – Andrew Myers, Cornell University 7 CS 611 Lecture 6 – Andrew Myers, Cornell University 8 Fixed points Monotonicity x ... x R ( A ) { x | 1 n is a rule inst. & x A } = ∀ ∈ • Inductively defined set A is a fixed point i 1 .. n i ∈ x of rule operator R • If applied to larger set, R yields at least • Applying R to A should give us no new as large a set (monotonic): elements: A = R ( A ) A B R ( A ) R ( B ) ⊆ ⊆ � • Recall: fixed point of function f : D → D • Consider ∅ , R ( ∅ ), R ( R ( ∅ )), R ( R ( R ( ∅ ))), … is x ∈ D such that x = f ( x ) = R 0 ( ∅ ), R 1 ( ∅ ), R 2 ( ∅ ), R 3 ( ∅ ), … • A = R ( A ) is equation , not definition • By induction: R n ( ∅ ) ⊆ R n +1 ( ∅ ) for all n • Which fixed point of R do we want? R 0 ( ∅ ) ⊆ R 1 ( ∅ ) ⊆ R 2 ( ∅ ) ⊆ R 3 ( ∅ ), … CS 611 Lecture 6 – Andrew Myers, Cornell University 9 CS 611 Lecture 6 – Andrew Myers, Cornell University 10 Inductive definition Properties of set • The set A defined by the rules is the union • The least fixed point operator fix : ( D → D ) → D of all sets R n ( ∅ ): • Why does � n ∈ω R n ( ∅ ) give fix ( R ) ? A = � n ∈ω R n ( ∅ ) • First must show A = R ( A ), i.e. A = � n ∈ω R n ( ∅ ) = R ( � n ∈ω R n ( ∅ )) • A is the least fixed point of function R • Step 1: A ⊇ R ( A ) x 1 … x m –if x in R ( A ), some rule was applied – smallest set A such that A = R ( A ) x –recall R n ( ∅ ) increasing with n –finite (but arbitrarily large) number of applications of R –must exist n such that x 1 … x m all in R n ( ∅ ) –therefore x in R n +1 ( ∅ ), x in A –elements whose proof trees have finite height CS 611 Lecture 6 – Andrew Myers, Cornell University 11 CS 611 Lecture 6 – Andrew Myers, Cornell University 12 2

LFP property Induction on proof height A = � n ∈ω R n ( ∅ ) • Goal: P( e ) for all e in set • Step 2: A ⊆ R ( A ) • Height of derivation of R 3 ( ∅ ) –Assume x in A. Then x in R n ( ∅ ) for some n. element e is n at which R 2 ( ∅ ) – x in R ( R n -1 ( ∅ )) ⊆ R ( A ) (by monotonicity) e ∈ R n ( ∅ ) R ( ∅ ) • Inductive step: prove that ∅ – x in R ( A ) P( e ) holds for all e in P • Step 3: A is no larger than any fixed pt R n ( ∅ ) assuming it holds –Suppose B = R ( B ) for all e � in R 1 ( ∅ )… R n -1 ( ∅ ) – R n ( ∅ ) ⊆ B � R n +1 ( ∅ ) ⊆ R ( B ) = B • Course-of-values: ∀ n . ∀ e ∈ R n ( ∅ ) . P( e ) –Induction: all R n ( ∅ ) ⊆ B , so A ⊆ B • Result: ∀ e ∈ fix ( R ) . P( e ) CS 611 Lecture 6 – Andrew Myers, Cornell University 13 CS 611 Lecture 6 – Andrew Myers, Cornell University 14 3