Proof Techniques for Operational Semantics. Questions? Why Bother? Mathematical Induction Well-Founded Induction Structural Induction

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1 Proof Techniques for Operational Semantics Announcements Homework 1 feedback/grades posted Homework 2 due tonight at 11:55pm Meeting 10, CSCI 5535, Spring Plan Questions? Why Bother? Mathematical Induction Well-Founded Induction Structural Induction By induction on the structure of the derivation D 4 Mathematical Induction Goal: prove n N. P(n) Base Case: prove P(0) Inductive Step: Prove For all n. if P(n), then P(n+1) Pick arbitrary n, assume P(n), prove P(n+1) Why does induction work? 5 Example (With IMP Eval. Semantics) Prove that if σ(x) 6 then <while x 5 do x := x + 1, σ> σ[x := 6] Reformulate the claim: Let W = while x 5 do x := x + 1 Let Claim: For all For all i N. <W, σ i > σ 0 Now looks provable by mathematical induction on i Most often, the hard part! 6 1

2 Evaluation Example (Base Case) Evaluation Example (Base Case) Base case: i = 0 or <W, σ 0 > σ 0 To prove an evaluation judgment, construct a derivation tree: Base case: i = 0 or <W, σ 0 > σ 0 To prove an evaluation judgment, construct a derivation tree: σ 0 (x) = 6 σ 0 (x) = 6 <x, σ 0 > 6 <6 5, σ 0 > false <x, σ 0 > 6 <6 5, σ 0 > false <x 5, σ 0 > false <x 5, σ 0 > false <while x 5 do x := x + 1, σ 0 > σ 0 <while x 5 do x := x + 1, σ 0 > σ Evaluation Example (Inductive Case) Evaluation Example (Inductive Case) Must prove for all i N. if <W, σ i > σ 0, then <W, σ i+1 > σ 0 Pick an arbitrary i N Assume that <W, σ i > σ 0 Now prove that <W, σ i+1 > σ 0 Must construct a derivation tree: <x, σ i+1 > 5 - i 5 - i 5 <x + 1, σ i+1 > 6 - i Must prove i N. <W, σ i > σ 0 <W, σ i+1 > σ 0 Pick an arbitrary i N Assume that <W, σ i > σ 0 Now prove that <W, σ i+1 > σ 0 Must construct a derivation tree: <x, σ i+1 > 5 - i 5 - i 5 <x + 1, σ i+1 > 6 - i <x 5, σ i+1 > true <x:=x+1, σ i+1 > σ i <W, σ i > σ 0 <x 5, σ i+1 > true <x:=x+1, σ i+1 > σ i <W, σ i > σ 0 <while x 5 do x := x + 1, σ i+1 > σ 0 <while x 5 do x := x + 1, σ i+1 > σ Well-Founded Induction Well-Founded Induction: Examples A relation A A is well-founded if there are no infinite descending chains in A Example: < 1 = { (x, x +1) x N} aka the predecessor relation Example: < = { (x, y) x, y N and x < y } Well-founded induction: To prove x A. P(x) it is enough to prove x A. ( y x P(y)) P(x) If is < 1 then we obtain mathematical induction as a special case 11 Consider p Z Zwith x p y iff (y < 0 Æ y = x - 1) Ç (y > 0 Æ y = x + 1). Induction principle: P(0) x 0. P(x) P(x - 1) x 0. P(x) P(x + 1) Consider p (N N) (N N) and (x 1, y 1 ) p (x 2, y 2 ) iff x 2 = x (x 1 = x 2 y 2 = y 1 + 1). Says what? Induction principle: P(0,0) x,y,y. (P(x,y) P(x + 1, y ) P(x, y+ 1)) This has a common name. Anyone see? Says what? 12 2

3 Structural Induction (on Expressions) For e ::= n x e 1 + e 2 e 1 * e 2 Define p Aexp Aexp such that e 1 p e 1 + e 2 e 2 p e 1 + e 2 e 1 p e 1 * e 2 e 2 p e 1 * e 2 no other elements of Aexp Aexp are related by p To prove e Aexp. P(e) prove n Z. P(n) prove x L. P(x) prove e 1, e 2 Aexp. P(e 1 ) P(e 2 ) P(e 1 + e 2 ) prove e 1, e 2 Aexp. P(e 1 ) P(e 2 ) P(e 1 * e 2 ) 13 Notes on Structural Induction Called structural induction because the proof is guided by the structure of the expression One proof case per form of expression Atomic expressions (with no subexpressions) are all base cases Composite expressions are the inductive cases Structural induction is the most useful form of induction in the study of PL 14 Example Proof Using Induction on the Structure of Expressions Let L(e) be the # of literals and variable occurrences in e O(e) be the # of operators in e Prove that for all e Aexp. L(e) = O(e) + 1 Proof: By induction on the structure of e. Case e = n: Example Proof Using Induction on the Structure of Expressions Case e = e 1 + e 2 : Case e = x: Case e = e 1 * e 2 : Example Proof Using Induction on the Structure of Expressions Let L(e) be the # of literals and variable occurrences in e O(e) be the # of operators in e Prove that e Aexp. L(e) = O(e) + 1 Proof: By induction on the structure of e. Case e = n: L(e) = 1 and O(e) = 0 Case e = x: L(e) = 1 and O(e) = 0 Case e = e 1 + e 2 : L(e) = L(e 1 ) + L(e 2 ) and O(e) = O(e 1 ) + O(e 2 ) + 1 By the induction hypothesis, L(e 1 ) = O(e 1 ) + 1 and L(e 2 ) = O(e 2 ) + 1 Thus, L(e) = O(e 1 ) + O(e 2 ) + 2 = O(e) + 1 Case e = e 1 * e 2 : Same as the case for + 17 Try it at home! Most proofs for the Aexp sublanguage of IMP can work by structural induction Small-step and big-step semantics obtain equivalent results: How do we state this formally? 18 3

4 Try it at home! Most proofs for the Aexp sublanguage of IMP can work by structural induction Small-step and big-step semantics obtain equivalent results: For all e Aexp. for all n Z. for all σ Σ. <e, σ> * <n, σ> iff <e, σ> n Structural induction on Aexp works here because all of the semantics are syntax-directed 19 Obvious, right? You are given a concrete state σ. You have <x + 1, σ> 5 You also have <x + 1, σ> 88 Is this possible? 20 Let s make sure Let s make sure For all e Aexp. for all σ Σ. for all n, n Z. if <e, σ> n and <e, σ> n then n = n For all b Bexp. for all σ Σ. for all t, t B. if <b, σ> t and <b, σ> t then t = t For all c Com. for all σ, σ,σ Σ. if <c, σ> σ and <c, σ> σ then σ = σ 21 For all e Aexp. for all σ Σ. for all n, n Z. if <e, σ> n and <e, σ> n then n = n For all b Bexp. for all σ Σ. for all t, t B. if <b, σ> t and <b, σ> t then t = t For all c Com. for all σ, σ,σ Σ. if <c, σ> σ and <c, σ> σ then σ = σ 22 How do we prove it? If <e, σ> n and <e, σ> n then n = n If <b, σ> t and <b, σ> t then t = t If <c, σ> σ and <c, σ> σ then σ = σ No immediate way to use mathematical induction For commands we cannot use induction on the structure of the command Can you imagine why? How do we prove it? If <e, σ> n and <e, σ> n then n = n If <b, σ> t and <b, σ> t then t = t If <c, σ> σ and <c, σ> σ then σ = σ No immediate way to use mathematical induction For commands we cannot use induction on the structure of the command Can you see why? <b, σ> true <c, σ> σ <while b do c, σ > σ <while b do c, σ> σ

5 How do we prove it? If <e, σ> n and <e, σ> n then n = n If <b, σ> t and <b, σ> t then t = t If <c, σ> σ and <c, σ> σ then σ = σ No immediate way to use mathematical induction For commands we cannot use induction on the structure of the command while s evaluation does not depend only on the evaluation of its strict subexpressions We need something new! Some more powerful form of induction With all the bells and whistles! <b, σ> true <c, σ> σ <while b do c, σ > σ <while b do c, σ> σ Recall Proof Systems Operational semantics assigns meanings to programs by listing rules of inference that allow to prove judgments by constructing derivations. A derivation is a tree-structured object made up of valid instances of inference rules. Find the keyword in the above. Ideas? 27 Induction on the Structure of Derivations Key idea: The hypothesis does not just assume a c Com but the existence of a derivation of <c, σ> σ Derivation trees are also defined inductively, just like expression trees A derivation is built of subderivations: <x, σ i+1 > 5 - i 5 - i 5 <x + 1, σ i+1 > 6 - i <x 5, σ i+1 > true <x:=x+1; W, σ i+1 > σ 0 <while x 5 do x := x + 1, σ i+1 > σ 0 <x:=x+1, σ i+1 > σ i <W, σ i > σ 0 Adapt the structural induction principle to work on the structure of derivations 28 Induction on the Structure of Derivations To prove that for all derivations D of a judgment, property P holds For each derivation rule of the form H 1 H n Assume P holds for derivations of H i (i = 1..n) Prove the the property holds for the derivation obtained from the derivations of H i using the given rule C 29 Notation: Naming Derivations Write D :: Judgment to mean D is the derivation that proves Judgment Example: D :: <e 1 + e 2, σ> n 1 + n

6 Proving Com Evaluation is Deterministic If D::<c, σ> σ and D :: <c, σ> σ, then σ = σ. Note: recall that meta-variables are universally-quantified (i.e., the above is for all c Com, σ, σ,σ Σ, derivations D and D. Start by picking each to be arbitrary. Proof: By induction on the structure of derivation D. 31 6