Programming Languages 2nd edition Tucker and Noonan

1. Programming Languages2nd editionTucker and Noonan Chapter 18 Program Correctness • To treat programming scientifically, it must be possible to specify the required properties of programs precisely. Formality is certainly not an end in itself. The importance of formal specifications must ultimately rest in their utility - in whether or not they are used to improve the quality of software or to reduce the cost of producing and maintaining software. • J. Horning

2. Contents 18.1 Axiomatic Semantics 18.2 Formal Methods Tools: JML 18.3 Correctness of Object-Oriented Programs 18.4 Correctness of Functional Programs 18.4.1 Recursion and Induction 18.4.2 Examples of Structural Induction

3. 18.4 Correctness of Functional Programs • Pure functional programs are more accessible to correctness proofs than imperative or OO programs. • Three major reasons: • Pure functional programs are state-free (no assignment), • Functions and variables mathematical ideas, and • Recursion aligns well with proof by induction.

4. 18.4.1 Recursion and Induction • Consider the Haskell function: > fact n > | n == 1 = 1 -- fact.1 > | n > 1 = n*fact(n-1) -- fact.2 Suppose we want to prove that this function correctly computes the factorial. I.e., that it computes: fact(1) = 1 fact(n) = 12…(n-1)n when n>1

5. Induction proof of a recursive function • The induction proof is straightforward. We use the definition of the function directly in the proof. • Basis step: The function computes the correct result for n = 1, using line fact.1 of the definition. • Induction step: Assume the hypothesis that the function computes the correct result for some n = k > 1. That is, it computes fact(k) = 12…(k-1)k. Then for n = k+1, it computes fact(k+1) =(k+1)*fact(k) using line fact.2 of the definition. Thus, it computes fact(k+1) = 12…(k-1)k(k+1), which completes the induction step.

6. 18.4.2 Examples of Structural Induction • List concatenation and reversal: > cat [] ys = ys -- cat.1 > cat (x:xs) ys = x : (cat xs ys) -- cat.2 > rev [] = [] -- rev.1 > rev (x:xs) = cat (rev (xs)) [x] -- rev.2 Suppose we want to prove the following property about the relationship between cat and rev: rev (cat xs ys) = cat (rev ys) (rev xs) E.g., rev (cat “hello ” “world”) = cat (rev “world”) (rev “hello ”) = “dlrow olleh”

7. The Proof • Basis step: • rev (cat [] ys) = rev (ys)from cat.1 • = cat (rev ys [])from rev.2 • = cat (rev ys rev []from rev.1 • Induction step: • Hypothesis: rev (cat xs ys) = cat (rev ys) (rev xs) • rev (cat (x:xs) ys) = rev x : (cat xs ys)from cat.2 • = rev (cat (xs ys) [x])from rev.2 • = cat (cat (rev ys) (rev xs)) [x]from hypothesis • = cat (cat (rev ys) (rev xs)) [x]cat associativity* • = cat (rev ys) (rev (x:xs)) from rev.2 • *Note: associativity of cat needs to be proved separately.

8. List Length and Concatenation > len [] = 0 -- len.1 > len (x:xs) = 1 + (len xs) -- len.2 E.g., len [1,3,4,7] = 1 + len [3,4,7] = 1 + (1 + len [4,7]) = 1 + (1 + (1 + len [7])) = 1 + (1 + (1 + (1 + len []))) = 1 + (1 + (1 + (1 + 0))) = 4 Suppose we want to prove the following property about the relationship between len and cat: len (cat xs ys) = len xs + len ys

9. The Proof • Basis step: • len (cat [] ys) = len (ys)from cat.1 • = 0 + len (ys) from arithmetic • = len [] + len ys from len.1 • Induction step: • Hypothesis: len (cat xs ys) = len xs + len ys • len (cat (x:xs) ys) = len x : (cat xs ys)from cat.2 • = 1 + len (cat xs ys) from len.2 • = 1 + len xs + len ys from hypothesis • = len x:xs + len ys from len.2