Formalizing Kruskal s Tree Theorem in Isabelle/HOL

Transcription

1 Formalizing Kruskal s Tree Theorem in Isabelle/HOL Christian Sternagel JAIST July 5, 2012 Kickoff Meeting of Austria-Japan Joint Project Gamagori, Japan

2 Kruskal s Tree Theorem If the set A is well-quasi-ordered then the set of finite trees over A is well-quasi-ordered by homeomorphic embedding. Comments proof structure as Nash-Williams 1963 which claims A new and simple proof is given... C. Sternagel (JAIST) AJJP Kickoff 2/20

3 Overview Motivation Preliminaries Kruskal s Tree Theorem - A Proof Sketch Formalization Challenges Conclusion C. Sternagel (JAIST) AJJP Kickoff 3/20

4 Bibliography G. Higman. Ordering by divisibility in abstract algebras. Proc. London Math. Soc., doi: /plms/s J. B. Kruskal. Well-quasi-ordering, the tree theorem, and vazsonyi s conjecture. Trans. Amer. Math. Soc., doi: / C. S. J. A. Nash-Williams. On well-quasi-ordering finite trees. Math. Proc. Cambridge, doi: /s C. Sternagel. Well-Quasi-Orders. In The Archive of Formal Proofs C. Sternagel (JAIST) AJJP Kickoff 4/20

5 Motivation Why? long-standing open problem in formalized mathematics Kruskal s Tree Theorem is main ingredient to prove well-foundedness of simplification orders for first-order rewriting ultimately, we want to strengthen termination library of IsaFoR (Isabelle Formalization of Rewriting) C. Sternagel (JAIST) AJJP Kickoff 5/20

6 Preliminaries Homeomorphic Embedding on Lists empty list, [] adding element x to finite list xs, x xs append lists xs and ys, ys set of finite lists over A, A : embedding relation w.r.t. : x A xs A [] A x xs A [] emb ys xs emb ys xs emb y ys x y xs emb ys x xs emb y ys C. Sternagel (JAIST) AJJP Kickoff 6/20

7 Example - List Embedding Preliminaries ? emb We can... ( ) xs emb ys drop elements xs emb y ys ( ) x y xs emb ys replace elements by smaller elements x xs emb y ys note that empty list is embedded in any list ([] emb ys) C. Sternagel (JAIST) AJJP Kickoff 7/20

8 Preliminaries Homeomorphic Embedding on Trees tree with node f and list of direct subtrees ts, f (ts) root of tree root(f (ts)) = f, direct subtrees args(f (ts)) = ts set of finite trees over A, T (A): f A t ts. t T (A) f (ts) T (A) homeomorphic embedding relation w.r.t. : t ts t emb f (ts) s emb t t emb u s emb u f g ss ( = emb ) emb ts f (ss) emb g(ts) s emb t f (ss s ss 2 ) emb f (ss t ss 2 ) C. Sternagel (JAIST) AJJP Kickoff 8/20

9 Preliminaries Homeomorphic Embedding on Trees (cont d) Embedding TRS let Emb( ) be the infinite TRS f (ts) t f (ts) g(ss) if t ts if g f and ss = emb ts Result s emb t iff t + Emb( ) s C. Sternagel (JAIST) AJJP Kickoff 9/20

10 Well-Quasi-Orders - Definitions Preliminaries let A be a set and a binary relation A is wqo by ( A is a wqo, or wqo( A )): (1) transitive: x A. y A. z A. x y y z x z (2) all infinite sequences over A are good: f. ( i. f (i) A) ( j k. j < k f (j) f (k)) f (1) f (2) f (3)... f (j)... f (k)... a sequence that is not good, is called bad Property strict part of is x y = x y y x let wqo( A ), then A is well-founded on A C. Sternagel (JAIST) AJJP Kickoff 10/20

11 Kruskal s Tree Theorem - A Proof Sketch C. Sternagel (JAIST) AJJP Kickoff 11/20

12 Kruskal s Tree Theorem - A Proof Sketch Proof Structure of wqo( F ) = wqo( T (F) ) assume F is wqo prove in what it! sense? assume T (F) is not wqo = exists minimal bad sequence t 1, t 2, t 3,... with t i T (F) = exists sequence f 1, f 2, f 3,... with root(t i ) = f i, args(t i ) = ts i = let T = i ( args(t Higman s i )) lemma = {fi } and T are wqo Dickson s lemma = {fi } T is wqo = exist j, k with j < k and (f j, ts j ) {fi } T (f k, ts k ) = t j T (F) t k = t 1, t 2, t 3,... is good contradiction! C. Sternagel (JAIST) AJJP Kickoff 12/20

13 Formalization Challenges C. Sternagel (JAIST) AJJP Kickoff 13/20

14 Formalization Challenges Existence of Minimal Bad Sequence - Nash-Williams 1963 Select an t 1 T (F) such that t 1 is the first term of a bad sequence of members of T (F) and t 1 is as small as possible. Then select an t 2 such that t 1, t 2 are the first two terms of a bad sequence of members of T (F) and t 2 is as small as possible [... ]. Assuming the Axiom of Choice, this process yields a bad sequence t 1, t 2, t 3,.... C. Sternagel (JAIST) AJJP Kickoff 14/20

15 The Axiom of Choice in Isabelle Formalization Challenges x. y. P x y = f. x. P x (f x) Minimal in What Sense? subtree relation, t is (proper) subtree of s, written t s t ts s t t ts (t s), iff: t f (ts) s f (ts) proper subtree relation is well-founded (allowing for induction) Auxiliary Definitions infinite sequence f is minimal at position n (min n (f )), iff: g. ( i < n. g(i) = f (i)) g(n) f (n) ( i n. j n. g(i) f (j)) = good emb (g) replace elements of sequence f by those of sequence g, starting at n: (f n g)(i) = if i n then g(i) else f (i) C. Sternagel (JAIST) AJJP Kickoff 15/20

16 Key Lemma (1) min n (f ) (2) bad emb (f ) = g. i n. g(i) = f (i) g(n + 1) f (n + 1) i n + 1. j n + 1. g(i) f (j) bad emb (f n + 1 g) min n+1 (f n + 1 g) Construct Minimal Bad Sequence Formalization Challenges from AC and key lemma obtain function ν, s.t., given sequence satisfying (1) and (2) and index n, returns sequence satisfying conclusion auxiliary sequence (of sequences) { m ν(f, n) if n = 0 (n) = m (n 1) n ν(m (n 1), n 1) otherwise minimal bad sequence m(i) = m (i)(i) C. Sternagel (JAIST) AJJP Kickoff 16/20

17 Idea... Formalization Challenges C. Sternagel (JAIST) AJJP Kickoff 17/20

18 Conclusion C. Sternagel (JAIST) AJJP Kickoff 18/20

19 Conclusion Related Work Murthy, Extracting Constructive Content from Classical Proofs, PhD, 1990 (Nuprl) Fridlender, Ramsey s Theorem in Type Theory, Tech. Report, 1993 (ALF) Herbelin, A Program from an A-Translated Impredicative Proof of Higman s Lemma, 1994 (2-letter alphabet, Coq) Fridlender, Higman s Lemma in Type Theory, PhD, 1997 (ALF) Seisenberger, On the Constructive Content of Proofs, PhD, 2003 (Minlog) Berghofer, A Constructive Proof of Higman s Lemma in Isabelle, TYPES 2003 (2-letter alphabet, Isabelle) Martín-Mateos et al., A Formal Proof of Higman s Lemma in ACL2, JAR 2011 (ACL2) C. Sternagel (JAIST) AJJP Kickoff 19/20

20 Conclusion Future Work investigate how Zorn s Lemma could be of help reformulate proof using open induction The End C. Sternagel (JAIST) AJJP Kickoff 20/20