KBO Orientability. Harald Zankl Nao Hirokawa Aart Middeldorp. Japan Advanced Institute of Science and Technology University of Innsbruck
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1 KBO Orientability Harald Zankl Nao Hirokawa Aart Middeldorp Japan Advanced Institute of Science and Technology University of Innsbruck KBO Orientability 1/32
2 Reference JAR 2009 Harald Zankl, Nao Hirokawa, and Aart Middeldorp KBO Orientability Journal of Automated Reasoning 43(2), pp , 2009 KBO Orientability 2/32
3 Term Rewriting Definition pair of terms l r is rewrite rule if l V Var(r) Var(l) term rewrite system (TRS) is set of rewrite rules (rewrite relation) s R t if l r R, context C, substitution σ. s = C[lσ] t = C[rσ] KBO Orientability 3/32
4 Term Rewriting Definition pair of terms l r is rewrite rule if l V Var(r) Var(l) term rewrite system (TRS) is set of rewrite rules (rewrite relation) s R t if l r R, context C, substitution σ. s = C[lσ] t = C[rσ] Example TRS R x + 0 x x + s(y) s(x + y) x 0 0 x s(y) x y + x KBO Orientability 3/32
5 Term Rewriting Definition pair of terms l r is rewrite rule if l V Var(r) Var(l) term rewrite system (TRS) is set of rewrite rules (rewrite relation) s R t if l r R, context C, substitution σ. s = C[lσ] t = C[rσ] Example TRS R x + 0 x x + s(y) s(x + y) x 0 0 x s(y) x y + x rewriting s(0) s(0) R s(0) 0 + s(0) R 0 + s(0) R s(0 + 0) R s(0) terminated KBO Orientability 3/32
6 Termination Definition TRS R is terminating if there is no infinite rewrite sequence t 1 R t 2 R KBO Orientability 4/32
7 Termination Definition TRS R is terminating if there is no infinite rewrite sequence t 1 R t 2 R QUESTION how to prove termination? KBO Orientability 4/32
8 Termination Definition TRS R is terminating if there is no infinite rewrite sequence t 1 R t 2 R QUESTION how to prove termination? Knuth-Bendix order (KBO) KBO Orientability 4/32
9 Termination Definition TRS R is terminating if there is no infinite rewrite sequence t 1 R t 2 R QUESTION how to prove termination? Knuth-Bendix order (KBO) introduced by Knuth and Bendix, 1970 KBO Orientability 4/32
10 Termination Definition TRS R is terminating if there is no infinite rewrite sequence t 1 R t 2 R QUESTION how to prove termination? Knuth-Bendix order (KBO) introduced by Knuth and Bendix, 1970 best studied termination methods KBO Orientability 4/32
11 Termination Definition TRS R is terminating if there is no infinite rewrite sequence t 1 R t 2 R QUESTION how to prove termination? Knuth-Bendix order (KBO) introduced by Knuth and Bendix, 1970 best studied termination methods great success in theorem provers (Waldmeister, Vampire,...) KBO Orientability 4/32
12 Knuth-Bendix Orders Definition precedence > is proper order on function symbols F KBO Orientability 5/32
13 Knuth-Bendix Orders Definition precedence > is proper order on function symbols F weight function (w, w 0 ) is pair in R 0 F R 0 KBO Orientability 5/32
14 Knuth-Bendix Orders Definition precedence > is proper order on function symbols F weight function (w, w 0 ) is pair in R 0 F R 0 weight of term t is { w 0 if t V w(t) = w(f) + w(t 1 ) + + w(t n ) if t = f(t 1,..., t n ) KBO Orientability 5/32
15 Knuth-Bendix Orders Definition precedence > is proper order on function symbols F weight function (w, w 0 ) is pair in R 0 F R 0 weight of term t is { w 0 if t V w(t) = w(f) + w(t 1 ) + + w(t n ) if t = f(t 1,..., t n ) weight function (w, w 0 ) is admissible for precedence > if w(f) > 0 or f g for all unary functions f and all functions g KBO Orientability 5/32
16 Definition Knuth-Bendix order > kbo on terms T (F, V): s > kbo t if s x t x for all x V and either KBO Orientability 6/32
17 Definition Knuth-Bendix order > kbo on terms T (F, V): s > kbo t if s x t x for all x V and either w(s) > w(t), or KBO Orientability 6/32
18 Definition Knuth-Bendix order > kbo on terms T (F, V): s > kbo t if s x t x for all x V and either w(s) > w(t), or w(s) = w(t) and KBO Orientability 6/32
19 Definition Knuth-Bendix order > kbo on terms T (F, V): s > kbo t if s x t x for all x V and either w(s) > w(t), or w(s) = w(t) and s = f n (t) and t V for some unary f and n 1; or KBO Orientability 6/32
20 Definition Knuth-Bendix order > kbo on terms T (F, V): s > kbo t if s x t x for all x V and either w(s) > w(t), or w(s) = w(t) and s = f n (t) and t V for some unary f and n 1; or s = f(s 1,..., s i 1, s i,..., s n), t = f(s 1,..., s i 1, t i,..., t n), and s i > kbo t i; or KBO Orientability 6/32
21 Definition Knuth-Bendix order > kbo on terms T (F, V): s > kbo t if s x t x for all x V and either w(s) > w(t), or w(s) = w(t) and s = f n (t) and t V for some unary f and n 1; or s = f(s 1,..., s i 1, s i,..., s n), t = f(s 1,..., s i 1, t i,..., t n), and s i > kbo t i; or s = f(s 1,..., s n), t = g(t 1,..., t m), and f > g KBO Orientability 6/32
22 Definition let X R 0. TRS R is KBO X terminating if precedence > admissible weight function (w, w 0 ) X F X such that l > kbo r for all l r R KBO Orientability 7/32
23 Definition let X R 0. TRS R is KBO X terminating if precedence > admissible weight function (w, w 0 ) X F X such that l > kbo r for all l r R Theorem Knuth and Bendix, 1970 TRS is terminating if it is KBO N terminating KBO Orientability 7/32
24 Quiz KBO Orientability 8/32
25 Example I a(a(x)) b(b(b(x))) b(b(b(b(b(x))))) a(a(a(x))) KBO Orientability 9/32
26 Example I a(a(x)) b(b(b(x))) b(b(b(b(b(x))))) a(a(a(x))) is KBO N terminating? KBO Orientability 9/32
27 Example I a(a(x)) b(b(b(x))) b(b(b(b(b(x))))) a(a(a(x))) is KBO N terminating? yes Proof take precedence a > b and weight function w(a) =? w(b) =? w 0 = 1 KBO Orientability 9/32
28 Example I a(a(x)) b(b(b(x))) b(b(b(b(b(x))))) a(a(a(x))) is KBO N terminating? yes Proof take precedence a > b and weight function w(a) = 3 w(b) = 2 w 0 = 1 KBO Orientability 9/32
29 Example I a(a(x)) b(b(b(x))) b(b(b(b(b(x))))) a(a(a(x))) is KBO N terminating? yes Proof take precedence a > b and weight function w(a) = 3 w(b) = 2 w 0 = 1 Proof another solution: take precedence > = and weight function w(a) =? w(b) =? w 0 = 1 KBO Orientability 9/32
30 Example I a(a(x)) b(b(b(x))) b(b(b(b(b(x))))) a(a(a(x))) is KBO N terminating? yes Proof take precedence a > b and weight function w(a) = 3 w(b) = 2 w 0 = 1 Proof another solution: take precedence > = and weight function w(a) = 13 w(b) = 8 w 0 = 1 KBO Orientability 9/32
31 Example II TRS R x + 0 x x + s(y) s(x + y) x 0 0 x s(y) x y + x is KBO N terminating? KBO Orientability 10/32
32 Example II TRS R x + 0 x x + s(y) s(x + y) x 0 0 x s(y) x y + x is KBO N terminating? no KBO Orientability 10/32
33 History of KBO KBO Orientability 11/32
34 Theorem Knuth and Bendix, 1970 TRS is terminating if it is KBO N terminating KBO Orientability 12/32
35 Theorem Knuth and Bendix, 1970 TRS is terminating if it is KBO N terminating Theorem Dershowitz, 1979 TRS is terminating if it is KBO R 0 terminating KBO Orientability 12/32
36 Theorem Knuth and Bendix, 1970 TRS is terminating if it is KBO N terminating Theorem Dershowitz, 1979 TRS is terminating if it is KBO R 0 terminating Theorem Dick, Kalmus, and Martin, 1990 KBO R 0 termination is decidable within exponential time KBO Orientability 12/32
37 Theorem Knuth and Bendix, 1970 TRS is terminating if it is KBO N terminating Theorem Dershowitz, 1979 TRS is terminating if it is KBO R 0 terminating Theorem Dick, Kalmus, and Martin, 1990 KBO R 0 termination is decidable within exponential time Theorem Korovin and Voronkov, 2001, 2003 TRS is KBO N terminating it is KBO R 0 terminating KBO R 0 termination is decidable within polynomial time KBO Orientability 12/32
38 Theorem Knuth and Bendix, 1970 TRS is terminating if it is KBO N terminating Theorem Dershowitz, 1979 TRS is terminating if it is KBO R 0 terminating Theorem Dick, Kalmus, and Martin, 1990 KBO R 0 termination is decidable within exponential time Theorem Korovin and Voronkov, 2001, 2003 TRS is KBO N terminating it is KBO R 0 terminating KBO R 0 termination is decidable within polynomial time Theorem Zankl and Middeldorp, 2007 KBO {0,1,...,B} termination (B N) can reduce to SAT and PBC KBO Orientability 12/32
39 This Talk MAIN RESULT for every TRS R and N = l r R ( l + r ) + 1 R is KBO N terminating R is KBO {0,1,...,B} terminating where, B = N 4N+1 KBO Orientability 13/32
40 This Talk MAIN RESULT for every TRS R and N = l r R ( l + r ) + 1 R is KBO N terminating R is KBO {0,1,...,B} terminating where, B = N 4N+1 OVERVIEW OF PROOF Principal Solutions KBO Orientability 13/32
41 This Talk MAIN RESULT for every TRS R and N = l r R ( l + r ) + 1 R is KBO N terminating R is KBO {0,1,...,B} terminating where, B = N 4N+1 OVERVIEW OF PROOF Principal Solutions MCD KBO Orientability 13/32
42 This Talk MAIN RESULT for every TRS R and N = l r R ( l + r ) + 1 R is KBO N terminating R is KBO {0,1,...,B} terminating where, B = N 4N+1 OVERVIEW OF PROOF Principal Solutions MCD Bound by Norm KBO Orientability 13/32
43 Principal Solutions KBO Orientability 14/32
44 Example given precedence, KBO N problem reduces to linear integral constraints Example a(a(x)) b(b(b(x))) b(b(b(b(b(x))))) a(a(a(x))) KBO Orientability 15/32
45 Example given precedence, KBO N problem reduces to linear integral constraints Example a(a(x)) b(b(b(x))) b(b(b(b(b(x))))) a(a(a(x))) for precedence a > b, solve wrt w(a), w(b) N 2 w(a) 3 w(b) 0 3 w(a) + 5 w(b) > 0 KBO Orientability 15/32
46 Example given precedence, KBO N problem reduces to linear integral constraints Example a(a(x)) b(b(b(x))) b(b(b(b(b(x))))) a(a(a(x))) for precedence a > b, solve wrt w(a), w(b) N 2 w(a) 3 w(b) 0 3 w(a) + 5 w(b) > 0 for precedence > = solve wrt w(a), w(b) N 2 w(a) 3 w(b) > 0 3 w(a) + 5 w(b) > 0 KBO Orientability 15/32
47 Principal Solutions Gale, 1960; Dick, Kalmus and Martin, 1990; Korovin and Voronkov, 2003 set I of KBO inequalities are of form { a i x i R i 0 } i with R i {, >}, a i Z n, x N KBO Orientability 16/32
48 Principal Solutions Gale, 1960; Dick, Kalmus and Martin, 1990; Korovin and Voronkov, 2003 set I of KBO inequalities are of form { a i x i R i 0 } i with R i {, >}, a i Z n, x N Definition write A I for ( a i ) 1,...,n KBO Orientability 16/32
49 Principal Solutions Gale, 1960; Dick, Kalmus and Martin, 1990; Korovin and Voronkov, 2003 set I of KBO inequalities are of form { a i x i R i 0 } i with R i {, >}, a i Z n, x N Definition write A I for ( a i ) 1,...,n x is solution of A I if A I x 0 and x N n KBO Orientability 16/32
50 Principal Solutions Gale, 1960; Dick, Kalmus and Martin, 1990; Korovin and Voronkov, 2003 set I of KBO inequalities are of form { a i x i R i 0 } i with R i {, >}, a i Z n, x N Definition write A I for ( a i ) 1,...,n x is solution of A I if A I x 0 and x N n solution x maximizing {i a i x > 0} wrt is principal KBO Orientability 16/32
51 Principal Solutions Gale, 1960; Dick, Kalmus and Martin, 1990; Korovin and Voronkov, 2003 set I of KBO inequalities are of form { a i x i R i 0 } i with R i {, >}, a i Z n, x N Definition write A I for ( a i ) 1,...,n x is solution of A I if A I x 0 and x N n solution x maximizing {i a i x > 0} wrt is principal Theorem principal solution of A always exists principal solution x : (I is solvable x satisfies I) KBO Orientability 16/32
52 { 2 w(a) 3 w(b) > 0 I = 3 w(a) + 5 w(b) > 0 } A I = ( 2 ) KBO Orientability 17/32
53 { 2 w(a) 3 w(b) > 0 I = 3 w(a) + 5 w(b) > 0 } A I = ( 2 ) since ( 3 2) ( ) ( 3 0 A I = 2 1) is not principal solution of A I ( ) ( 13 2 A I = 8 1) KBO Orientability 17/32
54 { 2 w(a) 3 w(b) > 0 I = 3 w(a) + 5 w(b) > 0 } A I = ( 2 ) since ( 3 2) ( ) ( 3 0 A I = 2 1) is not principal solution of A I ( ) ( 13 2 A I = 8 1) ( ) 13 is principal solution of A 8 I and satisfies I hence I is solvable KBO Orientability 17/32
55 Example I = 2 w(a) 3 w(b) > 0 3 w(a) + 5 w(b) > 0 4 w(a) + 3 w(b) 0 A I = KBO Orientability 18/32
56 Example I = 2 w(a) 3 w(b) > 0 3 w(a) + 5 w(b) > 0 4 w(a) + 3 w(b) 0 A I = principal solution of A I is ( ) 0 and does not satisfy I 0 KBO Orientability 18/32
57 Example I = 2 w(a) 3 w(b) > 0 3 w(a) + 5 w(b) > 0 4 w(a) + 3 w(b) 0 A I = principal solution of A I is hence I is not solvable ( ) 0 and does not satisfy I 0 KBO Orientability 18/32
58 Example I = 2 w(a) 3 w(b) > 0 3 w(a) + 5 w(b) > 0 4 w(a) + 3 w(b) 0 A I = principal solution of A I is hence I is not solvable ( ) 0 and does not satisfy I 0 QUESTION how to find principal solution? KBO Orientability 18/32
59 Example I = 2 w(a) 3 w(b) > 0 3 w(a) + 5 w(b) > 0 4 w(a) + 3 w(b) 0 A I = principal solution of A I is hence I is not solvable ( ) 0 and does not satisfy I 0 QUESTION how to find principal solution? MCD KBO Orientability 18/32
60 MCD: Method of Complete Description KBO Orientability 19/32
61 MCD Dick, Kalmus and Martin, 1990 (a 1,..., a n ) κ = ( e i a i 0) ++(a j e i a i e j a i < 0, a j > 0) KBO Orientability 20/32
62 MCD Dick, Kalmus and Martin, 1990 (a 1,..., a n ) κ = ( e i a i 0) ++(a j e i a i e j a i < 0, a j > 0) for m n matrix A and 0 i m { Si A E n if i = 0 = Si 1 A ( a isi 1 A )κ otherwise KBO Orientability 20/32
63 MCD Dick, Kalmus and Martin, 1990 (a 1,..., a n ) κ = ( e i a i 0) ++(a j e i a i e j a i < 0, a j > 0) for m n matrix A and 0 i m { Si A E n if i = 0 = Si 1 A ( a isi 1 A )κ otherwise sum of all column vectors of S A m is denoted by s A KBO Orientability 20/32
64 MCD Dick, Kalmus and Martin, 1990 (a 1,..., a n ) κ = ( e i a i 0) ++(a j e i a i e j a i < 0, a j > 0) for m n matrix A and 0 i m { Si A E n if i = 0 = Si 1 A ( a isi 1 A )κ otherwise sum of all column vectors of S A m is denoted by s A Theorem s A is principal solution of A KBO Orientability 20/32
65 Example I = { 2 w(a) 3 w(b) > 0 3 w(a) + 5 w(b) > 0 } A I = ( 2 ) KBO Orientability 21/32
66 Example I = { 2 w(a) 3 w(b) > 0 3 w(a) + 5 w(b) > 0 } A I = ( 2 ) performing MCD KBO Orientability 21/32
67 Example I = { 2 w(a) 3 w(b) > 0 3 w(a) + 5 w(b) > 0 } A I = ( 2 ) performing MCD S A I 0 = ( ) S A I 1 = S A I 0 ((2 3)SA I 0 )κ = S A I 2 = S A I 1 (( 3 5)SA I 1 )κ = ( ) ( ) KBO Orientability 21/32
68 Example I = { 2 w(a) 3 w(b) > 0 3 w(a) + 5 w(b) > 0 } A I = ( 2 ) performing MCD S A I 0 = we obtain principal solution ( ) S A I 1 = S A I 0 ((2 3)SA I 0 )κ = S A I 2 = S A I 1 (( 3 5)SA I 1 )κ = s A I = ( 3 2) + ( ) ( ) ( ) ( ) = 6 8 KBO Orientability 21/32
69 Example I = { 2 w(a) 3 w(b) > 0 3 w(a) + 5 w(b) > 0 } A I = ( 2 ) performing MCD S A I 0 = we obtain principal solution ( ) S A I 1 = S A I 0 ((2 3)SA I 0 )κ = S A I 2 = S A I 1 (( 3 5)SA I 1 )κ = s A I = ( 3 2) + which satisfies I. hence I is solvable KBO Orientability 21/32 ( ) ( ) ( ) ( ) = 6 8
70 Bound by Norm KBO Orientability 22/32
71 MCD (a 1,..., a n ) κ = ( e i a i 0) ++(a j e i a i e j a i < 0, a j > 0) for m n matrix A and 0 i m { Si A E n if i = 0 = Si 1 A ( a isi 1 A )κ otherwise sum of all column vectors of S A m is denoted by s A KBO Orientability 23/32
72 MCD (a 1,..., a n ) κ = ( e i a i 0) ++(a j e i a i e j a i < 0, a j > 0) for m n matrix A and 0 i m { Si A E n if i = 0 = Si 1 A ( a isi 1 A )κ otherwise sum of all column vectors of S A m is denoted by s A GOAL bound s A {0, 1,..., B} n KBO Orientability 23/32
73 MCD (a 1,..., a n ) κ = ( e i a i 0) ++(a j e i a i e j a i < 0, a j > 0) for m n matrix A and 0 i m { Si A E n if i = 0 = Si 1 A ( a isi 1 A )κ otherwise sum of all column vectors of S A m is denoted by s A GOAL bound s A {0, 1,..., B} n APPROACH define norm to calculate s A I with B = s A I KBO Orientability 23/32
74 Norm Definition A = max i,j a ij for m n matrix A = (a ij ) ij KBO Orientability 24/32
75 Norm Definition A = max i,j a ij for m n matrix A = (a ij ) ij Lemma for every m n matrix A and n p matrix B a κ = a KBO Orientability 24/32
76 Norm Definition A = max i,j a ij for m n matrix A = (a ij ) ij Lemma for every m n matrix A and n p matrix B a κ = a AB n A B KBO Orientability 24/32
77 Norm Definition A = max i,j a ij for m n matrix A = (a ij ) ij Lemma for every m n matrix A and n p matrix B a κ = a AB n A B (a 1,..., a n ) κ is n k matrix with k n 2. Si A is m k matrix with k n 2i. KBO Orientability 24/32
78 Norm Definition A = max i,j a ij for m n matrix A = (a ij ) ij Lemma for every m n matrix A and n p matrix B a κ = a AB n A B (a 1,..., a n ) κ is n k matrix with k n 2. Si A is m k matrix with k n 2i. Si A 1 (2m A )2i s A n 2m (2m A ) 2m 1 KBO Orientability 24/32
79 Lemma let R be TRS of size N I be set of inequalities induced by KBO with fixed precedence A I be of size m n m N n N therefore A I N s A I N 4N+1 := B thus, s A I {0, 1,..., B} n KBO Orientability 25/32
80 Main Result Theorem R is KBO N terminating R is KBO {0,1,...,B} terminating where, B = N 4N+1 KBO Orientability 26/32
81 Main Result Theorem R is KBO N terminating R is KBO {0,1,...,B} terminating where, B = N 4N+1 Corollary Zankl and Middeldorp s SAT and PBC encodings are complete for this B KBO Orientability 26/32
82 Summary let R be TRS, N = 4N+1 l r R ( l + r ) + 1, and B = N R is KBO R 0 terminating R is KBO N terminating Korovin and Voronkov R is KBO {0,1,...,B} terminating this talk KBO Orientability 27/32
83 Summary let R be TRS, N = 4N+1 l r R ( l + r ) + 1, and B = N R is KBO R 0 terminating R is KBO N terminating Korovin and Voronkov R is KBO {0,1,...,B} terminating this talk theoretical interest of decidability issue is more or less closed KBO Orientability 27/32
84 Summary let R be TRS, N = 4N+1 l r R ( l + r ) + 1, and B = N R is KBO R 0 terminating R is KBO N terminating Korovin and Voronkov R is KBO {0,1,...,B} terminating this talk theoretical interest of decidability issue is more or less closed how about automation? KBO Orientability 27/32
85 Automation KBO Orientability 28/32
86 Theorem Dick, Kalmus, and Martin, 1990 KBO R 0 termination is decidable by MCD KBO Orientability 29/32
87 Theorem Dick, Kalmus, and Martin, 1990 KBO R 0 termination is decidable by MCD Theorem Korovin and Voronkov, 2001, 2003 KBO R 0 termination is decidable by Karmarkar/Khachiyan algorithm KBO Orientability 29/32
88 Theorem Dick, Kalmus, and Martin, 1990 KBO R 0 termination is decidable by MCD Theorem Korovin and Voronkov, 2001, 2003 KBO R 0 termination is decidable by Karmarkar/Khachiyan algorithm Theorem Zankl and Middeldorp, 2007 KBO {0,1,...,B} termination (B N) is decidable via SAT/PBC encodings KBO Orientability 29/32
89 Theorem Dick, Kalmus, and Martin, 1990 KBO R 0 termination is decidable by MCD Theorem Korovin and Voronkov, 2001, 2003 KBO R 0 termination is decidable by Karmarkar/Khachiyan algorithm Theorem Zankl and Middeldorp, 2007 KBO {0,1,...,B} termination (B N) is decidable via SAT/PBC encodings Theorem KBO R 0 termination is decidable via SMT (linear arithmetic) encoding KBO Orientability 29/32
90 Experiments test-bed: 1381 TRSs from Termination Problem Data Base 4.0 KBO Orientability 30/32
91 Experiments test-bed: 1381 TRSs from Termination Problem Data Base 4.0 PC: 8 dual-core AMD Opteron 885, 2.6 GHz, 64 GB KBO Orientability 30/32
92 Experiments test-bed: 1381 TRSs from Termination Problem Data Base 4.0 PC: 8 dual-core AMD Opteron 885, 2.6 GHz, 64 GB 60 seconds timeout KBO Orientability 30/32
93 Experiments test-bed: 1381 TRSs from Termination Problem Data Base 4.0 PC: 8 dual-core AMD Opteron 885, 2.6 GHz, 64 GB 60 seconds timeout method(b) total time (sec) #success #timeout MCD KBO Orientability 30/32
94 Experiments test-bed: 1381 TRSs from Termination Problem Data Base 4.0 PC: 8 dual-core AMD Opteron 885, 2.6 GHz, 64 GB 60 seconds timeout method(b) total time (sec) #success #timeout MCD simplex KBO Orientability 30/32
95 Experiments test-bed: 1381 TRSs from Termination Problem Data Base 4.0 PC: 8 dual-core AMD Opteron 885, 2.6 GHz, 64 GB 60 seconds timeout method(b) total time (sec) #success #timeout MCD simplex SAT/PBC(4) 31/90 104/104 0/0 KBO Orientability 30/32
96 Experiments test-bed: 1381 TRSs from Termination Problem Data Base 4.0 PC: 8 dual-core AMD Opteron 885, 2.6 GHz, 64 GB 60 seconds timeout method(b) total time (sec) #success #timeout MCD simplex SAT/PBC(4) 31/90 104/104 0/0 SAT/PBC(8) 32/93 106/106 0/0 KBO Orientability 30/32
97 Experiments test-bed: 1381 TRSs from Termination Problem Data Base 4.0 PC: 8 dual-core AMD Opteron 885, 2.6 GHz, 64 GB 60 seconds timeout method(b) total time (sec) #success #timeout MCD simplex SAT/PBC(4) 31/90 104/104 0/0 SAT/PBC(8) 32/93 106/106 0/0 SAT/PBC(16) 34/ /107 0/0 KBO Orientability 30/32
98 Experiments test-bed: 1381 TRSs from Termination Problem Data Base 4.0 PC: 8 dual-core AMD Opteron 885, 2.6 GHz, 64 GB 60 seconds timeout method(b) total time (sec) #success #timeout MCD simplex SAT/PBC(4) 31/90 104/104 0/0 SAT/PBC(8) 32/93 106/106 0/0 SAT/PBC(16) 34/ /107 0/0 SAT/PBC(1024) 351/ /107 3/1 KBO Orientability 30/32
99 Experiments test-bed: 1381 TRSs from Termination Problem Data Base 4.0 PC: 8 dual-core AMD Opteron 885, 2.6 GHz, 64 GB 60 seconds timeout method(b) total time (sec) #success #timeout MCD simplex SAT/PBC(4) 31/90 104/104 0/0 SAT/PBC(8) 32/93 106/106 0/0 SAT/PBC(16) 34/ /107 0/0 SAT/PBC(1024) 351/ /107 3/1 SMT KBO Orientability 30/32
100 Conclusion let R be TRS, N = 4N+1 l r R ( l + r ) + 1, and B = N R is KBO R 0 terminating R is KBO N terminating Korovin and Voronkov R is KBO {0,1,...,B} terminating this talk CONCLUSION finite characterization of KBO orientability use SMT solver to automate KBO Orientability 31/32
101 Conclusion let R be TRS, N = 4N+1 l r R ( l + r ) + 1, and B = N R is KBO R 0 terminating R is KBO N terminating Korovin and Voronkov R is KBO {0,1,...,B} terminating this talk CONCLUSION finite characterization of KBO orientability use SMT solver to automate FUTURE WORK find optimal B KBO Orientability 31/32
102 Thank You for Your Attention KBO Orientability 32/32
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