[PPT] - KBO Orientability Harald Zankl Nao Hirokawa Aart Middeldorp PowerPoint Presentation

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KBO Orientability

Harald Zankl‡ Nao Hirokawa† Aart Middeldorp‡

† Japan Advanced Institute of Science and Technology ‡ University of Innsbruck

KBO Orientability 1/32

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Reference

JAR 2009

Harald Zankl, Nao Hirokawa, and Aart Middeldorp KBO Orientability Journal of Automated Reasoning 43(2), pp. 173-201, 2009

KBO Orientability 2/32

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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

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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

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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

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Termination

Definition

TRS R is terminating if there is no infinite rewrite sequence t1 →R t2 →R · · ·

KBO Orientability 4/32

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Termination

Definition

TRS R is terminating if there is no infinite rewrite sequence t1 →R t2 →R · · ·

QUESTION

how to prove termination?

KBO Orientability 4/32

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Termination

Definition

TRS R is terminating if there is no infinite rewrite sequence t1 →R t2 →R · · ·

QUESTION

how to prove termination? ☞ Knuth-Bendix order (KBO)

KBO Orientability 4/32

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Termination

Definition

TRS R is terminating if there is no infinite rewrite sequence t1 →R t2 →R · · ·

QUESTION

how to prove termination? ☞ Knuth-Bendix order (KBO) introduced by Knuth and Bendix, 1970

KBO Orientability 4/32

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Termination

Definition

TRS R is terminating if there is no infinite rewrite sequence t1 →R t2 →R · · ·

QUESTION

how to prove termination? ☞ Knuth-Bendix order (KBO) introduced by Knuth and Bendix, 1970 best studied termination methods

KBO Orientability 4/32

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Termination

Definition

TRS R is terminating if there is no infinite rewrite sequence t1 →R t2 →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

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Knuth-Bendix Orders

Definition

precedence > is proper order on function symbols F

KBO Orientability 5/32

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Knuth-Bendix Orders

Definition

precedence > is proper order on function symbols F weight function (w, w0) is pair in R0

F × R0

KBO Orientability 5/32

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Knuth-Bendix Orders

Definition

precedence > is proper order on function symbols F weight function (w, w0) is pair in R0

F × R0

weight of term t is w(t) =

w0

if t ∈ V w(f) + w(t1) + · · · + w(tn) if t = f(t1, . . . , tn)

KBO Orientability 5/32

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Knuth-Bendix Orders

Definition

precedence > is proper order on function symbols F weight function (w, w0) is pair in R0

F × R0

weight of term t is w(t) =

w0

if t ∈ V w(f) + w(t1) + · · · + w(tn) if t = f(t1, . . . , tn) weight function (w, w0) is admissible for precedence > if w(f) > 0

r

f g for all unary functions f and all functions g

KBO Orientability 5/32

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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

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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

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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

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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

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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(s1, . . . , si−1, si, . . . , sn), t = f(s1, . . . , si−1, ti, . . . , tn), and si >kbo ti; or

KBO Orientability 6/32

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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(s1, . . . , si−1, si, . . . , sn), t = f(s1, . . . , si−1, ti, . . . , tn), and si >kbo ti; or s = f(s1, . . . , sn), t = g(t1, . . . , tm), and f > g

KBO Orientability 6/32

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Definition

let X ⊆ R0. TRS R is KBOX terminating if ∃ precedence > ∃ admissible weight function (w, w0) ∈ XF × X such that l >kbo r for all l → r ∈ R

KBO Orientability 7/32

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Definition

let X ⊆ R0. TRS R is KBOX terminating if ∃ precedence > ∃ admissible weight function (w, w0) ∈ XF × X such that l >kbo r for all l → r ∈ R

Theorem Knuth and Bendix, 1970

TRS is terminating if it is KBON terminating

KBO Orientability 7/32

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Quiz

KBO Orientability 8/32

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Example I

a(a(x)) → b(b(b(x))) b(b(b(b(b(x))))) → a(a(a(x)))

KBO Orientability 9/32

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Example I

a(a(x)) → b(b(b(x))) b(b(b(b(b(x))))) → a(a(a(x))) is KBON terminating ?

KBO Orientability 9/32

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Example I

a(a(x)) → b(b(b(x))) b(b(b(b(b(x))))) → a(a(a(x))) is KBON terminating ? — yes

Proof

take precedence a > b and weight function w(a) = ? w(b) = ? w0 = 1

KBO Orientability 9/32

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Example I

a(a(x)) → b(b(b(x))) b(b(b(b(b(x))))) → a(a(a(x))) is KBON terminating ? — yes

Proof

take precedence a > b and weight function w(a) = 3 w(b) = 2 w0 = 1

KBO Orientability 9/32

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Example I

a(a(x)) → b(b(b(x))) b(b(b(b(b(x))))) → a(a(a(x))) is KBON terminating ? — yes

Proof

take precedence a > b and weight function w(a) = 3 w(b) = 2 w0 = 1

Proof

another solution: take precedence > = ∅ and weight function w(a) = ? w(b) = ? w0 = 1

KBO Orientability 9/32

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Example I

a(a(x)) → b(b(b(x))) b(b(b(b(b(x))))) → a(a(a(x))) is KBON terminating ? — yes

Proof

take precedence a > b and weight function w(a) = 3 w(b) = 2 w0 = 1

Proof

another solution: take precedence > = ∅ and weight function w(a) = 13 w(b) = 8 w0 = 1

KBO Orientability 9/32

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Example II

TRS R x + 0 → x x + s(y) → s(x + y) x × 0 → 0 x × s(y) → x × y + x is KBON terminating ?

KBO Orientability 10/32

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Example II

TRS R x + 0 → x x + s(y) → s(x + y) x × 0 → 0 x × s(y) → x × y + x is KBON terminating ? — no

KBO Orientability 10/32

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History of KBO

KBO Orientability 11/32

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Theorem Knuth and Bendix, 1970

TRS is terminating if it is KBON terminating

KBO Orientability 12/32

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Theorem Knuth and Bendix, 1970

TRS is terminating if it is KBON terminating

Theorem Dershowitz, 1979

TRS is terminating if it is KBOR0 terminating

KBO Orientability 12/32

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Theorem Knuth and Bendix, 1970

TRS is terminating if it is KBON terminating

Theorem Dershowitz, 1979

TRS is terminating if it is KBOR0 terminating

Theorem Dick, Kalmus, and Martin, 1990

KBOR0 termination is decidable within exponential time

KBO Orientability 12/32

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Theorem Knuth and Bendix, 1970

TRS is terminating if it is KBON terminating

Theorem Dershowitz, 1979

TRS is terminating if it is KBOR0 terminating

Theorem Dick, Kalmus, and Martin, 1990

KBOR0 termination is decidable within exponential time

Theorem Korovin and Voronkov, 2001, 2003

TRS is KBON terminating ⇐ ⇒ it is KBOR0 terminating KBOR0 termination is decidable within polynomial time

KBO Orientability 12/32

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Theorem Knuth and Bendix, 1970

TRS is terminating if it is KBON terminating

Theorem Dershowitz, 1979

TRS is terminating if it is KBOR0 terminating

Theorem Dick, Kalmus, and Martin, 1990

KBOR0 termination is decidable within exponential time

Theorem Korovin and Voronkov, 2001, 2003

TRS is KBON terminating ⇐ ⇒ it is KBOR0 terminating KBOR0 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

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This Talk

MAIN RESULT

for every TRS R and N =

l→r∈R(|l| + |r|) + 1

R is KBON terminating ⇐ ⇒ R is KBO{0,1,...,B} terminating where, B = N 4N+1

KBO Orientability 13/32

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This Talk

MAIN RESULT

for every TRS R and N =

l→r∈R(|l| + |r|) + 1

R is KBON terminating ⇐ ⇒ R is KBO{0,1,...,B} terminating where, B = N 4N+1

OVERVIEW OF PROOF

Principal Solutions

KBO Orientability 13/32

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This Talk

MAIN RESULT

for every TRS R and N =

l→r∈R(|l| + |r|) + 1

R is KBON terminating ⇐ ⇒ R is KBO{0,1,...,B} terminating where, B = N 4N+1

OVERVIEW OF PROOF

Principal Solutions MCD

KBO Orientability 13/32

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This Talk

MAIN RESULT

for every TRS R and N =

l→r∈R(|l| + |r|) + 1

R is KBON 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

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Principal Solutions

KBO Orientability 14/32

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Example

given precedence, KBON 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

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Example

given precedence, KBON 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

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Example

given precedence, KBON 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

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Principal Solutions

Gale, 1960; Dick, Kalmus and Martin, 1990; Korovin and Voronkov, 2003

set I of KBO inequalities are of form { ai · xi Ri 0 }i with Ri ∈ {, >}, ai ∈ Zn, x ∈ N

KBO Orientability 16/32

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Principal Solutions

Gale, 1960; Dick, Kalmus and Martin, 1990; Korovin and Voronkov, 2003

set I of KBO inequalities are of form { ai · xi Ri 0 }i with Ri ∈ {, >}, ai ∈ Zn, x ∈ N

Definition

write AI for ( ai)1,...,n

KBO Orientability 16/32

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Principal Solutions

Gale, 1960; Dick, Kalmus and Martin, 1990; Korovin and Voronkov, 2003

set I of KBO inequalities are of form { ai · xi Ri 0 }i with Ri ∈ {, >}, ai ∈ Zn, x ∈ N

Definition

write AI for ( ai)1,...,n

x is solution of AI if AI

x 0 and x ∈ Nn

KBO Orientability 16/32

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Principal Solutions

Gale, 1960; Dick, Kalmus and Martin, 1990; Korovin and Voronkov, 2003

set I of KBO inequalities are of form { ai · xi Ri 0 }i with Ri ∈ {, >}, ai ∈ Zn, x ∈ N

Definition

write AI for ( ai)1,...,n

x is solution of AI if AI

x 0 and x ∈ Nn solution x maximizing {i | ai · x > 0} wrt ⊆ is principal

KBO Orientability 16/32

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Principal Solutions

Gale, 1960; Dick, Kalmus and Martin, 1990; Korovin and Voronkov, 2003

set I of KBO inequalities are of form { ai · xi Ri 0 }i with Ri ∈ {, >}, ai ∈ Zn, x ∈ N

Definition

write AI for ( ai)1,...,n

x is solution of AI if AI

x 0 and x ∈ Nn solution x maximizing {i | ai · 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

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I =

2 · w(a) − 3 · w(b)

> −3 · w(a) + 5 · w(b) >

AI =
2

−3 −3 5

KBO Orientability

17/32

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I =

2 · w(a) − 3 · w(b)

> −3 · w(a) + 5 · w(b) >

AI =
2

−3 −3 5

since

AI 3 2

=

1

AI

13 8

=

2 1

3

2

is not principal solution of AI

KBO Orientability 17/32

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I =

2 · w(a) − 3 · w(b)

> −3 · w(a) + 5 · w(b) >

AI =
2

−3 −3 5

since

AI 3 2

=

1

AI

13 8

=

2 1

3

2

is not principal solution of AI
13

8

is principal solution of AI and satisfies I

hence I is solvable

KBO Orientability 17/32

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Example

I =    2 · w(a) − 3 · w(b) > −3 · w(a) + 5 · w(b) > −4 · w(a) + 3 · w(b)



  AI =   2 −3 −3 5 −4 3  

KBO Orientability 18/32

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Example

I =    2 · w(a) − 3 · w(b) > −3 · w(a) + 5 · w(b) > −4 · w(a) + 3 · w(b)



  AI =   2 −3 −3 5 −4 3   principal solution of AI is

and does not satisfy I

KBO Orientability 18/32

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Example

I =    2 · w(a) − 3 · w(b) > −3 · w(a) + 5 · w(b) > −4 · w(a) + 3 · w(b)



  AI =   2 −3 −3 5 −4 3   principal solution of AI is

and does not satisfy I

hence I is not solvable

KBO Orientability 18/32

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Example

I =    2 · w(a) − 3 · w(b) > −3 · w(a) + 5 · w(b) > −4 · w(a) + 3 · w(b)



  AI =   2 −3 −3 5 −4 3   principal solution of AI is

and does not satisfy I

hence I is not solvable

QUESTION

how to find principal solution?

KBO Orientability 18/32

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Example

I =    2 · w(a) − 3 · w(b) > −3 · w(a) + 5 · w(b) > −4 · w(a) + 3 · w(b)



  AI =   2 −3 −3 5 −4 3   principal solution of AI is

and does not satisfy I

hence I is not solvable

QUESTION

how to find principal solution? ☞ MCD

KBO Orientability 18/32

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MCD: Method of Complete Description

KBO Orientability 19/32

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MCD

Dick, Kalmus and Martin, 1990

(a1, . . . , an)κ = ( ei | ai 0) + +(aj ei − ai ej | ai < 0, aj > 0)

KBO Orientability 20/32

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MCD

Dick, Kalmus and Martin, 1990

(a1, . . . , an)κ = ( ei | ai 0) + +(aj ei − ai ej | ai < 0, aj > 0) for m × n matrix A and 0 i m SA

i =

En

if i = 0 SA

i−1(

aiSA

i−1)κ

therwise

KBO Orientability 20/32

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MCD

Dick, Kalmus and Martin, 1990

(a1, . . . , an)κ = ( ei | ai 0) + +(aj ei − ai ej | ai < 0, aj > 0) for m × n matrix A and 0 i m SA

i =

En

if i = 0 SA

i−1(

aiSA

i−1)κ

therwise

sum of all column vectors of SA

m is denoted by

sA

KBO Orientability 20/32

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MCD

Dick, Kalmus and Martin, 1990

(a1, . . . , an)κ = ( ei | ai 0) + +(aj ei − ai ej | ai < 0, aj > 0) for m × n matrix A and 0 i m SA

i =

En

if i = 0 SA

i−1(

aiSA

i−1)κ

therwise

sum of all column vectors of SA

m is denoted by

sA

Theorem

sA is principal solution of A

KBO Orientability 20/32

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Example

I =

2 · w(a) − 3 · w(b)

> −3 · w(a) + 5 · w(b) >

AI =

2 −3 −3 5

KBO Orientability

21/32

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Example

I =

2 · w(a) − 3 · w(b)

> −3 · w(a) + 5 · w(b) >

AI =

2 −3 −3 5

performing MCD

KBO Orientability 21/32

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Example

I =

2 · w(a) − 3 · w(b)

> −3 · w(a) + 5 · w(b) >

AI =

2 −3 −3 5

performing MCD

SAI =

1

1

SAI

1

= SAI

0 ((2 − 3)SAI 0 )κ =

1 3 2

SAI

2

= SAI

1 ((−3 5)SAI 1 )κ =

3 10 2 6

KBO Orientability

21/32

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Example

I =

2 · w(a) − 3 · w(b)

> −3 · w(a) + 5 · w(b) >

AI =

2 −3 −3 5

performing MCD

SAI =

1

1

SAI

1

= SAI

0 ((2 − 3)SAI 0 )κ =

1 3 2

SAI

2

= SAI

1 ((−3 5)SAI 1 )κ =

3 10 2 6

we obtain principal solution
sAI =
3

2

+
10

6

=
13

8

KBO Orientability

21/32

SLIDE 69

Example

I =

2 · w(a) − 3 · w(b)

> −3 · w(a) + 5 · w(b) >

AI =

2 −3 −3 5

performing MCD

SAI =

1

1

SAI

1

= SAI

0 ((2 − 3)SAI 0 )κ =

1 3 2

SAI

2

= SAI

1 ((−3 5)SAI 1 )κ =

3 10 2 6

we obtain principal solution
sAI =
3

2

+
10

6

=
13

8

which satisfies I. hence I is solvable

KBO Orientability 21/32

SLIDE 70

Bound by Norm

KBO Orientability 22/32

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MCD

(a1, . . . , an)κ = ( ei | ai 0) + +(aj ei − ai ej | ai < 0, aj > 0) for m × n matrix A and 0 i m SA

i =

En

if i = 0 SA

i−1(

aiSA

i−1)κ

therwise

sum of all column vectors of SA

m is denoted by

sA

KBO Orientability 23/32

SLIDE 72

MCD

(a1, . . . , an)κ = ( ei | ai 0) + +(aj ei − ai ej | ai < 0, aj > 0) for m × n matrix A and 0 i m SA

i =

En

if i = 0 SA

i−1(

aiSA

i−1)κ

therwise

sum of all column vectors of SA

m is denoted by

sA

GOAL

bound sA ∈ {0, 1, . . . , B}n

KBO Orientability 23/32

SLIDE 73

MCD

(a1, . . . , an)κ = ( ei | ai 0) + +(aj ei − ai ej | ai < 0, aj > 0) for m × n matrix A and 0 i m SA

i =

En

if i = 0 SA

i−1(

aiSA

i−1)κ

therwise

sum of all column vectors of SA

m is denoted by

sA

GOAL

bound sA ∈ {0, 1, . . . , B}n

APPROACH

define norm · to calculate sAI with B = sAI

KBO Orientability 23/32

SLIDE 74

Norm

Definition

A = maxi,j |aij| for m × n matrix A = (aij)ij

KBO Orientability 24/32

SLIDE 75

Norm

Definition

A = maxi,j |aij| for m × n matrix A = (aij)ij

Lemma

for every m × n matrix A and n × p matrix B aκ = a

KBO Orientability 24/32

SLIDE 76

Norm

Definition

A = maxi,j |aij| for m × n matrix A = (aij)ij

Lemma

for every m × n matrix A and n × p matrix B aκ = a AB nAB

KBO Orientability 24/32

SLIDE 77

Norm

Definition

A = maxi,j |aij| for m × n matrix A = (aij)ij

Lemma

for every m × n matrix A and n × p matrix B aκ = a AB nAB (a1, . . . , an)κ is n × k matrix with k n2. SA

i is m × k matrix with k n2i.

KBO Orientability 24/32

SLIDE 78

Norm

Definition

A = maxi,j |aij| for m × n matrix A = (aij)ij

Lemma

for every m × n matrix A and n × p matrix B aκ = a AB nAB (a1, . . . , an)κ is n × k matrix with k n2. SA

i is m × k matrix with k n2i.

SA

i (2mA)2i−1

sA n2m · (2mA)2m−1

KBO Orientability 24/32

SLIDE 79

Lemma

let R be TRS of size N I be set of inequalities induced by KBO with fixed precedence AI be of size m × n m N n N AI N therefore

sAI N 4N+1 := B

thus,

sAI ∈ {0, 1, . . . , B}n

KBO Orientability 25/32

SLIDE 80

Main Result

Theorem

R is KBON terminating ⇐ ⇒ R is KBO{0,1,...,B} terminating where, B = N 4N+1

KBO Orientability 26/32

SLIDE 81

Main Result

Theorem

R is KBON 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

SLIDE 82

Summary

let R be TRS, N =

l→r∈R(|l| + |r|) + 1, and B = N 4N+1

R is KBOR0 terminating ⇐ ⇒ R is KBON terminating

Korovin and Voronkov

⇐ ⇒ R is KBO{0,1,...,B} terminating

this talk

KBO Orientability 27/32

SLIDE 83

Summary

let R be TRS, N =

l→r∈R(|l| + |r|) + 1, and B = N 4N+1

R is KBOR0 terminating ⇐ ⇒ R is KBON 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

SLIDE 84

Summary

let R be TRS, N =

l→r∈R(|l| + |r|) + 1, and B = N 4N+1

R is KBOR0 terminating ⇐ ⇒ R is KBON 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? ☞

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SLIDE 85

Automation

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SLIDE 86

Theorem Dick, Kalmus, and Martin, 1990

KBOR0 termination is decidable by MCD

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SLIDE 87

Theorem Dick, Kalmus, and Martin, 1990

KBOR0 termination is decidable by MCD

Theorem Korovin and Voronkov, 2001, 2003

KBOR0 termination is decidable by Karmarkar/Khachiyan algorithm

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SLIDE 88

Theorem Dick, Kalmus, and Martin, 1990

KBOR0 termination is decidable by MCD

Theorem Korovin and Voronkov, 2001, 2003

KBOR0 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

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SLIDE 89

Theorem Dick, Kalmus, and Martin, 1990

KBOR0 termination is decidable by MCD

Theorem Korovin and Voronkov, 2001, 2003

KBOR0 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

KBOR0 termination is decidable via SMT (linear arithmetic) encoding

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SLIDE 90

Experiments

test-bed: 1381 TRSs from Termination Problem Data Base 4.0

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SLIDE 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

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SLIDE 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

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SLIDE 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 444 102 7

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SLIDE 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 444 102 7 simplex 370 105 4

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SLIDE 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 444 102 7 simplex 370 105 4 SAT/PBC(4) 31/90 104/104 0/0

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SLIDE 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 444 102 7 simplex 370 105 4 SAT/PBC(4) 31/90 104/104 0/0 SAT/PBC(8) 32/93 106/106 0/0

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SLIDE 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 444 102 7 simplex 370 105 4 SAT/PBC(4) 31/90 104/104 0/0 SAT/PBC(8) 32/93 106/106 0/0 SAT/PBC(16) 34/100 107/107 0/0

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SLIDE 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 444 102 7 simplex 370 105 4 SAT/PBC(4) 31/90 104/104 0/0 SAT/PBC(8) 32/93 106/106 0/0 SAT/PBC(16) 34/100 107/107 0/0 SAT/PBC(1024) 351/187 107/107 3/1

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SLIDE 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 444 102 7 simplex 370 105 4 SAT/PBC(4) 31/90 104/104 0/0 SAT/PBC(8) 32/93 106/106 0/0 SAT/PBC(16) 34/100 107/107 0/0 SAT/PBC(1024) 351/187 107/107 3/1 SMT 26 107

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SLIDE 100

Conclusion

let R be TRS, N =

l→r∈R(|l| + |r|) + 1, and B = N 4N+1

R is KBOR0 terminating ⇐ ⇒ R is KBON terminating

Korovin and Voronkov

⇐ ⇒ R is KBO{0,1,...,B} terminating

this talk

CONCLUSION

finite characterization of KBO orientability use SMT solver to automate

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SLIDE 101

Conclusion

let R be TRS, N =

l→r∈R(|l| + |r|) + 1, and B = N 4N+1

R is KBOR0 terminating ⇐ ⇒ R is KBON 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

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SLIDE 102

Thank You for Your Attention

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