Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/96119)

The rewrite relation of the following TRS is considered.

0(1(2(x1)))	→	0(3(3(x1)))	(1)
0(4(2(5(3(2(x1))))))	→	0(1(2(4(0(x1)))))	(2)
4(4(1(4(2(1(1(x1)))))))	→	5(0(1(1(1(5(x1))))))	(3)
4(5(0(3(2(5(3(x1)))))))	→	4(1(1(0(2(3(x1))))))	(4)
0(0(1(0(0(3(5(3(x1))))))))	→	0(0(0(2(3(2(2(3(x1))))))))	(5)
5(1(1(3(1(5(4(2(x1))))))))	→	5(5(3(4(4(2(0(2(x1))))))))	(6)
1(1(0(0(5(2(5(2(2(x1)))))))))	→	1(3(1(1(5(3(3(5(1(x1)))))))))	(7)
4(3(2(4(0(0(2(2(3(5(0(x1)))))))))))	→	1(0(5(2(2(5(2(1(5(5(x1))))))))))	(8)
5(1(2(5(5(2(4(1(1(5(2(x1)))))))))))	→	5(1(5(2(0(3(5(0(1(4(1(x1)))))))))))	(9)
1(0(4(3(2(1(4(1(5(1(1(2(x1))))))))))))	→	4(2(0(3(3(5(3(1(4(5(3(3(5(x1)))))))))))))	(10)
5(2(0(5(0(1(1(1(2(4(4(2(x1))))))))))))	→	5(2(0(2(0(3(3(0(0(4(2(4(x1))))))))))))	(11)
4(1(5(5(5(3(3(1(3(3(3(4(5(x1)))))))))))))	→	4(1(4(3(4(5(1(1(1(1(2(5(x1))))))))))))	(12)
1(1(5(5(0(3(1(4(0(0(3(5(5(4(x1))))))))))))))	→	0(1(3(1(3(0(3(4(5(0(4(1(5(4(x1))))))))))))))	(13)
3(1(3(0(1(4(5(2(5(0(2(2(4(3(x1))))))))))))))	→	3(0(5(2(2(0(4(5(2(1(5(5(5(x1)))))))))))))	(14)
5(3(1(5(4(5(4(5(1(1(1(2(2(1(5(2(x1))))))))))))))))	→	5(0(0(5(2(3(2(4(4(1(1(1(5(5(5(x1)))))))))))))))	(15)
0(3(0(2(1(3(4(2(1(5(2(4(3(0(1(0(2(x1)))))))))))))))))	→	2(0(4(5(5(2(0(0(1(0(3(2(5(4(1(3(3(x1)))))))))))))))))	(16)
1(2(0(4(4(0(3(0(5(2(4(0(3(3(1(1(4(x1)))))))))))))))))	→	4(5(4(0(4(3(5(0(2(5(0(5(3(0(3(1(4(x1)))))))))))))))))	(17)
4(0(1(5(4(1(3(5(5(0(0(5(4(5(4(0(5(4(x1))))))))))))))))))	→	2(1(2(5(0(0(4(1(0(1(5(5(4(5(2(4(1(4(x1))))))))))))))))))	(18)
4(4(4(4(4(1(3(4(1(5(1(4(0(1(4(0(4(5(x1))))))))))))))))))	→	4(2(2(3(1(2(3(5(4(3(3(4(2(1(2(0(3(4(5(x1)))))))))))))))))))	(19)
4(3(3(3(5(5(1(2(5(0(4(1(4(4(2(3(0(4(2(x1)))))))))))))))))))	→	4(3(4(5(5(0(2(5(5(0(2(2(5(2(0(0(3(3(x1))))))))))))))))))	(20)
0(3(2(2(5(4(0(2(4(0(1(3(3(1(0(3(4(3(1(5(x1))))))))))))))))))))	→	0(3(1(0(5(4(3(2(0(3(3(4(1(0(4(3(3(3(1(1(x1))))))))))))))))))))	(21)
1(0(5(2(2(3(2(0(5(5(3(4(4(3(1(0(0(2(1(2(x1))))))))))))))))))))	→	1(3(1(4(5(3(3(3(3(0(2(5(1(4(5(4(3(1(5(1(x1))))))))))))))))))))	(22)
5(4(4(2(2(2(3(0(1(5(3(5(1(0(2(3(1(5(4(1(x1))))))))))))))))))))	→	5(1(0(5(1(1(3(0(5(0(3(5(5(2(4(0(1(5(5(x1)))))))))))))))))))	(23)
3(0(2(2(2(0(0(4(3(2(2(5(5(0(3(2(1(2(2(2(0(x1)))))))))))))))))))))	→	5(2(2(5(1(3(5(0(3(2(1(1(1(0(4(5(3(4(0(5(x1))))))))))))))))))))	(24)
4(4(1(2(0(1(2(3(1(0(1(0(0(3(3(4(4(1(1(1(2(x1)))))))))))))))))))))	→	4(1(0(1(0(3(1(4(1(5(3(3(3(0(1(2(3(1(2(5(1(x1)))))))))))))))))))))	(25)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Closure Under Flat Contexts

Using the flat contexts

{5(☐), 4(☐), 3(☐), 2(☐), 1(☐), 0(☐)}

We obtain the transformed TRS

There are 150 ruless (increase limit for explicit display).

1.1 Semantic Labeling

The following interpretations form a model of the rules.

As carrier we take the set {0,...,5}. Symbols are labeled by the interpretation of their arguments using the interpretations (modulo 6):

[5(x₁)]	=	6x₁ + 0
[4(x₁)]	=	6x₁ + 1
[3(x₁)]	=	6x₁ + 2
[2(x₁)]	=	6x₁ + 3
[1(x₁)]	=	6x₁ + 4
[0(x₁)]	=	6x₁ + 5

We obtain the labeled TRS

There are 900 ruless (increase limit for explicit display).

1.1.1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1

[5₀(x₁)]

x₁ +

1471385/6297

[5₁(x₁)]

x₁ +

2003747/6297

[5₂(x₁)]

x₁ +

1997450/6297

[5₃(x₁)]

x₁ +

[5₄(x₁)]

x₁ +

[5₅(x₁)]

x₁ +

148210/6297

[4₀(x₁)]

x₁ +

[4₁(x₁)]

x₁ +

2028935/6297

[4₂(x₁)]

x₁ +

1997450/6297

[4₃(x₁)]

x₁ +

2028935/6297

[4₄(x₁)]

x₁ +

714980/6297

[4₅(x₁)]

x₁ +

1510340/6297

[3₀(x₁)]

x₁ +

[3₁(x₁)]

x₁ +

[3₂(x₁)]

x₁ +

[3₃(x₁)]

x₁ +

2028935/6297

[3₄(x₁)]

x₁ +

415515/2099

[3₅(x₁)]

x₁ +

659645/2099

[2₀(x₁)]

x₁ +

309919/2099

[2₁(x₁)]

x₁ +

1997450/6297

[2₂(x₁)]

x₁ +

883120/6297

[2₃(x₁)]

x₁ +

2003747/6297

[2₄(x₁)]

x₁ +

1997450/6297

[2₅(x₁)]

x₁ +

584920/6297

[1₀(x₁)]

x₁ +

1997450/6297

[1₁(x₁)]

x₁ +

1059925/6297

[1₂(x₁)]

x₁ +

508760/2099

[1₃(x₁)]

x₁ +

1997450/6297

[1₄(x₁)]

x₁ +

1271357/6297

[1₅(x₁)]

x₁ +

1997450/6297

[0₀(x₁)]

x₁ +

607419/2099

[0₁(x₁)]

x₁ +

474215/2099

[0₂(x₁)]

x₁ +

1807105/6297

[0₃(x₁)]

x₁ +

1950095/6297

[0₄(x₁)]

x₁ +

646905/2099

[0₅(x₁)]

x₁ +

1515202/6297

all of the following rules can be deleted.

There are 900 ruless (increase limit for explicit display).

1.1.1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.