Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/96403)

The rewrite relation of the following TRS is considered.

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

80/11

[5₁(x₁)]

x₁ +

940/11

[5₂(x₁)]

x₁ +

6331/11

[5₃(x₁)]

x₁ +

6439/11

[5₄(x₁)]

x₁ +

6439/11

[5₅(x₁)]

x₁ +

6298/11

[4₀(x₁)]

x₁ +

4498/11

[4₁(x₁)]

x₁ +

6439/11

[4₂(x₁)]

x₁ +

6450/11

[4₃(x₁)]

x₁ +

1410/11

[4₄(x₁)]

x₁ +

[4₅(x₁)]

x₁ +

3995/11

[3₀(x₁)]

x₁ +

6560/11

[3₁(x₁)]

x₁ +

6450/11

[3₂(x₁)]

x₁ +

6472/11

[3₃(x₁)]

x₁ +

1421/11

[3₄(x₁)]

x₁ +

6439/11

[3₅(x₁)]

x₁ +

4028/11

[2₀(x₁)]

x₁ +

6439/11

[2₁(x₁)]

x₁ +

[2₂(x₁)]

x₁ +

[2₃(x₁)]

x₁ +

[2₄(x₁)]

x₁ +

6560/11

[2₅(x₁)]

x₁ +

[1₀(x₁)]

x₁ +

[1₁(x₁)]

x₁ +

[1₂(x₁)]

x₁ +

6439/11

[1₃(x₁)]

x₁ +

6439/11

[1₄(x₁)]

x₁ +

[1₅(x₁)]

x₁ +

[0₀(x₁)]

x₁ +

[0₁(x₁)]

x₁ +

4451/11

[0₂(x₁)]

x₁ +

6439/11

[0₃(x₁)]

x₁ +

[0₄(x₁)]

x₁ +

[0₅(x₁)]

x₁ +

6439/11

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.