Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/96198)

The rewrite relation of the following TRS is considered.

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

64527/472

[5₁(x₁)]

x₁ +

849325/708

[5₂(x₁)]

x₁ +

[5₃(x₁)]

x₁ +

119/24

[5₄(x₁)]

x₁ +

568557/472

[5₅(x₁)]

x₁ +

329087/1416

[4₀(x₁)]

x₁ +

70108/177

[4₁(x₁)]

x₁ +

9275/177

[4₂(x₁)]

x₁ +

849325/708

[4₃(x₁)]

x₁ +

849325/708

[4₄(x₁)]

x₁ +

[4₅(x₁)]

x₁ +

1725613/1416

[3₀(x₁)]

x₁ +

529205/708

[3₁(x₁)]

x₁ +

568557/472

[3₂(x₁)]

x₁ +

1690831/1416

[3₃(x₁)]

x₁ +

119/24

[3₄(x₁)]

x₁ +

1701541/1416

[3₅(x₁)]

x₁ +

1593311/1416

[2₀(x₁)]

x₁ +

119/24

[2₁(x₁)]

x₁ +

1699063/1416

[2₂(x₁)]

x₁ +

[2₃(x₁)]

x₁ +

148371/236

[2₄(x₁)]

x₁ +

89701/1416

[2₅(x₁)]

x₁ +

742265/708

[1₀(x₁)]

x₁ +

[1₁(x₁)]

x₁ +

77293/708

[1₂(x₁)]

x₁ +

145351/1416

[1₃(x₁)]

x₁ +

1594727/1416

[1₄(x₁)]

x₁ +

53397/472

[1₅(x₁)]

x₁ +

47831/1416

[0₀(x₁)]

x₁ +

504369/472

[0₁(x₁)]

x₁ +

336815/708

[0₂(x₁)]

x₁ +

568557/472

[0₃(x₁)]

x₁ +

143/24

[0₄(x₁)]

x₁ +

614807/1416

[0₅(x₁)]

x₁ +

7/2

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.