Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/212043)

The rewrite relation of the following TRS is considered.

0(1(2(x1)))	→	1(0(0(2(x1))))	(1)
0(1(2(x1)))	→	1(0(3(2(x1))))	(2)
0(1(2(x1)))	→	1(0(0(3(2(x1)))))	(3)
0(1(2(x1)))	→	4(5(1(0(2(x1)))))	(4)
0(5(2(x1)))	→	5(0(0(2(x1))))	(5)
0(5(2(x1)))	→	5(5(0(2(x1))))	(6)
0(5(2(x1)))	→	5(0(3(3(2(x1)))))	(7)
0(5(3(x1)))	→	5(0(0(3(x1))))	(8)
0(5(3(x1)))	→	5(5(0(3(x1))))	(9)
0(0(5(2(x1))))	→	0(2(0(3(5(5(x1))))))	(10)
0(1(2(3(x1))))	→	1(3(2(0(3(x1)))))	(11)
0(1(2(4(x1))))	→	4(5(1(0(2(x1)))))	(12)
0(1(4(2(x1))))	→	4(4(1(0(2(x1)))))	(13)
0(4(1(2(x1))))	→	0(4(5(5(1(2(x1))))))	(14)
0(5(2(3(x1))))	→	5(0(3(2(3(x1)))))	(15)
1(2(1(2(x1))))	→	1(1(5(2(2(x1)))))	(16)
4(0(1(2(x1))))	→	4(1(0(0(2(x1)))))	(17)
4(3(0(2(x1))))	→	4(0(0(3(2(x1)))))	(18)
4(3(1(2(x1))))	→	3(2(5(4(1(1(x1))))))	(19)
0(0(1(2(3(x1)))))	→	3(2(1(0(0(3(x1))))))	(20)
0(0(1(3(2(x1)))))	→	1(0(2(0(0(3(x1))))))	(21)
0(0(1(3(3(x1)))))	→	0(0(3(1(0(3(x1))))))	(22)
0(0(1(5(2(x1)))))	→	1(5(0(0(3(2(x1))))))	(23)
0(1(2(1(2(x1)))))	→	1(1(0(2(2(2(x1))))))	(24)
0(1(4(5(2(x1)))))	→	4(1(0(3(2(5(x1))))))	(25)
0(5(0(1(2(x1)))))	→	1(2(0(1(5(0(x1))))))	(26)
0(5(1(0(2(x1)))))	→	1(5(0(0(3(2(x1))))))	(27)
0(5(1(4(3(x1)))))	→	1(0(3(5(4(5(x1))))))	(28)
0(5(1(4(3(x1)))))	→	4(5(5(1(0(3(x1))))))	(29)
0(5(3(1(2(x1)))))	→	0(1(5(0(2(3(x1))))))	(30)
0(5(3(1(2(x1)))))	→	5(0(1(4(3(2(x1))))))	(31)
0(5(3(4(2(x1)))))	→	3(2(0(3(5(4(x1))))))	(32)
0(5(4(3(2(x1)))))	→	0(0(4(3(2(5(x1))))))	(33)
1(2(5(2(3(x1)))))	→	5(1(2(3(3(2(x1))))))	(34)
1(3(0(5(2(x1)))))	→	0(5(1(0(3(2(x1))))))	(35)
1(3(0(5(2(x1)))))	→	1(3(0(0(2(5(x1))))))	(36)
1(3(3(4(2(x1)))))	→	5(1(3(3(2(4(x1))))))	(37)
4(3(0(2(3(x1)))))	→	0(3(3(3(2(4(x1))))))	(38)
4(3(0(5(3(x1)))))	→	5(4(3(5(0(3(x1))))))	(39)
4(3(3(1(2(x1)))))	→	1(3(0(4(3(2(x1))))))	(40)
5(2(3(0(2(x1)))))	→	4(5(0(2(3(2(x1))))))	(41)
5(2(3(1(2(x1)))))	→	2(3(2(4(1(5(x1))))))	(42)
5(3(0(2(2(x1)))))	→	5(3(3(2(0(2(x1))))))	(43)
5(3(0(5(2(x1)))))	→	5(5(3(0(2(4(x1))))))	(44)
5(3(1(2(2(x1)))))	→	5(1(3(2(2(2(x1))))))	(45)
5(3(1(5(2(x1)))))	→	2(5(5(1(5(3(x1))))))	(46)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Closure Under Flat Contexts

Using the flat contexts

{0(☐), 1(☐), 2(☐), 3(☐), 4(☐), 5(☐)}

We obtain the transformed TRS

There are 215 ruless (increase limit for explicit display).

1.1 Semantic Labeling

Root-labeling is applied.

We obtain the labeled TRS

There are 1290 ruless (increase limit for explicit display).

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals

[0₀(x₁)]	=	1 · x₁
[0₅(x₁)]	=	1 · x₁ + 16
[5₂(x₁)]	=	1 · x₁ + 72
[2₀(x₁)]	=	1 · x₁
[0₂(x₁)]	=	1 · x₁
[0₃(x₁)]	=	1 · x₁
[3₅(x₁)]	=	1 · x₁
[5₅(x₁)]	=	1 · x₁ + 6
[5₀(x₁)]	=	1 · x₁
[2₅(x₁)]	=	1 · x₁ + 4
[2₂(x₁)]	=	1 · x₁ + 61
[2₃(x₁)]	=	1 · x₁ + 38
[5₃(x₁)]	=	1 · x₁ + 48
[2₄(x₁)]	=	1 · x₁ + 52
[5₄(x₁)]	=	1 · x₁
[2₁(x₁)]	=	1 · x₁ + 4
[5₁(x₁)]	=	1 · x₁
[0₄(x₁)]	=	1 · x₁
[4₁(x₁)]	=	1 · x₁ + 7
[1₂(x₁)]	=	1 · x₁ + 65
[4₅(x₁)]	=	1 · x₁
[1₁(x₁)]	=	1 · x₁
[1₅(x₁)]	=	1 · x₁
[4₀(x₁)]	=	1 · x₁ + 4
[0₁(x₁)]	=	1 · x₁ + 3
[1₀(x₁)]	=	1 · x₁ + 8
[4₃(x₁)]	=	1 · x₁
[3₀(x₁)]	=	1 · x₁ + 37
[3₂(x₁)]	=	1 · x₁ + 19
[1₃(x₁)]	=	1 · x₁ + 46
[3₃(x₁)]	=	1 · x₁ + 1
[3₁(x₁)]	=	1 · x₁ + 41
[3₄(x₁)]	=	1 · x₁ + 89
[4₂(x₁)]	=	1 · x₁ + 18
[4₄(x₁)]	=	1 · x₁ + 3
[1₄(x₁)]	=	1 · x₁ + 50

all of the following rules can be deleted.

There are 1273 ruless (increase limit for explicit display).

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals

[2₀(x₁)]	=	1 · x₁
[0₅(x₁)]	=	1 · x₁
[5₀(x₁)]	=	1 · x₁
[0₁(x₁)]	=	1 · x₁
[1₂(x₁)]	=	1 · x₁
[2₅(x₁)]	=	1 · x₁
[2₁(x₁)]	=	1 · x₁
[1₅(x₁)]	=	1 · x₁
[3₀(x₁)]	=	1 · x₁ + 1
[3₁(x₁)]	=	1 · x₁ + 1
[5₁(x₁)]	=	1 · x₁
[1₄(x₁)]	=	1 · x₁ + 1
[4₃(x₁)]	=	1 · x₁
[2₄(x₁)]	=	1 · x₁
[4₅(x₁)]	=	1 · x₁
[5₅(x₁)]	=	1 · x₁
[1₀(x₁)]	=	1 · x₁
[0₃(x₁)]	=	1 · x₁
[3₅(x₁)]	=	1 · x₁
[3₂(x₁)]	=	1 · x₁
[3₃(x₁)]	=	1 · x₁
[3₄(x₁)]	=	1 · x₁
[5₃(x₁)]	=	1 · x₁
[5₂(x₁)]	=	1 · x₁

all of the following rules can be deleted.

2₀(0₅(5₁(1₄(4₃(3₀(x1))))))	→	2₄(4₅(5₅(5₁(1₀(0₃(3₀(x1)))))))	(1120)
2₀(0₅(5₁(1₄(4₃(3₅(x1))))))	→	2₄(4₅(5₅(5₁(1₀(0₃(3₅(x1)))))))	(1121)
2₀(0₅(5₁(1₄(4₃(3₂(x1))))))	→	2₄(4₅(5₅(5₁(1₀(0₃(3₂(x1)))))))	(1122)
2₀(0₅(5₁(1₄(4₃(3₃(x1))))))	→	2₄(4₅(5₅(5₁(1₀(0₃(3₃(x1)))))))	(1123)
2₀(0₅(5₁(1₄(4₃(3₄(x1))))))	→	2₄(4₅(5₅(5₁(1₀(0₃(3₄(x1)))))))	(1124)
2₀(0₅(5₁(1₄(4₃(3₁(x1))))))	→	2₄(4₅(5₅(5₁(1₀(0₃(3₁(x1)))))))	(1125)
3₀(0₅(5₁(1₄(4₃(3₀(x1))))))	→	3₄(4₅(5₅(5₁(1₀(0₃(3₀(x1)))))))	(1126)
3₀(0₅(5₁(1₄(4₃(3₅(x1))))))	→	3₄(4₅(5₅(5₁(1₀(0₃(3₅(x1)))))))	(1127)
3₀(0₅(5₁(1₄(4₃(3₂(x1))))))	→	3₄(4₅(5₅(5₁(1₀(0₃(3₂(x1)))))))	(1128)
3₀(0₅(5₁(1₄(4₃(3₃(x1))))))	→	3₄(4₅(5₅(5₁(1₀(0₃(3₃(x1)))))))	(1129)
3₀(0₅(5₁(1₄(4₃(3₄(x1))))))	→	3₄(4₅(5₅(5₁(1₀(0₃(3₄(x1)))))))	(1130)
3₀(0₅(5₁(1₄(4₃(3₁(x1))))))	→	3₄(4₅(5₅(5₁(1₀(0₃(3₁(x1)))))))	(1131)
5₅(5₃(3₁(1₅(5₂(2₄(x1))))))	→	5₂(2₅(5₅(5₁(1₅(5₃(3₄(x1)))))))	(1538)

1.1.1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

2₀^#(0₅(5₀(0₁(1₂(2₅(x1))))))	→	2₀^#(0₁(1₅(5₀(0₅(x1)))))	(1540)
3₀^#(0₅(5₀(0₁(1₂(2₅(x1))))))	→	2₀^#(0₁(1₅(5₀(0₅(x1)))))	(1541)
5₅^#(5₃(3₁(1₅(5₂(2₀(x1))))))	→	5₅^#(5₁(1₅(5₃(3₀(x1)))))	(1542)
5₅^#(5₃(3₁(1₅(5₂(2₀(x1))))))	→	3₀^#(x1)	(1543)
5₅^#(5₃(3₁(1₅(5₂(2₁(x1))))))	→	5₅^#(5₁(1₅(5₃(3₁(x1)))))	(1544)

1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.