Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/96370)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS

2(0(0(1(0(0(x1))))))	→	4(4(3(2(0(0(x1))))))	(26)
3(1(4(4(2(3(x1))))))	→	3(1(2(3(5(5(x1))))))	(27)
0(2(3(1(0(5(x1))))))	→	1(0(1(5(5(x1)))))	(28)
1(1(2(0(1(5(1(x1)))))))	→	4(0(1(1(2(3(1(x1)))))))	(29)
0(5(5(5(0(0(1(0(x1))))))))	→	0(0(4(0(1(2(2(0(x1))))))))	(30)
3(1(0(2(1(5(2(1(x1))))))))	→	3(5(3(4(5(2(4(1(x1))))))))	(31)
0(2(3(2(4(0(5(4(x1))))))))	→	0(2(3(2(4(1(1(x1)))))))	(32)
3(2(4(5(1(0(4(2(0(x1)))))))))	→	3(3(0(0(4(3(0(2(2(x1)))))))))	(33)
0(0(2(5(0(3(1(5(1(x1)))))))))	→	5(2(2(0(5(2(3(1(1(x1)))))))))	(34)
0(1(2(4(1(4(2(0(1(5(x1))))))))))	→	3(2(3(0(4(2(1(5(3(1(x1))))))))))	(35)
1(0(0(5(0(2(0(5(2(1(0(x1)))))))))))	→	1(2(1(3(0(2(1(4(1(0(x1))))))))))	(36)
0(2(1(4(1(5(1(1(3(2(3(3(1(x1)))))))))))))	→	3(4(2(3(5(5(4(4(4(5(5(3(1(x1)))))))))))))	(37)
0(1(0(5(2(4(4(2(5(1(4(1(2(x1)))))))))))))	→	0(3(0(1(3(4(4(1(0(1(2(0(x1))))))))))))	(38)
1(1(1(4(2(2(5(0(0(0(1(2(2(x1)))))))))))))	→	5(2(4(2(2(0(2(5(2(0(4(1(1(x1)))))))))))))	(39)
1(1(2(4(1(5(0(0(1(3(5(5(5(3(0(x1)))))))))))))))	→	1(2(5(4(2(3(1(2(1(0(1(2(5(3(0(x1)))))))))))))))	(40)
5(1(0(3(2(3(4(1(5(5(2(4(0(0(2(x1)))))))))))))))	→	4(2(5(4(4(2(5(4(2(3(1(0(1(2(x1))))))))))))))	(41)
3(2(1(3(5(5(4(5(3(3(2(0(1(5(2(2(x1))))))))))))))))	→	3(3(5(3(4(0(1(5(3(1(1(1(2(4(2(x1)))))))))))))))	(42)
0(0(4(2(3(3(2(4(4(4(1(4(2(0(2(1(3(x1)))))))))))))))))	→	0(0(2(3(1(4(5(1(2(0(4(4(1(2(1(5(2(x1)))))))))))))))))	(43)
0(2(1(4(0(5(0(4(4(3(0(0(2(4(0(0(5(0(x1))))))))))))))))))	→	4(2(0(4(2(1(5(2(3(2(2(4(0(3(5(5(3(0(x1))))))))))))))))))	(44)
4(5(5(5(4(1(4(2(2(2(4(0(4(3(2(5(5(5(x1))))))))))))))))))	→	4(4(3(4(1(2(0(5(2(2(3(3(2(1(1(0(3(x1)))))))))))))))))	(45)
0(1(2(0(1(4(2(2(2(0(2(5(4(5(2(3(4(5(1(4(x1))))))))))))))))))))	→	0(0(2(3(1(3(4(0(4(1(0(2(4(0(2(3(2(5(5(4(x1))))))))))))))))))))	(46)
2(0(1(5(0(3(5(1(4(4(3(5(3(5(3(2(2(1(4(0(2(x1)))))))))))))))))))))	→	2(3(0(4(0(3(5(3(3(3(5(4(1(0(0(2(4(4(2(2(2(x1)))))))))))))))))))))	(47)
0(5(3(5(0(4(0(0(3(0(4(3(2(4(0(5(2(4(2(2(3(x1)))))))))))))))))))))	→	4(4(0(2(5(3(1(5(5(1(2(3(1(4(5(1(4(1(1(x1)))))))))))))))))))	(48)
3(5(1(3(4(2(5(2(5(3(4(5(5(1(2(5(1(1(0(1(5(x1)))))))))))))))))))))	→	3(0(5(2(0(3(4(1(0(4(5(2(2(3(4(5(4(3(0(1(1(x1)))))))))))))))))))))	(49)
2(4(3(1(3(3(1(3(4(1(2(3(2(0(3(2(5(4(0(1(5(x1)))))))))))))))))))))	→	3(1(0(0(1(3(4(4(0(3(2(2(5(4(0(5(3(1(1(0(x1))))))))))))))))))))	(50)

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals

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

all of the following rules can be deleted.

0(2(3(1(0(5(x1))))))	→	1(0(1(5(5(x1)))))	(28)
0(2(3(2(4(0(5(4(x1))))))))	→	0(2(3(2(4(1(1(x1)))))))	(32)
1(0(0(5(0(2(0(5(2(1(0(x1)))))))))))	→	1(2(1(3(0(2(1(4(1(0(x1))))))))))	(36)
0(1(0(5(2(4(4(2(5(1(4(1(2(x1)))))))))))))	→	0(3(0(1(3(4(4(1(0(1(2(0(x1))))))))))))	(38)
5(1(0(3(2(3(4(1(5(5(2(4(0(0(2(x1)))))))))))))))	→	4(2(5(4(4(2(5(4(2(3(1(0(1(2(x1))))))))))))))	(41)
3(2(1(3(5(5(4(5(3(3(2(0(1(5(2(2(x1))))))))))))))))	→	3(3(5(3(4(0(1(5(3(1(1(1(2(4(2(x1)))))))))))))))	(42)
4(5(5(5(4(1(4(2(2(2(4(0(4(3(2(5(5(5(x1))))))))))))))))))	→	4(4(3(4(1(2(0(5(2(2(3(3(2(1(1(0(3(x1)))))))))))))))))	(45)
0(5(3(5(0(4(0(0(3(0(4(3(2(4(0(5(2(4(2(2(3(x1)))))))))))))))))))))	→	4(4(0(2(5(3(1(5(5(1(2(3(1(4(5(1(4(1(1(x1)))))))))))))))))))	(48)
2(4(3(1(3(3(1(3(4(1(2(3(2(0(3(2(5(4(0(1(5(x1)))))))))))))))))))))	→	3(1(0(0(1(3(4(4(0(3(2(2(5(4(0(5(3(1(1(0(x1))))))))))))))))))))	(50)

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

There are 129 ruless (increase limit for explicit display).

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

3^#(2(4(5(1(0(4(2(0(x1)))))))))	→	3^#(3(0(0(4(3(0(2(2(x1)))))))))	(71)
3^#(2(4(5(1(0(4(2(0(x1)))))))))	→	3^#(0(0(4(3(0(2(2(x1))))))))	(72)
3^#(2(4(5(1(0(4(2(0(x1)))))))))	→	0^#(0(4(3(0(2(2(x1)))))))	(73)
0^#(0(4(2(3(3(2(4(4(4(1(4(2(0(2(1(3(x1)))))))))))))))))	→	2^#(x1)	(126)
2^#(0(0(1(0(0(x1))))))	→	3^#(2(0(0(x1))))	(51)
3^#(2(4(5(1(0(4(2(0(x1)))))))))	→	3^#(0(2(2(x1))))	(75)
3^#(2(4(5(1(0(4(2(0(x1)))))))))	→	0^#(2(2(x1)))	(76)
0^#(5(5(5(0(0(1(0(x1))))))))	→	0^#(0(4(0(1(2(2(0(x1))))))))	(62)
0^#(5(5(5(0(0(1(0(x1))))))))	→	0^#(1(2(2(0(x1)))))	(64)
0^#(5(5(5(0(0(1(0(x1))))))))	→	1^#(2(2(0(x1))))	(65)
1^#(1(2(0(1(5(1(x1)))))))	→	0^#(1(1(2(3(1(x1))))))	(57)
0^#(5(5(5(0(0(1(0(x1))))))))	→	2^#(2(0(x1)))	(66)
2^#(0(0(1(0(0(x1))))))	→	2^#(0(0(x1)))	(52)
2^#(0(1(5(0(3(5(1(4(4(3(5(3(5(3(2(2(1(4(0(2(x1)))))))))))))))))))))	→	2^#(2(2(x1)))	(164)
2^#(0(1(5(0(3(5(1(4(4(3(5(3(5(3(2(2(1(4(0(2(x1)))))))))))))))))))))	→	2^#(2(x1))	(165)
0^#(5(5(5(0(0(1(0(x1))))))))	→	2^#(0(x1))	(67)
0^#(0(2(5(0(3(1(5(1(x1)))))))))	→	2^#(2(0(5(2(3(1(1(x1))))))))	(79)
0^#(0(2(5(0(3(1(5(1(x1)))))))))	→	2^#(0(5(2(3(1(1(x1)))))))	(80)
0^#(0(2(5(0(3(1(5(1(x1)))))))))	→	0^#(5(2(3(1(1(x1))))))	(81)
0^#(0(2(5(0(3(1(5(1(x1)))))))))	→	2^#(3(1(1(x1))))	(82)
0^#(0(2(5(0(3(1(5(1(x1)))))))))	→	3^#(1(1(x1)))	(83)
3^#(2(4(5(1(0(4(2(0(x1)))))))))	→	2^#(2(x1))	(77)
3^#(2(4(5(1(0(4(2(0(x1)))))))))	→	2^#(x1)	(78)
3^#(5(1(3(4(2(5(2(5(3(4(5(5(1(2(5(1(1(0(1(5(x1)))))))))))))))))))))	→	3^#(0(1(1(x1))))	(176)
3^#(5(1(3(4(2(5(2(5(3(4(5(5(1(2(5(1(1(0(1(5(x1)))))))))))))))))))))	→	0^#(1(1(x1)))	(177)
0^#(0(2(5(0(3(1(5(1(x1)))))))))	→	1^#(1(x1))	(84)
1^#(1(2(0(1(5(1(x1)))))))	→	1^#(1(2(3(1(x1)))))	(58)
1^#(1(2(0(1(5(1(x1)))))))	→	1^#(2(3(1(x1))))	(59)
1^#(1(2(0(1(5(1(x1)))))))	→	2^#(3(1(x1)))	(60)
1^#(1(2(0(1(5(1(x1)))))))	→	3^#(1(x1))	(61)
3^#(5(1(3(4(2(5(2(5(3(4(5(5(1(2(5(1(1(0(1(5(x1)))))))))))))))))))))	→	1^#(1(x1))	(178)
1^#(1(1(4(2(2(5(0(0(0(1(2(2(x1)))))))))))))	→	1^#(1(x1))	(103)
1^#(1(1(4(2(2(5(0(0(0(1(2(2(x1)))))))))))))	→	1^#(x1)	(104)
3^#(5(1(3(4(2(5(2(5(3(4(5(5(1(2(5(1(1(0(1(5(x1)))))))))))))))))))))	→	1^#(x1)	(179)
0^#(1(2(4(1(4(2(0(1(5(x1))))))))))	→	3^#(1(x1))	(91)
0^#(1(2(4(1(4(2(0(1(5(x1))))))))))	→	1^#(x1)	(92)
0^#(2(1(4(0(5(0(4(4(3(0(0(2(4(0(0(5(0(x1))))))))))))))))))	→	3^#(0(x1))	(137)

1.1.1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals

[3^#(x₁)]	=	1 · x₁
[2(x₁)]	=	1 + 1 · x₁
[4(x₁)]	=	1 + 1 · x₁
[5(x₁)]	=	1 + 1 · x₁
[1(x₁)]	=	1 + 1 · x₁
[0(x₁)]	=	1 + 1 · x₁
[3(x₁)]	=	1 + 1 · x₁
[0^#(x₁)]	=	1 + 1 · x₁
[2^#(x₁)]	=	1 · x₁
[1^#(x₁)]	=	1 · x₁

the pairs

3^#(2(4(5(1(0(4(2(0(x1)))))))))	→	3^#(0(0(4(3(0(2(2(x1))))))))	(72)
3^#(2(4(5(1(0(4(2(0(x1)))))))))	→	0^#(0(4(3(0(2(2(x1)))))))	(73)
0^#(0(4(2(3(3(2(4(4(4(1(4(2(0(2(1(3(x1)))))))))))))))))	→	2^#(x1)	(126)
2^#(0(0(1(0(0(x1))))))	→	3^#(2(0(0(x1))))	(51)
3^#(2(4(5(1(0(4(2(0(x1)))))))))	→	3^#(0(2(2(x1))))	(75)
3^#(2(4(5(1(0(4(2(0(x1)))))))))	→	0^#(2(2(x1)))	(76)
0^#(5(5(5(0(0(1(0(x1))))))))	→	0^#(1(2(2(0(x1)))))	(64)
0^#(5(5(5(0(0(1(0(x1))))))))	→	1^#(2(2(0(x1))))	(65)
0^#(5(5(5(0(0(1(0(x1))))))))	→	2^#(2(0(x1)))	(66)
2^#(0(0(1(0(0(x1))))))	→	2^#(0(0(x1)))	(52)
2^#(0(1(5(0(3(5(1(4(4(3(5(3(5(3(2(2(1(4(0(2(x1)))))))))))))))))))))	→	2^#(2(2(x1)))	(164)
2^#(0(1(5(0(3(5(1(4(4(3(5(3(5(3(2(2(1(4(0(2(x1)))))))))))))))))))))	→	2^#(2(x1))	(165)
0^#(5(5(5(0(0(1(0(x1))))))))	→	2^#(0(x1))	(67)
0^#(0(2(5(0(3(1(5(1(x1)))))))))	→	2^#(2(0(5(2(3(1(1(x1))))))))	(79)
0^#(0(2(5(0(3(1(5(1(x1)))))))))	→	2^#(0(5(2(3(1(1(x1)))))))	(80)
0^#(0(2(5(0(3(1(5(1(x1)))))))))	→	0^#(5(2(3(1(1(x1))))))	(81)
0^#(0(2(5(0(3(1(5(1(x1)))))))))	→	2^#(3(1(1(x1))))	(82)
0^#(0(2(5(0(3(1(5(1(x1)))))))))	→	3^#(1(1(x1)))	(83)
3^#(2(4(5(1(0(4(2(0(x1)))))))))	→	2^#(2(x1))	(77)
3^#(2(4(5(1(0(4(2(0(x1)))))))))	→	2^#(x1)	(78)
3^#(5(1(3(4(2(5(2(5(3(4(5(5(1(2(5(1(1(0(1(5(x1)))))))))))))))))))))	→	3^#(0(1(1(x1))))	(176)
3^#(5(1(3(4(2(5(2(5(3(4(5(5(1(2(5(1(1(0(1(5(x1)))))))))))))))))))))	→	0^#(1(1(x1)))	(177)
0^#(0(2(5(0(3(1(5(1(x1)))))))))	→	1^#(1(x1))	(84)
1^#(1(2(0(1(5(1(x1)))))))	→	1^#(1(2(3(1(x1)))))	(58)
1^#(1(2(0(1(5(1(x1)))))))	→	1^#(2(3(1(x1))))	(59)
1^#(1(2(0(1(5(1(x1)))))))	→	2^#(3(1(x1)))	(60)
1^#(1(2(0(1(5(1(x1)))))))	→	3^#(1(x1))	(61)
3^#(5(1(3(4(2(5(2(5(3(4(5(5(1(2(5(1(1(0(1(5(x1)))))))))))))))))))))	→	1^#(1(x1))	(178)
1^#(1(1(4(2(2(5(0(0(0(1(2(2(x1)))))))))))))	→	1^#(1(x1))	(103)
1^#(1(1(4(2(2(5(0(0(0(1(2(2(x1)))))))))))))	→	1^#(x1)	(104)
3^#(5(1(3(4(2(5(2(5(3(4(5(5(1(2(5(1(1(0(1(5(x1)))))))))))))))))))))	→	1^#(x1)	(179)
0^#(1(2(4(1(4(2(0(1(5(x1))))))))))	→	3^#(1(x1))	(91)
0^#(1(2(4(1(4(2(0(1(5(x1))))))))))	→	1^#(x1)	(92)
0^#(2(1(4(0(5(0(4(4(3(0(0(2(4(0(0(5(0(x1))))))))))))))))))	→	3^#(0(x1))	(137)

could be deleted.

1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

3^#(2(4(5(1(0(4(2(0(x1)))))))))

→

3^#(3(0(0(4(3(0(2(2(x1)))))))))

(71)

1.1.1.1.1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

[3^#(x₁)]	=	1 · x₁
[2(x₁)]	=	1 + 1 · x₁
[4(x₁)]	=	1 + 1 · x₁
[5(x₁)]	=	1 · x₁
[1(x₁)]	=	1 + 1 · x₁
[0(x₁)]	=	1
[3(x₁)]	=	0

together with the usable rules

3(1(4(4(2(3(x1))))))	→	3(1(2(3(5(5(x1))))))	(27)
3(1(0(2(1(5(2(1(x1))))))))	→	3(5(3(4(5(2(4(1(x1))))))))	(31)
3(2(4(5(1(0(4(2(0(x1)))))))))	→	3(3(0(0(4(3(0(2(2(x1)))))))))	(33)
3(5(1(3(4(2(5(2(5(3(4(5(5(1(2(5(1(1(0(1(5(x1)))))))))))))))))))))	→	3(0(5(2(0(3(4(1(0(4(5(2(2(3(4(5(4(3(0(1(1(x1)))))))))))))))))))))	(49)

(w.r.t. the implicit argument filter of the reduction pair), the pair

3^#(2(4(5(1(0(4(2(0(x1)))))))))

→

3^#(3(0(0(4(3(0(2(2(x1)))))))))

(71)

could be deleted.

1.1.1.1.1.1.1.1 P is empty

There are no pairs anymore.