Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/213407)

The rewrite relation of the following TRS is considered.

0(1(1(x1)))	→	1(0(2(1(x1))))	(1)
0(1(1(x1)))	→	1(1(0(3(2(x1)))))	(2)
0(1(1(x1)))	→	1(4(0(0(2(1(x1))))))	(3)
0(1(4(x1)))	→	1(4(0(3(x1))))	(4)
0(1(4(x1)))	→	4(0(2(1(x1))))	(5)
0(1(4(x1)))	→	4(0(2(1(3(x1)))))	(6)
0(4(1(x1)))	→	0(2(4(1(x1))))	(7)
0(4(1(x1)))	→	4(0(2(1(x1))))	(8)
0(4(1(x1)))	→	4(0(3(1(x1))))	(9)
0(4(1(x1)))	→	1(4(4(0(2(x1)))))	(10)
0(4(1(x1)))	→	2(1(4(0(2(x1)))))	(11)
0(5(4(x1)))	→	1(4(0(0(2(5(x1))))))	(12)
0(5(4(x1)))	→	4(0(2(5(2(5(x1))))))	(13)
0(5(4(x1)))	→	5(0(4(0(3(3(x1))))))	(14)
4(2(1(x1)))	→	4(0(2(1(x1))))	(15)
4(2(1(x1)))	→	1(2(4(0(2(x1)))))	(16)
0(1(0(4(x1))))	→	0(0(2(2(1(4(x1))))))	(17)
0(1(1(2(x1))))	→	5(1(0(2(1(x1)))))	(18)
0(1(1(3(x1))))	→	1(1(5(0(3(x1)))))	(19)
0(1(4(1(x1))))	→	1(0(2(4(1(x1)))))	(20)
0(1(4(3(x1))))	→	1(0(2(4(3(x1)))))	(21)
0(1(4(3(x1))))	→	4(0(2(1(3(x1)))))	(22)
0(4(1(2(x1))))	→	2(0(3(1(4(x1)))))	(23)
0(4(2(1(x1))))	→	0(4(3(0(2(1(x1))))))	(24)
0(5(0(4(x1))))	→	1(5(4(0(0(2(x1))))))	(25)
0(5(1(3(x1))))	→	1(1(5(0(3(x1)))))	(26)
0(5(1(3(x1))))	→	5(3(1(0(3(x1)))))	(27)
0(5(4(1(x1))))	→	4(1(5(4(0(2(x1))))))	(28)
0(5(4(3(x1))))	→	0(2(5(0(3(4(x1))))))	(29)
0(5(4(3(x1))))	→	1(5(3(4(0(3(x1))))))	(30)
1(0(5(4(x1))))	→	1(4(5(0(3(3(x1))))))	(31)
1(0(5(4(x1))))	→	5(5(1(0(2(4(x1))))))	(32)
1(4(2(1(x1))))	→	4(0(2(1(2(1(x1))))))	(33)
4(1(2(1(x1))))	→	4(1(0(2(1(x1)))))	(34)
4(1(2(1(x1))))	→	3(4(0(2(1(1(x1))))))	(35)
4(3(2(1(x1))))	→	0(3(4(0(2(1(x1))))))	(36)
0(0(1(2(3(x1)))))	→	1(0(3(0(0(2(x1))))))	(37)
0(0(5(1(2(x1)))))	→	0(0(2(5(0(1(x1))))))	(38)
0(0(5(1(3(x1)))))	→	4(5(0(0(3(1(x1))))))	(39)
0(1(3(4(2(x1)))))	→	1(3(4(3(0(2(x1))))))	(40)
0(4(5(3(4(x1)))))	→	0(3(2(5(4(4(x1))))))	(41)
0(5(0(2(2(x1)))))	→	0(0(2(5(2(4(x1))))))	(42)
0(5(1(1(3(x1)))))	→	0(3(2(1(1(5(x1))))))	(43)
0(5(1(4(3(x1)))))	→	3(0(2(1(5(4(x1))))))	(44)
0(5(5(1(2(x1)))))	→	1(5(3(0(2(5(x1))))))	(45)
1(0(1(2(4(x1)))))	→	1(1(0(2(3(4(x1))))))	(46)
1(4(2(1(2(x1)))))	→	0(2(2(1(1(4(x1))))))	(47)
4(0(0(5(4(x1)))))	→	5(0(0(4(4(5(x1))))))	(48)
4(2(5(4(1(x1)))))	→	4(4(1(5(3(2(x1))))))	(49)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

There are 140 ruless (increase limit for explicit display).

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair

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

→

0^#(1(x1))

(153)

1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

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

together with the usable rules

1(0(5(4(x1))))	→	1(4(5(0(3(3(x1))))))	(31)
1(0(5(4(x1))))	→	5(5(1(0(2(4(x1))))))	(32)
1(4(2(1(x1))))	→	4(0(2(1(2(1(x1))))))	(33)
1(0(1(2(4(x1)))))	→	1(1(0(2(3(4(x1))))))	(46)
1(4(2(1(2(x1)))))	→	0(2(2(1(1(4(x1))))))	(47)
4(2(1(x1)))	→	4(0(2(1(x1))))	(15)
4(2(1(x1)))	→	1(2(4(0(2(x1)))))	(16)
4(1(2(1(x1))))	→	4(1(0(2(1(x1)))))	(34)
4(1(2(1(x1))))	→	3(4(0(2(1(1(x1))))))	(35)
4(3(2(1(x1))))	→	0(3(4(0(2(1(x1))))))	(36)
4(0(0(5(4(x1)))))	→	5(0(0(4(4(5(x1))))))	(48)
4(2(5(4(1(x1)))))	→	4(4(1(5(3(2(x1))))))	(49)

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

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

→

0^#(1(x1))

(153)

could be deleted.

1.1.1.1 P is empty

There are no pairs anymore.

The 2^nd component contains the pair

4^#(1(2(1(x1))))	→	1^#(1(x1))	(142)
1^#(4(2(1(2(x1)))))	→	1^#(1(4(x1)))	(181)
1^#(4(2(1(2(x1)))))	→	1^#(4(x1))	(180)
1^#(4(2(1(2(x1)))))	→	4^#(x1)	(179)

1.1.2 Reduction Pair Processor with Usable Rules

Using the

prec(4^#)	=	stat(4^#)	=	lex
prec(1^#)	=	stat(1^#)	=	lex
prec(5)	=	stat(5)	=	lex
prec(4)	=	stat(4)	=	lex
prec(3)	=	stat(3)	=	lex
prec(2)	=	stat(2)	=	lex
prec(0)	=	stat(0)	=	lex
prec(1)	=	stat(1)	=	lex

π(4^#)	=	1
π(1^#)	=	1
π(5)	=	1
π(4)	=	[1]
π(3)	=	1
π(2)	=	1
π(0)	=	1
π(1)	=	1

together with the usable rules

1(0(5(4(x1))))	→	1(4(5(0(3(3(x1))))))	(31)
1(0(5(4(x1))))	→	5(5(1(0(2(4(x1))))))	(32)
1(4(2(1(x1))))	→	4(0(2(1(2(1(x1))))))	(33)
1(0(1(2(4(x1)))))	→	1(1(0(2(3(4(x1))))))	(46)
1(4(2(1(2(x1)))))	→	0(2(2(1(1(4(x1))))))	(47)
4(2(1(x1)))	→	4(0(2(1(x1))))	(15)
4(2(1(x1)))	→	1(2(4(0(2(x1)))))	(16)
4(1(2(1(x1))))	→	4(1(0(2(1(x1)))))	(34)
4(1(2(1(x1))))	→	3(4(0(2(1(1(x1))))))	(35)
4(3(2(1(x1))))	→	0(3(4(0(2(1(x1))))))	(36)
4(0(0(5(4(x1)))))	→	5(0(0(4(4(5(x1))))))	(48)
4(2(5(4(1(x1)))))	→	4(4(1(5(3(2(x1))))))	(49)

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

1^#(4(2(1(2(x1)))))

→

4^#(x1)

(179)

could be deleted.

1.1.2.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

1^#(4(2(1(2(x1)))))	→	1^#(1(4(x1)))	(181)
1^#(4(2(1(2(x1)))))	→	1^#(4(x1))	(180)

1.1.2.1.1 Reduction Pair Processor with Usable Rules

Using the

prec(1^#)	=	0	stat(1^#)	=	lex
prec(5)	=	0	stat(5)	=	lex
prec(4)	=	0	stat(4)	=	lex
prec(3)	=	0	stat(3)	=	lex
prec(2)	=	0	stat(2)	=	lex
prec(0)	=	0	stat(0)	=	lex
prec(1)	=	1	stat(1)	=	lex

π(1^#)	=	[1]
π(5)	=	1
π(4)	=	1
π(3)	=	1
π(2)	=	1
π(0)	=	1
π(1)	=	[1]

together with the usable rules

4(2(1(x1)))	→	4(0(2(1(x1))))	(15)
4(2(1(x1)))	→	1(2(4(0(2(x1)))))	(16)
4(1(2(1(x1))))	→	4(1(0(2(1(x1)))))	(34)
4(1(2(1(x1))))	→	3(4(0(2(1(1(x1))))))	(35)
4(3(2(1(x1))))	→	0(3(4(0(2(1(x1))))))	(36)
4(0(0(5(4(x1)))))	→	5(0(0(4(4(5(x1))))))	(48)
4(2(5(4(1(x1)))))	→	4(4(1(5(3(2(x1))))))	(49)
1(0(5(4(x1))))	→	1(4(5(0(3(3(x1))))))	(31)
1(0(5(4(x1))))	→	5(5(1(0(2(4(x1))))))	(32)
1(4(2(1(x1))))	→	4(0(2(1(2(1(x1))))))	(33)
1(0(1(2(4(x1)))))	→	1(1(0(2(3(4(x1))))))	(46)
1(4(2(1(2(x1)))))	→	0(2(2(1(1(4(x1))))))	(47)

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

1^#(4(2(1(2(x1)))))

→

1^#(4(x1))

(180)

could be deleted.

1.1.2.1.1.1 Reduction Pair Processor with Usable Rules

Using the

prec(1^#)	=	0	stat(1^#)	=	lex
prec(5)	=	0	stat(5)	=	lex
prec(4)	=	3	stat(4)	=	lex
prec(3)	=	2	stat(3)	=	lex
prec(2)	=	0	stat(2)	=	lex
prec(0)	=	0	stat(0)	=	lex
prec(1)	=	5	stat(1)	=	lex

π(1^#)	=	1
π(5)	=	1
π(4)	=	[1]
π(3)	=	[]
π(2)	=	1
π(0)	=	[]
π(1)	=	[]

together with the usable rules

4(2(1(x1)))	→	4(0(2(1(x1))))	(15)
4(2(1(x1)))	→	1(2(4(0(2(x1)))))	(16)
4(1(2(1(x1))))	→	4(1(0(2(1(x1)))))	(34)
4(1(2(1(x1))))	→	3(4(0(2(1(1(x1))))))	(35)
4(3(2(1(x1))))	→	0(3(4(0(2(1(x1))))))	(36)
4(0(0(5(4(x1)))))	→	5(0(0(4(4(5(x1))))))	(48)
4(2(5(4(1(x1)))))	→	4(4(1(5(3(2(x1))))))	(49)
1(0(5(4(x1))))	→	1(4(5(0(3(3(x1))))))	(31)
1(0(5(4(x1))))	→	5(5(1(0(2(4(x1))))))	(32)
1(4(2(1(x1))))	→	4(0(2(1(2(1(x1))))))	(33)
1(0(1(2(4(x1)))))	→	1(1(0(2(3(4(x1))))))	(46)
1(4(2(1(2(x1)))))	→	0(2(2(1(1(4(x1))))))	(47)

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

1^#(4(2(1(2(x1)))))

→

1^#(1(4(x1)))

(181)

could be deleted.

1.1.2.1.1.1.1 P is empty

There are no pairs anymore.