Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/85477)

The rewrite relation of the following TRS is considered.

0(0(1(2(2(x1)))))	→	2(3(3(2(2(x1)))))	(1)
0(4(1(0(4(5(x1))))))	→	5(2(5(3(4(5(x1))))))	(2)
0(5(2(4(2(0(0(5(2(x1)))))))))	→	1(5(2(0(2(3(5(2(x1))))))))	(3)
3(4(2(2(1(5(3(3(2(x1)))))))))	→	3(1(0(3(5(3(3(0(2(x1)))))))))	(4)
0(5(5(3(1(3(3(0(2(0(x1))))))))))	→	3(3(3(2(5(0(4(0(5(0(x1))))))))))	(5)
5(0(3(5(1(4(2(1(0(2(x1))))))))))	→	5(2(0(1(3(1(1(0(2(x1)))))))))	(6)
2(4(5(2(4(3(4(3(1(2(3(x1)))))))))))	→	5(3(1(3(3(3(2(2(4(1(x1))))))))))	(7)
0(2(1(5(3(5(3(0(5(5(5(0(x1))))))))))))	→	2(2(5(4(4(2(3(3(0(3(5(4(x1))))))))))))	(8)
3(2(0(0(4(0(0(0(2(5(1(0(x1))))))))))))	→	1(5(2(4(0(4(2(0(0(5(0(2(x1))))))))))))	(9)
3(3(3(2(3(3(1(2(2(4(3(3(x1))))))))))))	→	1(2(2(1(3(2(3(5(2(5(5(x1)))))))))))	(10)
5(5(4(0(3(0(2(3(3(2(3(3(x1))))))))))))	→	5(2(2(3(0(1(1(3(4(4(5(x1)))))))))))	(11)
0(0(3(1(0(2(2(3(2(5(3(0(1(x1)))))))))))))	→	0(1(3(4(3(1(0(0(3(3(4(0(x1))))))))))))	(12)
2(1(3(3(4(3(0(2(0(4(0(3(1(x1)))))))))))))	→	2(4(1(3(0(1(3(3(2(1(3(4(4(x1)))))))))))))	(13)
2(4(2(3(3(5(0(4(5(5(2(5(5(x1)))))))))))))	→	3(5(4(3(3(2(0(1(1(4(3(5(x1))))))))))))	(14)
5(2(3(1(2(1(5(4(5(1(1(0(0(x1)))))))))))))	→	5(2(0(5(4(3(4(3(3(0(5(0(0(x1)))))))))))))	(15)
2(5(0(2(2(0(3(3(1(4(2(5(0(0(0(x1)))))))))))))))	→	5(3(1(3(1(4(4(2(5(4(5(4(4(1(0(x1)))))))))))))))	(16)
2(5(1(2(0(0(5(1(2(2(1(0(2(1(5(x1)))))))))))))))	→	0(3(2(5(5(2(4(4(3(5(3(4(3(4(1(x1)))))))))))))))	(17)
2(5(3(5(2(4(2(1(3(0(5(0(3(1(3(x1)))))))))))))))	→	5(3(2(3(5(1(5(3(5(3(5(1(5(x1)))))))))))))	(18)
0(3(5(5(0(3(2(4(1(1(4(5(1(5(5(3(x1))))))))))))))))	→	0(5(1(2(4(0(2(0(1(2(4(3(3(3(0(5(x1))))))))))))))))	(19)
0(4(4(1(2(2(1(3(5(3(1(3(4(1(1(0(x1))))))))))))))))	→	0(1(5(2(2(5(0(5(3(3(2(3(4(1(5(2(x1))))))))))))))))	(20)
3(3(2(0(1(5(4(5(4(2(4(2(3(4(4(3(x1))))))))))))))))	→	5(2(5(5(5(5(2(3(3(1(2(5(0(4(0(x1)))))))))))))))	(21)
3(0(0(2(0(3(5(3(0(2(5(3(5(5(2(3(4(x1)))))))))))))))))	→	5(3(4(3(0(4(2(2(1(0(4(1(0(0(3(4(x1))))))))))))))))	(22)
0(1(1(3(5(1(1(1(3(0(0(4(1(2(2(3(1(5(0(x1)))))))))))))))))))	→	2(4(5(5(5(4(0(2(5(0(5(4(4(5(5(0(5(0(0(x1)))))))))))))))))))	(23)
1(0(3(2(0(2(4(1(3(2(5(0(0(4(5(0(3(4(3(x1)))))))))))))))))))	→	4(2(2(0(0(0(0(4(1(4(0(5(3(5(0(2(4(3(x1))))))))))))))))))	(24)
1(2(1(3(4(2(4(4(1(2(5(2(3(3(3(5(1(0(0(x1)))))))))))))))))))	→	1(5(3(2(5(0(5(2(3(0(0(3(1(5(3(4(4(3(0(x1)))))))))))))))))))	(25)
3(3(5(4(1(1(3(2(1(4(4(0(1(1(0(4(3(1(0(x1)))))))))))))))))))	→	3(4(0(2(3(1(3(0(4(4(3(4(4(4(3(2(1(0(x1))))))))))))))))))	(26)
2(5(0(4(0(3(4(3(4(0(0(2(4(2(4(1(0(1(2(3(x1))))))))))))))))))))	→	0(1(2(0(5(4(3(2(0(2(3(3(3(0(1(5(5(5(1(x1)))))))))))))))))))	(27)
0(3(4(2(2(0(3(4(1(1(5(0(3(5(2(1(3(3(4(1(0(x1)))))))))))))))))))))	→	2(4(4(3(0(5(5(5(5(2(2(0(2(5(1(1(0(1(0(0(1(x1)))))))))))))))))))))	(28)
2(4(5(2(1(3(5(5(1(1(1(0(0(4(5(0(1(2(0(2(5(x1)))))))))))))))))))))	→	2(0(0(2(2(1(2(1(2(2(2(0(5(3(3(2(5(2(2(5(2(x1)))))))))))))))))))))	(29)
4(3(1(4(3(3(3(3(4(4(2(5(1(4(5(1(4(3(2(3(3(x1)))))))))))))))))))))	→	2(5(5(1(1(1(4(5(3(2(3(3(0(4(2(3(1(5(4(5(x1))))))))))))))))))))	(30)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over the naturals

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

all of the following rules can be deleted.

0(5(2(4(2(0(0(5(2(x1)))))))))	→	1(5(2(0(2(3(5(2(x1))))))))	(3)
5(0(3(5(1(4(2(1(0(2(x1))))))))))	→	5(2(0(1(3(1(1(0(2(x1)))))))))	(6)
2(4(5(2(4(3(4(3(1(2(3(x1)))))))))))	→	5(3(1(3(3(3(2(2(4(1(x1))))))))))	(7)
3(3(3(2(3(3(1(2(2(4(3(3(x1))))))))))))	→	1(2(2(1(3(2(3(5(2(5(5(x1)))))))))))	(10)
5(5(4(0(3(0(2(3(3(2(3(3(x1))))))))))))	→	5(2(2(3(0(1(1(3(4(4(5(x1)))))))))))	(11)
0(0(3(1(0(2(2(3(2(5(3(0(1(x1)))))))))))))	→	0(1(3(4(3(1(0(0(3(3(4(0(x1))))))))))))	(12)
2(4(2(3(3(5(0(4(5(5(2(5(5(x1)))))))))))))	→	3(5(4(3(3(2(0(1(1(4(3(5(x1))))))))))))	(14)
2(5(3(5(2(4(2(1(3(0(5(0(3(1(3(x1)))))))))))))))	→	5(3(2(3(5(1(5(3(5(3(5(1(5(x1)))))))))))))	(18)
3(3(2(0(1(5(4(5(4(2(4(2(3(4(4(3(x1))))))))))))))))	→	5(2(5(5(5(5(2(3(3(1(2(5(0(4(0(x1)))))))))))))))	(21)
3(0(0(2(0(3(5(3(0(2(5(3(5(5(2(3(4(x1)))))))))))))))))	→	5(3(4(3(0(4(2(2(1(0(4(1(0(0(3(4(x1))))))))))))))))	(22)
1(0(3(2(0(2(4(1(3(2(5(0(0(4(5(0(3(4(3(x1)))))))))))))))))))	→	4(2(2(0(0(0(0(4(1(4(0(5(3(5(0(2(4(3(x1))))))))))))))))))	(24)
3(3(5(4(1(1(3(2(1(4(4(0(1(1(0(4(3(1(0(x1)))))))))))))))))))	→	3(4(0(2(3(1(3(0(4(4(3(4(4(4(3(2(1(0(x1))))))))))))))))))	(26)
2(5(0(4(0(3(4(3(4(0(0(2(4(2(4(1(0(1(2(3(x1))))))))))))))))))))	→	0(1(2(0(5(4(3(2(0(2(3(3(3(0(1(5(5(5(1(x1)))))))))))))))))))	(27)
4(3(1(4(3(3(3(3(4(4(2(5(1(4(5(1(4(3(2(3(3(x1)))))))))))))))))))))	→	2(5(5(1(1(1(4(5(3(2(3(3(0(4(2(3(1(5(4(5(x1))))))))))))))))))))	(30)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

There are 180 ruless (increase limit for explicit display).

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair
2^#(4(5(2(1(3(5(5(1(1(1(0(0(4(5(0(1(2(0(2(5(x1))))))))))))))))))))) → 2^#(x1) (210)

1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals

[4(x₁)] = 1 · x₁

[5(x₁)] = 1 · x₁

[2(x₁)] = 1 · x₁

[1(x₁)] = 1 · x₁

[3(x₁)] = 1 · x₁

[0(x₁)] = 1 · x₁

[2^#(x₁)] = 1 · x₁

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the rule could be deleted.
1.1.1.1.1 Size-Change Termination

Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.

2^#(4(5(2(1(3(5(5(1(1(1(0(0(4(5(0(1(2(0(2(5(x1))))))))))))))))))))) → 2^#(x1) (210)

1 > 1

As there is no critical graph in the transitive closure, there are no infinite chains.

The 2^nd component contains the pair

2^#(5(1(2(0(0(5(1(2(2(1(0(2(1(5(x1)))))))))))))))	→	3^#(4(3(4(1(x1)))))	(109)
3^#(4(2(2(1(5(3(3(2(x1)))))))))	→	0^#(3(5(3(3(0(2(x1)))))))	(40)
0^#(0(1(2(2(x1)))))	→	2^#(3(3(2(2(x1)))))	(31)
2^#(5(1(2(0(0(5(1(2(2(1(0(2(1(5(x1)))))))))))))))	→	3^#(4(1(x1)))	(110)
3^#(4(2(2(1(5(3(3(2(x1)))))))))	→	0^#(2(x1))	(45)
0^#(5(5(3(1(3(3(0(2(0(x1))))))))))	→	2^#(5(0(4(0(5(0(x1)))))))	(49)
0^#(5(5(3(1(3(3(0(2(0(x1))))))))))	→	0^#(5(0(x1)))	(52)

1.1.1.2 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

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

together with the usable rules

1(2(1(3(4(2(4(4(1(2(5(2(3(3(3(5(1(0(0(x1)))))))))))))))))))	→	1(5(3(2(5(0(5(2(3(0(0(3(1(5(3(4(4(3(0(x1)))))))))))))))))))	(25)
3(4(2(2(1(5(3(3(2(x1)))))))))	→	3(1(0(3(5(3(3(0(2(x1)))))))))	(4)
3(2(0(0(4(0(0(0(2(5(1(0(x1))))))))))))	→	1(5(2(4(0(4(2(0(0(5(0(2(x1))))))))))))	(9)
5(2(3(1(2(1(5(4(5(1(1(0(0(x1)))))))))))))	→	5(2(0(5(4(3(4(3(3(0(5(0(0(x1)))))))))))))	(15)

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

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

→

2^#(3(3(2(2(x1)))))

(31)

could be deleted.

1.1.1.2.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

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

together with the usable rules

1(2(1(3(4(2(4(4(1(2(5(2(3(3(3(5(1(0(0(x1)))))))))))))))))))	→	1(5(3(2(5(0(5(2(3(0(0(3(1(5(3(4(4(3(0(x1)))))))))))))))))))	(25)
3(4(2(2(1(5(3(3(2(x1)))))))))	→	3(1(0(3(5(3(3(0(2(x1)))))))))	(4)
2(1(3(3(4(3(0(2(0(4(0(3(1(x1)))))))))))))	→	2(4(1(3(0(1(3(3(2(1(3(4(4(x1)))))))))))))	(13)
2(5(0(2(2(0(3(3(1(4(2(5(0(0(0(x1)))))))))))))))	→	5(3(1(3(1(4(4(2(5(4(5(4(4(1(0(x1)))))))))))))))	(16)
2(5(1(2(0(0(5(1(2(2(1(0(2(1(5(x1)))))))))))))))	→	0(3(2(5(5(2(4(4(3(5(3(4(3(4(1(x1)))))))))))))))	(17)
2(4(5(2(1(3(5(5(1(1(1(0(0(4(5(0(1(2(0(2(5(x1)))))))))))))))))))))	→	2(0(0(2(2(1(2(1(2(2(2(0(5(3(3(2(5(2(2(5(2(x1)))))))))))))))))))))	(29)
3(2(0(0(4(0(0(0(2(5(1(0(x1))))))))))))	→	1(5(2(4(0(4(2(0(0(5(0(2(x1))))))))))))	(9)
5(2(3(1(2(1(5(4(5(1(1(0(0(x1)))))))))))))	→	5(2(0(5(4(3(4(3(3(0(5(0(0(x1)))))))))))))	(15)

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

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

→

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

(40)

could be deleted.

1.1.1.2.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

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

together with the usable rules

1(2(1(3(4(2(4(4(1(2(5(2(3(3(3(5(1(0(0(x1)))))))))))))))))))	→	1(5(3(2(5(0(5(2(3(0(0(3(1(5(3(4(4(3(0(x1)))))))))))))))))))	(25)
3(4(2(2(1(5(3(3(2(x1)))))))))	→	3(1(0(3(5(3(3(0(2(x1)))))))))	(4)
3(2(0(0(4(0(0(0(2(5(1(0(x1))))))))))))	→	1(5(2(4(0(4(2(0(0(5(0(2(x1))))))))))))	(9)

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

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

→

0^#(2(x1))

(45)

could be deleted.

1.1.1.2.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair
0^#(5(5(3(1(3(3(0(2(0(x1)))))))))) → 0^#(5(0(x1))) (52)

1.1.1.2.1.1.1.1 Reduction Pair Processor
Using the linear polynomial interpretation over the naturals

[0^#(x₁)] = 1 · x₁

[5(x₁)] = 1 + 1 · x₁

[3(x₁)] = 1 + 1 · x₁

[1(x₁)] = 1 + 1 · x₁

[0(x₁)] = 1 + 1 · x₁

[2(x₁)] = 1 + 1 · x₁

[4(x₁)] = 1 + 1 · x₁

the pair
0^#(5(5(3(1(3(3(0(2(0(x1)))))))))) → 0^#(5(0(x1))) (52)
could be deleted.
1.1.1.2.1.1.1.1.1 P is empty

There are no pairs anymore.

2^#(4(5(2(1(3(5(5(1(1(1(0(0(4(5(0(1(2(0(2(5(x1)))))))))))))))))))))	→	2^#(x1)	(210)

1	>	1

Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/85477)

Property / Task

Answer / Result

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

1.1 Dependency Pair Transformation

1.1.1 Dependency Graph Processor

1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules

1.1.1.1.1 Size-Change Termination

1.1.1.2 Reduction Pair Processor with Usable Rules

1.1.1.2.1 Reduction Pair Processor with Usable Rules

1.1.1.2.1.1 Reduction Pair Processor with Usable Rules

1.1.1.2.1.1.1 Dependency Graph Processor

1.1.1.2.1.1.1.1 Reduction Pair Processor

1.1.1.2.1.1.1.1.1 P is empty