Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/157388)

The rewrite relation of the following TRS is considered.

There are 151 ruless (increase limit for explicit display).

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

There are 151 ruless (increase limit for explicit display).

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

There are 199 ruless (increase limit for explicit display).

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

1^#(0(3(4(x1))))	→	1^#(x1)	(333)
1^#(0(1(4(x1))))	→	1^#(x1)	(309)
1^#(1(0(1(4(x1)))))	→	1^#(x1)	(382)
1^#(0(2(1(4(x1)))))	→	1^#(x1)	(388)
1^#(0(5(1(4(x1)))))	→	1^#(1(x1))	(393)
1^#(0(5(1(4(x1)))))	→	1^#(x1)	(394)
1^#(1(5(1(4(x1)))))	→	1^#(x1)	(397)
1^#(1(0(3(4(x1)))))	→	1^#(x1)	(399)
1^#(0(1(3(4(x1)))))	→	1^#(x1)	(408)
1^#(1(2(3(4(x1)))))	→	1^#(x1)	(415)
1^#(0(5(3(4(x1)))))	→	1^#(x1)	(425)
1^#(1(5(3(4(x1)))))	→	1^#(x1)	(433)
1^#(0(1(5(4(x1)))))	→	1^#(x1)	(452)
1^#(0(3(5(4(x1)))))	→	1^#(x1)	(468)

1.1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals

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

the pairs

1^#(0(3(4(x1))))	→	1^#(x1)	(333)
1^#(0(1(4(x1))))	→	1^#(x1)	(309)
1^#(1(0(1(4(x1)))))	→	1^#(x1)	(382)
1^#(0(2(1(4(x1)))))	→	1^#(x1)	(388)
1^#(0(5(1(4(x1)))))	→	1^#(1(x1))	(393)
1^#(0(5(1(4(x1)))))	→	1^#(x1)	(394)
1^#(1(0(3(4(x1)))))	→	1^#(x1)	(399)
1^#(0(1(3(4(x1)))))	→	1^#(x1)	(408)
1^#(0(5(3(4(x1)))))	→	1^#(x1)	(425)
1^#(0(1(5(4(x1)))))	→	1^#(x1)	(452)
1^#(0(3(5(4(x1)))))	→	1^#(x1)	(468)

could be deleted.

1.1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

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

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

1.1.1.1.1.1 Size-Change Termination

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

1^#(1(5(1(4(x1)))))	→	1^#(x1)	(397)

1	>	1
1^#(1(2(3(4(x1)))))	→	1^#(x1)	(415)

1	>	1
1^#(1(5(3(4(x1)))))	→	1^#(x1)	(433)

1	>	1

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