Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/26069)

The rewrite relation of the following TRS is considered.

0(1(2(1(x1))))	→	1(2(1(1(0(1(2(0(1(2(x1))))))))))	(1)
0(1(2(1(x1))))	→	1(2(1(1(0(1(2(0(1(2(0(1(2(x1)))))))))))))	(2)
0(1(2(1(x1))))	→	1(2(1(1(0(1(2(0(1(2(0(1(2(0(1(2(x1))))))))))))))))	(3)
0(1(2(1(x1))))	→	1(2(1(1(0(1(2(0(1(2(0(1(2(0(1(2(0(1(2(x1)))))))))))))))))))	(4)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by NaTT @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(x1))))))	(5)
0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(x1))))))	(5)
0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(x1))))))	(5)
0^#(1(2(1(x1))))	→	0^#(1(2(x1)))	(6)
0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(0(1(2(0(1(2(x1))))))))))))	(7)
0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(0(1(2(x1)))))))))	(8)
0^#(1(2(1(x1))))	→	0^#(1(2(x1)))	(6)
0^#(1(2(1(x1))))	→	0^#(1(2(x1)))	(6)
0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(0(1(2(0(1(2(0(1(2(x1)))))))))))))))	(9)
0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(0(1(2(x1)))))))))	(8)
0^#(1(2(1(x1))))	→	0^#(1(2(x1)))	(6)
0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(0(1(2(x1)))))))))	(8)
0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(x1))))))	(5)
0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(0(1(2(0(1(2(x1))))))))))))	(7)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(0(1(2(0(1(2(x1))))))))))))	(7)
0^#(1(2(1(x1))))	→	0^#(1(2(x1)))	(6)
0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(0(1(2(x1)))))))))	(8)
0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(x1))))))	(5)
0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(0(1(2(x1)))))))))	(8)
0^#(1(2(1(x1))))	→	0^#(1(2(x1)))	(6)
0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(0(1(2(0(1(2(x1))))))))))))	(7)
0^#(1(2(1(x1))))	→	0^#(1(2(x1)))	(6)
0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(x1))))))	(5)
0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(x1))))))	(5)
0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(0(1(2(x1)))))))))	(8)
0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(0(1(2(0(1(2(0(1(2(x1)))))))))))))))	(9)
0^#(1(2(1(x1))))	→	0^#(1(2(x1)))	(6)
0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(x1))))))	(5)

1.1.1 Reduction Pair Processor with Usable Rules

Using the matrix interpretations of dimension 2 with strict dimension 1 over the naturals

[0^#(x₁)]

=

0	1
0	1

· x₁ +

0

0

[1(x₁)]

=

x₁ +

165269

1

[0(x₁)]

=

0	1
0	1

· x₁ +

0

371854

[2(x₁)]

=

0	0
1	0

· x₁ +

1

41316

together with the usable rules

0(1(2(1(x1))))	→	1(2(1(1(0(1(2(0(1(2(0(1(2(0(1(2(0(1(2(x1)))))))))))))))))))	(4)
0(1(2(1(x1))))	→	1(2(1(1(0(1(2(0(1(2(x1))))))))))	(1)
0(1(2(1(x1))))	→	1(2(1(1(0(1(2(0(1(2(0(1(2(0(1(2(x1))))))))))))))))	(3)
0(1(2(1(x1))))	→	1(2(1(1(0(1(2(0(1(2(0(1(2(x1)))))))))))))	(2)

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

0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(0(1(2(0(1(2(x1))))))))))))	(7)
0^#(1(2(1(x1))))	→	0^#(1(2(x1)))	(6)
0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(0(1(2(x1)))))))))	(8)
0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(x1))))))	(5)
0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(0(1(2(x1)))))))))	(8)
0^#(1(2(1(x1))))	→	0^#(1(2(x1)))	(6)
0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(0(1(2(0(1(2(x1))))))))))))	(7)
0^#(1(2(1(x1))))	→	0^#(1(2(x1)))	(6)
0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(x1))))))	(5)
0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(x1))))))	(5)
0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(0(1(2(x1)))))))))	(8)
0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(0(1(2(0(1(2(0(1(2(x1)))))))))))))))	(9)
0^#(1(2(1(x1))))	→	0^#(1(2(x1)))	(6)
0^#(1(2(1(x1))))	→	0^#(1(2(0(1(2(x1))))))	(5)

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.