Certification Problem

Input (TPDB TRS_Standard/SK90/4.07)

The rewrite relation of the following TRS is considered.

*(i(x),x)	→	1	(1)
*(1,y)	→	y	(2)
*(x,0)	→	0	(3)
((x,y),z)	→	(x,(y,z))	(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.

^#((x,y),z)	→	*^#(y,z)	(5)
^#((x,y),z)	→	^#(x,(y,z))	(6)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

*^#(*(x,y),z) → *^#(x,*(y,z)) (6)

*^#(*(x,y),z) → *^#(y,z) (5)

1.1.1 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation

[1] = 23678

[*^#(x₁, x₂)] = x₁ + 0

[0] = 1

[i(x₁)] = 23676

[*(x₁, x₂)] = x₁ + x₂ + 1

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pairs

*^#(*(x,y),z) → *^#(x,*(y,z)) (6)

*^#(*(x,y),z) → *^#(y,z) (5)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.