Certification Problem

Input (TPDB TRS_Standard/SK90/2.09)

The rewrite relation of the following TRS is considered.

+(0,y)	→	y	(1)
+(s(x),y)	→	s(+(x,y))	(2)
+(s(x),y)	→	+(x,s(y))	(3)

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.

+^#(s(x),y)	→	+^#(x,s(y))	(4)
+^#(s(x),y)	→	+^#(x,y)	(5)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

+^#(s(x),y) → +^#(x,y) (5)

+^#(s(x),y) → +^#(x,s(y)) (4)

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

[s(x₁)] = x₁ + 1

[0] = 0

[+(x₁, x₂)] = 0

[+^#(x₁, x₂)] = x₁ + 0

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

+^#(s(x),y) → +^#(x,y) (5)

+^#(s(x),y) → +^#(x,s(y)) (4)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.