Certification Problem

Input (TPDB TRS_Standard/SK90/4.12)

The rewrite relation of the following TRS is considered.

+(0,y)	→	y	(1)
+(s(x),0)	→	s(x)	(2)
+(s(x),s(y))	→	s(+(s(x),+(y,0)))	(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),s(y))	→	+^#(y,0)	(4)
+^#(s(x),s(y))	→	+^#(s(x),+(y,0))	(5)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair
+^#(s(x),s(y)) → +^#(s(x),+(y,0)) (5)

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

[s(x₁)] = x₁ + 7720

[0] = 0

[+(x₁, x₂)] = x₁ + x₂ + 0

[+^#(x₁, x₂)] = x₂ + 0

together with the usable rules

+(0,y) → y (1)

+(s(x),s(y)) → s(+(s(x),+(y,0))) (3)

+(s(x),0) → s(x) (2)

(w.r.t. the implicit argument filter of the reduction pair), the pair
+^#(s(x),s(y)) → +^#(s(x),+(y,0)) (5)
could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.