Certification Problem

Input (TPDB TRS_Standard/SK90/2.28)

The rewrite relation of the following TRS is considered.

dfib(s(s(x)),y)

→

dfib(s(x),dfib(x,y))

(1)

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.

dfib^#(s(s(x)),y)	→	dfib^#(x,y)	(2)
dfib^#(s(s(x)),y)	→	dfib^#(s(x),dfib(x,y))	(3)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

dfib^#(s(s(x)),y) → dfib^#(s(x),dfib(x,y)) (3)

dfib^#(s(s(x)),y) → dfib^#(x,y) (2)

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

[s(x₁)] = x₁ + 2

[dfib(x₁, x₂)] = 1

[dfib^#(x₁, x₂)] = x₁ + x₂ + 0

together with the usable rule
dfib(s(s(x)),y) → dfib(s(x),dfib(x,y)) (1)
(w.r.t. the implicit argument filter of the reduction pair), the pairs

dfib^#(s(s(x)),y) → dfib^#(s(x),dfib(x,y)) (3)

dfib^#(s(s(x)),y) → dfib^#(x,y) (2)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.