Certification Problem

Input (TPDB TRS_Standard/AG01/#3.26)

The rewrite relation of the following TRS is considered.

f(x)	→	s(x)	(1)
f(s(s(x)))	→	s(f(f(x)))	(2)

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.

f^#(s(s(x)))	→	f^#(x)	(3)
f^#(s(s(x)))	→	f^#(f(x))	(4)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

f^#(s(s(x))) → f^#(f(x)) (4)

f^#(s(s(x))) → f^#(x) (3)

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

[s(x₁)] = x₁ + 21239

[f(x₁)] = x₁ + 21239

[f^#(x₁)] = x₁ + 0

together with the usable rules

f(x) → s(x) (1)

f(s(s(x))) → s(f(f(x))) (2)

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

f^#(s(s(x))) → f^#(f(x)) (4)

f^#(s(s(x))) → f^#(x) (3)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.