Certification Problem

Input (TPDB TRS_Standard/AG01/#3.24)

The rewrite relation of the following TRS is considered.

f(0)	→	s(0)	(1)
f(s(0))	→	s(0)	(2)
f(s(s(x)))	→	f(f(s(x)))	(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.

f^#(s(s(x)))	→	f^#(s(x))	(4)
f^#(s(s(x)))	→	f^#(f(s(x)))	(5)

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(s(x))) (5)

f^#(s(s(x))) → f^#(s(x)) (4)

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

[s(x₁)] = x₁ + 2

[f(x₁)] = 3

[0] = 1

[f^#(x₁)] = x₁ + 0

together with the usable rules

f(0) → s(0) (1)

f(s(s(x))) → f(f(s(x))) (3)

f(s(0)) → s(0) (2)

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

f^#(s(s(x))) → f^#(f(s(x))) (5)

f^#(s(s(x))) → f^#(s(x)) (4)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.