Certification Problem

Input (TPDB TRS_Standard/SK90/2.15)

The rewrite relation of the following TRS is considered.

f(0)	→	1	(1)
f(s(x))	→	g(f(x))	(2)
g(x)	→	+(x,s(x))	(3)
f(s(x))	→	+(f(x),s(f(x)))	(4)

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(x))	→	f^#(x)	(5)
f^#(s(x))	→	f^#(x)	(5)
f^#(s(x))	→	g^#(f(x))	(6)
f^#(s(x))	→	f^#(x)	(5)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

f^#(s(x)) → f^#(x) (5)

f^#(s(x)) → f^#(x) (5)

f^#(s(x)) → f^#(x) (5)

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

[1] = 0

[s(x₁)] = x₁ + 1

[f(x₁)] = 0

[0] = 0

[f^#(x₁)] = x₁ + 0

[g^#(x₁)] = 0

[+(x₁, x₂)] = 0

[g(x₁)] = 0

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

f^#(s(x)) → f^#(x) (5)

f^#(s(x)) → f^#(x) (5)

f^#(s(x)) → f^#(x) (5)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.