Certification Problem

Input (TPDB TRS_Standard/SK90/4.16)

The rewrite relation of the following TRS is considered.

f(0)	→	s(0)	(1)
f(s(0))	→	s(s(0))	(2)
f(s(0))	→	*(s(s(0)),f(0))	(3)
f(+(x,s(0)))	→	+(s(s(0)),f(x))	(4)
f(+(x,y))	→	*(f(x),f(y))	(5)

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^#(+(x,y))	→	f^#(y)	(6)
f^#(s(0))	→	f^#(0)	(7)
f^#(+(x,y))	→	f^#(x)	(8)
f^#(+(x,s(0)))	→	f^#(x)	(9)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

f^#(+(x,s(0))) → f^#(x) (9)

f^#(+(x,y)) → f^#(x) (8)

f^#(+(x,y)) → f^#(y) (6)

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

[s(x₁)] = x₁ + 0

[f(x₁)] = 0

[0] = 39

[f^#(x₁)] = x₁ + 0

[+(x₁, x₂)] = x₁ + x₂ + 1

[*(x₁, x₂)] = 0

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

f^#(+(x,s(0))) → f^#(x) (9)

f^#(+(x,y)) → f^#(x) (8)

f^#(+(x,y)) → f^#(y) (6)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.