Certification Problem

Input (TPDB TRS_Standard/SK90/2.24)

The rewrite relation of the following TRS is considered.

fib(0)	→	0	(1)
fib(s(0))	→	s(0)	(2)
fib(s(s(x)))	→	+(fib(s(x)),fib(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.

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

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

fib^#(s(s(x))) → fib^#(s(x)) (5)

fib^#(s(s(x))) → fib^#(x) (4)

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

[s(x₁)] = x₁ + 1

[fib(x₁)] = 0

[0] = 0

[+(x₁, x₂)] = 0

[fib^#(x₁)] = x₁ + 0

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

fib^#(s(s(x))) → fib^#(s(x)) (5)

fib^#(s(s(x))) → fib^#(x) (4)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.