Certification Problem

Input (TPDB TRS_Standard/SK90/4.25)

The rewrite relation of the following TRS is considered.

rev(a)	→	a	(1)
rev(b)	→	b	(2)
rev(++(x,y))	→	++(rev(y),rev(x))	(3)
rev(++(x,x))	→	rev(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.

rev^#(++(x,y))	→	rev^#(x)	(5)
rev^#(++(x,x))	→	rev^#(x)	(6)
rev^#(++(x,y))	→	rev^#(y)	(7)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

rev^#(++(x,y)) → rev^#(y) (7)

rev^#(++(x,x)) → rev^#(x) (6)

rev^#(++(x,y)) → rev^#(x) (5)

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

[a] = 0

[rev^#(x₁)] = x₁ + 0

[b] = 0

[++(x₁, x₂)] = x₁ + x₂ + 1

[rev(x₁)] = 0

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

rev^#(++(x,y)) → rev^#(y) (7)

rev^#(++(x,x)) → rev^#(x) (6)

rev^#(++(x,y)) → rev^#(x) (5)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.