Certification Problem

Input (TPDB TRS_Standard/Mixed_TRS/jones1)

The rewrite relation of the following TRS is considered.

rev(ls)	→	r1(ls,empty)	(1)
r1(empty,a)	→	a	(2)
r1(cons(x,k),a)	→	r1(k,cons(x,a))	(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.

rev^#(ls)	→	r1^#(ls,empty)	(4)
r1^#(cons(x,k),a)	→	r1^#(k,cons(x,a))	(5)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair
r1^#(cons(x,k),a) → r1^#(k,cons(x,a)) (5)

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

[rev^#(x₁)] = 0

[rev(x₁)] = 0

[r1^#(x₁, x₂)] = x₁ + 0

[r1(x₁, x₂)] = 0

[cons(x₁, x₂)] = x₂ + 1

[empty] = 0

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
r1^#(cons(x,k),a) → r1^#(k,cons(x,a)) (5)
could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.