Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/Ex3_12_Luc96a_Z)

The rewrite relation of the following TRS is considered.

from(X)	→	cons(X,n__from(s(X)))	(1)
sel(0,cons(X,Y))	→	X	(2)
sel(s(X),cons(Y,Z))	→	sel(X,activate(Z))	(3)
from(X)	→	n__from(X)	(4)
activate(n__from(X))	→	from(X)	(5)
activate(X)	→	X	(6)

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.

sel^#(s(X),cons(Y,Z))	→	sel^#(X,activate(Z))	(7)
sel^#(s(X),cons(Y,Z))	→	activate^#(Z)	(8)
activate^#(n__from(X))	→	from^#(X)	(9)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

sel^#(s(X),cons(Y,Z))

→

sel^#(X,activate(Z))

(7)

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[s(x₁)]	=	x₁ + 1
[activate(x₁)]	=	1
[n__from(x₁)]	=	x₁ + 3
[activate^#(x₁)]	=	0
[0]	=	0
[sel^#(x₁, x₂)]	=	x₁ + 0
[from(x₁)]	=	x₁ + 2
[sel(x₁, x₂)]	=	0
[from^#(x₁)]	=	0
[cons(x₁, x₂)]	=	x₁ + 3

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

sel^#(s(X),cons(Y,Z))

→

sel^#(X,activate(Z))

(7)

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.