Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/Ex6_Luc98_Z)

The rewrite relation of the following TRS is considered.

first(0,X)	→	nil	(1)
first(s(X),cons(Y,Z))	→	cons(Y,n__first(X,activate(Z)))	(2)
from(X)	→	cons(X,n__from(s(X)))	(3)
first(X1,X2)	→	n__first(X1,X2)	(4)
from(X)	→	n__from(X)	(5)
activate(n__first(X1,X2))	→	first(X1,X2)	(6)
activate(n__from(X))	→	from(X)	(7)
activate(X)	→	X	(8)

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.

activate^#(n__first(X1,X2))	→	first^#(X1,X2)	(9)
activate^#(n__from(X))	→	from^#(X)	(10)
first^#(s(X),cons(Y,Z))	→	activate^#(Z)	(11)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

first^#(s(X),cons(Y,Z))	→	activate^#(Z)	(11)
activate^#(n__first(X1,X2))	→	first^#(X1,X2)	(9)

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[s(x₁)]	=	1
[n__first(x₁, x₂)]	=	x₁ + x₂ + 1
[activate(x₁)]	=	0
[n__from(x₁)]	=	0
[activate^#(x₁)]	=	x₁ + 0
[0]	=	0
[from(x₁)]	=	0
[first^#(x₁, x₂)]	=	x₁ + x₂ + 0
[nil]	=	0
[first(x₁, x₂)]	=	0
[from^#(x₁)]	=	0
[cons(x₁, x₂)]	=	x₂ + 28958

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

first^#(s(X),cons(Y,Z))	→	activate^#(Z)	(11)
activate^#(n__first(X1,X2))	→	first^#(X1,X2)	(9)

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.