Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/Ex4_7_15_Bor03_Z)

The rewrite relation of the following TRS is considered.

f(0)	→	cons(0,n__f(s(0)))	(1)
f(s(0))	→	f(p(s(0)))	(2)
p(s(0))	→	0	(3)
f(X)	→	n__f(X)	(4)
activate(n__f(X))	→	f(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.

f^#(s(0))	→	p^#(s(0))	(7)
activate^#(n__f(X))	→	f^#(X)	(8)
f^#(s(0))	→	f^#(p(s(0)))	(9)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

f^#(s(0))

→

f^#(p(s(0)))

(9)

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[s(x₁)]	=	x₁ + 2438
[activate(x₁)]	=	0
[activate^#(x₁)]	=	0
[p^#(x₁)]	=	0
[f(x₁)]	=	0
[p(x₁)]	=	2438
[0]	=	1
[n__f(x₁)]	=	0
[f^#(x₁)]	=	x₁ + 0
[cons(x₁, x₂)]	=	0

together with the usable rule

p(s(0))

→

0

(3)

(w.r.t. the implicit argument filter of the reduction pair), the pair

f^#(s(0))

→

f^#(p(s(0)))

(9)

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.