Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/Ex25_Luc06_Z)

The rewrite relation of the following TRS is considered.

f(f(X))	→	c(n__f(g(n__f(X))))	(1)
c(X)	→	d(activate(X))	(2)
h(X)	→	c(n__d(X))	(3)
f(X)	→	n__f(X)	(4)
d(X)	→	n__d(X)	(5)
activate(n__f(X))	→	f(X)	(6)
activate(n__d(X))	→	d(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.

c^#(X)	→	activate^#(X)	(9)
f^#(f(X))	→	c^#(n__f(g(n__f(X))))	(10)
h^#(X)	→	c^#(n__d(X))	(11)
activate^#(n__f(X))	→	f^#(X)	(12)
c^#(X)	→	d^#(activate(X))	(13)
activate^#(n__d(X))	→	d^#(X)	(14)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

activate^#(n__f(X))	→	f^#(X)	(12)
f^#(f(X))	→	c^#(n__f(g(n__f(X))))	(10)
c^#(X)	→	activate^#(X)	(9)

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[h(x₁)]	=	0
[d(x₁)]	=	0
[activate(x₁)]	=	0
[activate^#(x₁)]	=	x₁ + 0
[c(x₁)]	=	0
[f(x₁)]	=	x₁ + 11296
[h^#(x₁)]	=	0
[n__f(x₁)]	=	x₁ + 1
[d^#(x₁)]	=	0
[f^#(x₁)]	=	x₁ + 0
[c^#(x₁)]	=	x₁ + 1
[n__d(x₁)]	=	0
[g(x₁)]	=	11293

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

activate^#(n__f(X))	→	f^#(X)	(12)
f^#(f(X))	→	c^#(n__f(g(n__f(X))))	(10)
c^#(X)	→	activate^#(X)	(9)

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.