Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/Ex5_Zan97_Z)

The rewrite relation of the following TRS is considered.

f(X)	→	if(X,c,n__f(true))	(1)
if(true,X,Y)	→	X	(2)
if(false,X,Y)	→	activate(Y)	(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.

if^#(false,X,Y)	→	activate^#(Y)	(7)
f^#(X)	→	if^#(X,c,n__f(true))	(8)
activate^#(n__f(X))	→	f^#(X)	(9)

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)	(9)
f^#(X)	→	if^#(X,c,n__f(true))	(8)
if^#(false,X,Y)	→	activate^#(Y)	(7)

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[activate(x₁)]	=	0
[activate^#(x₁)]	=	x₁ + 30097
[false]	=	30098
[c]	=	1
[true]	=	30094
[f(x₁)]	=	0
[if(x₁, x₂, x₃)]	=	0
[n__f(x₁)]	=	x₁ + 14235
[f^#(x₁)]	=	x₁ + 44331
[if^#(x₁, x₂, x₃)]	=	x₁ + x₂ + x₃ + 0

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

activate^#(n__f(X))	→	f^#(X)	(9)
f^#(X)	→	if^#(X,c,n__f(true))	(8)
if^#(false,X,Y)	→	activate^#(Y)	(7)

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.