Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/Ex25_Luc06_GM)

The rewrite relation of the following TRS is considered.

a__f(f(X))	→	a__c(f(g(f(X))))	(1)
a__c(X)	→	d(X)	(2)
a__h(X)	→	a__c(d(X))	(3)
mark(f(X))	→	a__f(mark(X))	(4)
mark(c(X))	→	a__c(X)	(5)
mark(h(X))	→	a__h(mark(X))	(6)
mark(g(X))	→	g(X)	(7)
mark(d(X))	→	d(X)	(8)
a__f(X)	→	f(X)	(9)
a__c(X)	→	c(X)	(10)
a__h(X)	→	h(X)	(11)

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.

mark^#(h(X))	→	mark^#(X)	(12)
mark^#(f(X))	→	a__f^#(mark(X))	(13)
mark^#(f(X))	→	mark^#(X)	(14)
a__f^#(f(X))	→	a__c^#(f(g(f(X))))	(15)
mark^#(c(X))	→	a__c^#(X)	(16)
mark^#(h(X))	→	a__h^#(mark(X))	(17)
a__h^#(X)	→	a__c^#(d(X))	(18)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

mark^#(f(X))	→	mark^#(X)	(14)
mark^#(h(X))	→	mark^#(X)	(12)

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[h(x₁)]	=	x₁ + 1
[a__c^#(x₁)]	=	0
[d(x₁)]	=	0
[a__h(x₁)]	=	0
[a__f(x₁)]	=	0
[c(x₁)]	=	0
[a__h^#(x₁)]	=	0
[f(x₁)]	=	x₁ + 1
[mark^#(x₁)]	=	x₁ + 0
[mark(x₁)]	=	0
[a__f^#(x₁)]	=	0
[g(x₁)]	=	0
[a__c(x₁)]	=	0

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

mark^#(f(X))	→	mark^#(X)	(14)
mark^#(h(X))	→	mark^#(X)	(12)

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.