Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/Ex4_4_Luc96b_GM)

The rewrite relation of the following TRS is considered.

a__f(g(X),Y)	→	a__f(mark(X),f(g(X),Y))	(1)
mark(f(X1,X2))	→	a__f(mark(X1),X2)	(2)
mark(g(X))	→	g(mark(X))	(3)
a__f(X1,X2)	→	f(X1,X2)	(4)

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^#(f(X1,X2))	→	mark^#(X1)	(5)
a__f^#(g(X),Y)	→	mark^#(X)	(6)
mark^#(g(X))	→	mark^#(X)	(7)
mark^#(f(X1,X2))	→	a__f^#(mark(X1),X2)	(8)
a__f^#(g(X),Y)	→	a__f^#(mark(X),f(g(X),Y))	(9)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

a__f^#(g(X),Y)	→	a__f^#(mark(X),f(g(X),Y))	(9)
mark^#(f(X1,X2))	→	a__f^#(mark(X1),X2)	(8)
mark^#(g(X))	→	mark^#(X)	(7)
a__f^#(g(X),Y)	→	mark^#(X)	(6)
mark^#(f(X1,X2))	→	mark^#(X1)	(5)

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[a__f(x₁, x₂)]	=	x₁ + 1
[f(x₁, x₂)]	=	x₁ + 1
[mark^#(x₁)]	=	x₁ + 23676
[mark(x₁)]	=	x₁ + 23676
[a__f^#(x₁, x₂)]	=	x₁ + 0
[g(x₁)]	=	x₁ + 23677

together with the usable rules

a__f(X1,X2)	→	f(X1,X2)	(4)
a__f(g(X),Y)	→	a__f(mark(X),f(g(X),Y))	(1)
mark(g(X))	→	g(mark(X))	(3)
mark(f(X1,X2))	→	a__f(mark(X1),X2)	(2)

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

a__f^#(g(X),Y)	→	a__f^#(mark(X),f(g(X),Y))	(9)
mark^#(f(X1,X2))	→	a__f^#(mark(X1),X2)	(8)
mark^#(g(X))	→	mark^#(X)	(7)
a__f^#(g(X),Y)	→	mark^#(X)	(6)
mark^#(f(X1,X2))	→	mark^#(X1)	(5)

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.