Certification Problem

Input (TPDB TRS_Standard/Der95/08)

The rewrite relation of the following TRS is considered.

D(t)	→	1	(1)
D(constant)	→	0	(2)
D(+(x,y))	→	+(D(x),D(y))	(3)
D(*(x,y))	→	+((y,D(x)),(x,D(y)))	(4)
D(-(x,y))	→	-(D(x),D(y))	(5)

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.

D^#(-(x,y))	→	D^#(y)	(6)
D^#(*(x,y))	→	D^#(y)	(7)
D^#(*(x,y))	→	D^#(x)	(8)
D^#(+(x,y))	→	D^#(x)	(9)
D^#(-(x,y))	→	D^#(x)	(10)
D^#(+(x,y))	→	D^#(y)	(11)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

D^#(+(x,y))	→	D^#(y)	(11)
D^#(-(x,y))	→	D^#(x)	(10)
D^#(+(x,y))	→	D^#(x)	(9)
D^#(*(x,y))	→	D^#(y)	(7)
D^#(*(x,y))	→	D^#(x)	(8)
D^#(-(x,y))	→	D^#(y)	(6)

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[1]	=	0
[constant]	=	0
[t]	=	0
[D(x₁)]	=	0
[0]	=	0
[D^#(x₁)]	=	x₁ + 0
[-(x₁, x₂)]	=	x₁ + x₂ + 1
[+(x₁, x₂)]	=	x₁ + x₂ + 1
[*(x₁, x₂)]	=	x₁ + x₂ + 1

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

D^#(+(x,y))	→	D^#(y)	(11)
D^#(-(x,y))	→	D^#(x)	(10)
D^#(+(x,y))	→	D^#(x)	(9)
D^#(*(x,y))	→	D^#(y)	(7)
D^#(*(x,y))	→	D^#(x)	(8)
D^#(-(x,y))	→	D^#(y)	(6)

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.