Certification Problem

Input (TPDB TRS_Standard/SK90/2.39)

The rewrite relation of the following TRS is considered.

rev(nil)	→	nil	(1)
rev(.(x,y))	→	++(rev(y),.(x,nil))	(2)
car(.(x,y))	→	x	(3)
cdr(.(x,y))	→	y	(4)
null(nil)	→	true	(5)
null(.(x,y))	→	false	(6)
++(nil,y)	→	y	(7)
++(.(x,y),z)	→	.(x,++(y,z))	(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.

rev^#(.(x,y))	→	rev^#(y)	(9)
++^#(.(x,y),z)	→	++^#(y,z)	(10)
rev^#(.(x,y))	→	++^#(rev(y),.(x,nil))	(11)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair

rev^#(.(x,y))

→

rev^#(y)

(9)

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[rev^#(x₁)]	=	x₁ + 0
[cdr(x₁)]	=	0
[++(x₁, x₂)]	=	0
[false]	=	0
[true]	=	0
[null(x₁)]	=	0
[++^#(x₁, x₂)]	=	0
[nil]	=	0
[rev(x₁)]	=	0
[.(x₁, x₂)]	=	x₂ + 1
[null^#(x₁)]	=	0
[cdr^#(x₁)]	=	0
[car^#(x₁)]	=	0
[car(x₁)]	=	0

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

rev^#(.(x,y))

→

rev^#(y)

(9)

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.

The 2^nd component contains the pair

++^#(.(x,y),z)

→

++^#(y,z)

(10)

1.1.2 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[rev^#(x₁)]	=	0
[cdr(x₁)]	=	0
[++(x₁, x₂)]	=	0
[false]	=	0
[true]	=	0
[null(x₁)]	=	0
[++^#(x₁, x₂)]	=	x₁ + 0
[nil]	=	0
[rev(x₁)]	=	0
[.(x₁, x₂)]	=	x₂ + 1
[null^#(x₁)]	=	0
[cdr^#(x₁)]	=	0
[car^#(x₁)]	=	0
[car(x₁)]	=	0

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

++^#(.(x,y),z)

→

++^#(y,z)

(10)

could be deleted.

1.1.2.1 Dependency Graph Processor

The dependency pairs are split into 0 components.