Certification Problem

Input (TPDB TRS_Standard/SK90/4.26)

The rewrite relation of the following TRS is considered.

rev(nil)	→	nil	(1)
rev(rev(x))	→	x	(2)
rev(++(x,y))	→	++(rev(y),rev(x))	(3)
++(nil,y)	→	y	(4)
++(x,nil)	→	x	(5)
++(.(x,y),z)	→	.(x,++(y,z))	(6)
++(x,++(y,z))	→	++(++(x,y),z)	(7)
make(x)	→	.(x,nil)	(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.

++^#(x,++(y,z))	→	++^#(++(x,y),z)	(9)
rev^#(++(x,y))	→	rev^#(x)	(10)
rev^#(++(x,y))	→	rev^#(y)	(11)
++^#(x,++(y,z))	→	++^#(x,y)	(12)
++^#(.(x,y),z)	→	++^#(y,z)	(13)
rev^#(++(x,y))	→	++^#(rev(y),rev(x))	(14)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair

rev^#(++(x,y))	→	rev^#(x)	(10)
rev^#(++(x,y))	→	rev^#(y)	(11)

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[rev^#(x₁)]	=	x₁ + 0
[++(x₁, x₂)]	=	x₁ + x₂ + 1
[make(x₁)]	=	0
[++^#(x₁, x₂)]	=	0
[nil]	=	0
[rev(x₁)]	=	0
[.(x₁, x₂)]	=	0
[make^#(x₁)]	=	0

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

rev^#(++(x,y))	→	rev^#(x)	(10)
rev^#(++(x,y))	→	rev^#(y)	(11)

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)	(13)
++^#(x,++(y,z))	→	++^#(x,y)	(12)
++^#(x,++(y,z))	→	++^#(++(x,y),z)	(9)

1.1.2 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[rev^#(x₁)]	=	0
[++(x₁, x₂)]	=	x₁ + x₂ + 1
[make(x₁)]	=	0
[++^#(x₁, x₂)]	=	x₂ + 0
[nil]	=	1
[rev(x₁)]	=	0
[.(x₁, x₂)]	=	x₂ + 1
[make^#(x₁)]	=	0

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

++^#(x,++(y,z))	→	++^#(x,y)	(12)
++^#(x,++(y,z))	→	++^#(++(x,y),z)	(9)

could be deleted.

1.1.2.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

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

→

++^#(y,z)

(13)

1.1.2.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[rev^#(x₁)]	=	0
[++(x₁, x₂)]	=	x₁ + x₂ + 1
[make(x₁)]	=	0
[++^#(x₁, x₂)]	=	x₁ + 0
[nil]	=	10157
[rev(x₁)]	=	0
[.(x₁, x₂)]	=	x₂ + 21239
[make^#(x₁)]	=	0

together with the usable rules

++(nil,y)	→	y	(4)
++(x,nil)	→	x	(5)
++(x,++(y,z))	→	++(++(x,y),z)	(7)
++(.(x,y),z)	→	.(x,++(y,z))	(6)

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

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

→

++^#(y,z)

(13)

could be deleted.

1.1.2.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.