Certification Problem

Input (TPDB TRS_Standard/SK90/2.43)

The rewrite relation of the following TRS is considered.

merge(nil,y)	→	y	(1)
merge(x,nil)	→	x	(2)
merge(.(x,y),.(u,v))	→	if(<(x,u),.(x,merge(y,.(u,v))),.(u,merge(.(x,y),v)))	(3)
++(nil,y)	→	y	(4)
++(.(x,y),z)	→	.(x,++(y,z))	(5)
if(true,x,y)	→	x	(6)
if(false,x,y)	→	x	(7)

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.

merge^#(.(x,y),.(u,v))	→	if^#(<(x,u),.(x,merge(y,.(u,v))),.(u,merge(.(x,y),v)))	(8)
merge^#(.(x,y),.(u,v))	→	merge^#(y,.(u,v))	(9)
++^#(.(x,y),z)	→	++^#(y,z)	(10)
merge^#(.(x,y),.(u,v))	→	merge^#(.(x,y),v)	(11)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair

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

→

++^#(y,z)

(10)

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[merge(x₁, x₂)]	=	0
[<(x₁, x₂)]	=	0
[++(x₁, x₂)]	=	0
[false]	=	0
[merge^#(x₁, x₂)]	=	0
[true]	=	0
[if(x₁, x₂, x₃)]	=	0
[++^#(x₁, x₂)]	=	x₁ + 0
[nil]	=	0
[.(x₁, x₂)]	=	x₂ + 1
[if^#(x₁, x₂, x₃)]	=	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.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.

The 2^nd component contains the pair

merge^#(.(x,y),.(u,v))	→	merge^#(.(x,y),v)	(11)
merge^#(.(x,y),.(u,v))	→	merge^#(y,.(u,v))	(9)

1.1.2 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[merge(x₁, x₂)]	=	0
[<(x₁, x₂)]	=	0
[++(x₁, x₂)]	=	0
[false]	=	0
[merge^#(x₁, x₂)]	=	x₁ + 0
[true]	=	0
[if(x₁, x₂, x₃)]	=	0
[++^#(x₁, x₂)]	=	0
[nil]	=	0
[.(x₁, x₂)]	=	x₂ + 1
[if^#(x₁, x₂, x₃)]	=	0

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

merge^#(.(x,y),.(u,v))

→

merge^#(y,.(u,v))

(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

merge^#(.(x,y),.(u,v))

→

merge^#(.(x,y),v)

(11)

1.1.2.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[merge(x₁, x₂)]	=	0
[<(x₁, x₂)]	=	0
[++(x₁, x₂)]	=	0
[false]	=	0
[merge^#(x₁, x₂)]	=	x₂ + 0
[true]	=	0
[if(x₁, x₂, x₃)]	=	0
[++^#(x₁, x₂)]	=	0
[nil]	=	0
[.(x₁, x₂)]	=	x₂ + 1
[if^#(x₁, x₂, x₃)]	=	0

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

merge^#(.(x,y),.(u,v))

→

merge^#(.(x,y),v)

(11)

could be deleted.

1.1.2.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.