Certification Problem

Input (TPDB TRS_Standard/Der95/32)

The rewrite relation of the following TRS is considered.

sort(nil)	→	nil	(1)
sort(cons(x,y))	→	insert(x,sort(y))	(2)
insert(x,nil)	→	cons(x,nil)	(3)
insert(x,cons(v,w))	→	choose(x,cons(v,w),x,v)	(4)
choose(x,cons(v,w),y,0)	→	cons(x,cons(v,w))	(5)
choose(x,cons(v,w),0,s(z))	→	cons(v,insert(x,w))	(6)
choose(x,cons(v,w),s(y),s(z))	→	choose(x,cons(v,w),y,z)	(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.

sort^#(cons(x,y))	→	sort^#(y)	(8)
insert^#(x,cons(v,w))	→	choose^#(x,cons(v,w),x,v)	(9)
choose^#(x,cons(v,w),0,s(z))	→	insert^#(x,w)	(10)
sort^#(cons(x,y))	→	insert^#(x,sort(y))	(11)
choose^#(x,cons(v,w),s(y),s(z))	→	choose^#(x,cons(v,w),y,z)	(12)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair

sort^#(cons(x,y))

→

sort^#(y)

(8)

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[insert^#(x₁, x₂)]	=	0
[s(x₁)]	=	0
[insert(x₁, x₂)]	=	0
[choose^#(x₁,...,x₄)]	=	0
[sort^#(x₁)]	=	x₁ + 0
[0]	=	0
[nil]	=	0
[sort(x₁)]	=	0
[choose(x₁,...,x₄)]	=	0
[cons(x₁, x₂)]	=	x₂ + 1

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

sort^#(cons(x,y))

→

sort^#(y)

(8)

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

choose^#(x,cons(v,w),s(y),s(z))	→	choose^#(x,cons(v,w),y,z)	(12)
choose^#(x,cons(v,w),0,s(z))	→	insert^#(x,w)	(10)
insert^#(x,cons(v,w))	→	choose^#(x,cons(v,w),x,v)	(9)

1.1.2 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[insert^#(x₁, x₂)]	=	x₁ + x₂ + 1
[s(x₁)]	=	1
[insert(x₁, x₂)]	=	0
[choose^#(x₁,...,x₄)]	=	x₁ + x₂ + 0
[sort^#(x₁)]	=	0
[0]	=	1
[nil]	=	0
[sort(x₁)]	=	0
[choose(x₁,...,x₄)]	=	0
[cons(x₁, x₂)]	=	x₁ + x₂ + 2

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

choose^#(x,cons(v,w),0,s(z))	→	insert^#(x,w)	(10)
insert^#(x,cons(v,w))	→	choose^#(x,cons(v,w),x,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

choose^#(x,cons(v,w),s(y),s(z))

→

choose^#(x,cons(v,w),y,z)

(12)

1.1.2.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[insert^#(x₁, x₂)]	=	1
[s(x₁)]	=	x₁ + 1
[insert(x₁, x₂)]	=	0
[choose^#(x₁,...,x₄)]	=	x₄ + 0
[sort^#(x₁)]	=	0
[0]	=	1
[nil]	=	0
[sort(x₁)]	=	0
[choose(x₁,...,x₄)]	=	0
[cons(x₁, x₂)]	=	2

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

choose^#(x,cons(v,w),s(y),s(z))

→

choose^#(x,cons(v,w),y,z)

(12)

could be deleted.

1.1.2.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.