Certification Problem

Input (TPDB TRS_Standard/Rubio_04/revlist)

The rewrite relation of the following TRS is considered.

rev1(0,nil)	→	0	(1)
rev1(s(X),nil)	→	s(X)	(2)
rev1(X,cons(Y,L))	→	rev1(Y,L)	(3)
rev(nil)	→	nil	(4)
rev(cons(X,L))	→	cons(rev1(X,L),rev2(X,L))	(5)
rev2(X,nil)	→	nil	(6)
rev2(X,cons(Y,L))	→	rev(cons(X,rev(rev2(Y,L))))	(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.

rev2^#(X,cons(Y,L))	→	rev^#(rev2(Y,L))	(8)
rev1^#(X,cons(Y,L))	→	rev1^#(Y,L)	(9)
rev^#(cons(X,L))	→	rev2^#(X,L)	(10)
rev2^#(X,cons(Y,L))	→	rev2^#(Y,L)	(11)
rev2^#(X,cons(Y,L))	→	rev^#(cons(X,rev(rev2(Y,L))))	(12)
rev^#(cons(X,L))	→	rev1^#(X,L)	(13)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair

rev2^#(X,cons(Y,L))	→	rev^#(cons(X,rev(rev2(Y,L))))	(12)
rev2^#(X,cons(Y,L))	→	rev2^#(Y,L)	(11)
rev^#(cons(X,L))	→	rev2^#(X,L)	(10)
rev2^#(X,cons(Y,L))	→	rev^#(rev2(Y,L))	(8)

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[rev^#(x₁)]	=	x₁ + 0
[s(x₁)]	=	x₁ + 0
[rev1(x₁, x₂)]	=	1
[rev1^#(x₁, x₂)]	=	0
[rev2^#(x₁, x₂)]	=	x₂ + 1
[0]	=	0
[nil]	=	1
[rev(x₁)]	=	x₁ + 0
[cons(x₁, x₂)]	=	x₂ + 10452
[rev2(x₁, x₂)]	=	x₂ + 0

together with the usable rules

rev(nil)	→	nil	(4)
rev(cons(X,L))	→	cons(rev1(X,L),rev2(X,L))	(5)
rev2(X,cons(Y,L))	→	rev(cons(X,rev(rev2(Y,L))))	(7)
rev2(X,nil)	→	nil	(6)

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

rev2^#(X,cons(Y,L))	→	rev^#(cons(X,rev(rev2(Y,L))))	(12)
rev2^#(X,cons(Y,L))	→	rev2^#(Y,L)	(11)
rev^#(cons(X,L))	→	rev2^#(X,L)	(10)
rev2^#(X,cons(Y,L))	→	rev^#(rev2(Y,L))	(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

rev1^#(X,cons(Y,L))

→

rev1^#(Y,L)

(9)

1.1.2 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[rev^#(x₁)]	=	x₁ + 0
[s(x₁)]	=	x₁ + 0
[rev1(x₁, x₂)]	=	1
[rev1^#(x₁, x₂)]	=	x₂ + 0
[rev2^#(x₁, x₂)]	=	1
[0]	=	0
[nil]	=	1
[rev(x₁)]	=	x₁ + 0
[cons(x₁, x₂)]	=	x₂ + 1
[rev2(x₁, x₂)]	=	x₂ + 0

together with the usable rules

rev(nil)	→	nil	(4)
rev(cons(X,L))	→	cons(rev1(X,L),rev2(X,L))	(5)
rev2(X,cons(Y,L))	→	rev(cons(X,rev(rev2(Y,L))))	(7)
rev2(X,nil)	→	nil	(6)

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

rev1^#(X,cons(Y,L))

→

rev1^#(Y,L)

(9)

could be deleted.

1.1.2.1 Dependency Graph Processor

The dependency pairs are split into 0 components.