Certification Problem

Input (TPDB TRS_Standard/AG01/#3.38)

The rewrite relation of the following TRS is considered.

rev(nil)	→	nil	(1)
rev(cons(x,l))	→	cons(rev1(x,l),rev2(x,l))	(2)
rev1(0,nil)	→	0	(3)
rev1(s(x),nil)	→	s(x)	(4)
rev1(x,cons(y,l))	→	rev1(y,l)	(5)
rev2(x,nil)	→	nil	(6)
rev2(x,cons(y,l))	→	rev(cons(x,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.

rev^#(cons(x,l))	→	rev2^#(x,l)	(8)
rev1^#(x,cons(y,l))	→	rev1^#(y,l)	(9)
rev2^#(x,cons(y,l))	→	rev^#(cons(x,rev2(y,l)))	(10)
rev^#(cons(x,l))	→	rev1^#(x,l)	(11)
rev2^#(x,cons(y,l))	→	rev2^#(y,l)	(12)

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))	→	rev2^#(y,l)	(12)
rev2^#(x,cons(y,l))	→	rev^#(cons(x,rev2(y,l)))	(10)
rev^#(cons(x,l))	→	rev2^#(x,l)	(8)

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[rev^#(x₁)]	=	x₁ + 0
[s(x₁)]	=	3
[rev1(x₁, x₂)]	=	x₂ + 1
[rev1^#(x₁, x₂)]	=	0
[rev2^#(x₁, x₂)]	=	x₂ + 1
[0]	=	3
[nil]	=	1
[rev(x₁)]	=	x₁ + 0
[cons(x₁, x₂)]	=	x₂ + 8857
[rev2(x₁, x₂)]	=	x₂ + 0

together with the usable rules

rev(nil)	→	nil	(1)
rev2(x,cons(y,l))	→	rev(cons(x,rev2(y,l)))	(7)
rev2(x,nil)	→	nil	(6)
rev(cons(x,l))	→	cons(rev1(x,l),rev2(x,l))	(2)

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

rev2^#(x,cons(y,l))	→	rev2^#(y,l)	(12)
rev2^#(x,cons(y,l))	→	rev^#(cons(x,rev2(y,l)))	(10)
rev^#(cons(x,l))	→	rev2^#(x,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)]	=	3
[rev1(x₁, x₂)]	=	x₂ + 1
[rev1^#(x₁, x₂)]	=	x₂ + 0
[rev2^#(x₁, x₂)]	=	1
[0]	=	3
[nil]	=	1
[rev(x₁)]	=	x₁ + 0
[cons(x₁, x₂)]	=	x₂ + 1
[rev2(x₁, x₂)]	=	x₂ + 0

together with the usable rules

rev(nil)	→	nil	(1)
rev2(x,cons(y,l))	→	rev(cons(x,rev2(y,l)))	(7)
rev2(x,nil)	→	nil	(6)
rev(cons(x,l))	→	cons(rev1(x,l),rev2(x,l))	(2)

(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.