Certification Problem

Input (TPDB TRS_Standard/SK90/4.24)

The rewrite relation of the following TRS is considered.

rev(nil)	→	nil	(1)
rev(++(x,y))	→	++(rev1(x,y),rev2(x,y))	(2)
rev1(x,nil)	→	x	(3)
rev1(x,++(y,z))	→	rev1(y,z)	(4)
rev2(x,nil)	→	nil	(5)
rev2(x,++(y,z))	→	rev(++(x,rev(rev2(y,z))))	(6)

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^#(++(x,y))	→	rev2^#(x,y)	(7)
rev1^#(x,++(y,z))	→	rev1^#(y,z)	(8)
rev2^#(x,++(y,z))	→	rev2^#(y,z)	(9)
rev2^#(x,++(y,z))	→	rev^#(++(x,rev(rev2(y,z))))	(10)
rev^#(++(x,y))	→	rev1^#(x,y)	(11)
rev2^#(x,++(y,z))	→	rev^#(rev2(y,z))	(12)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair

rev2^#(x,++(y,z))	→	rev^#(rev2(y,z))	(12)
rev2^#(x,++(y,z))	→	rev^#(++(x,rev(rev2(y,z))))	(10)
rev2^#(x,++(y,z))	→	rev2^#(y,z)	(9)
rev^#(++(x,y))	→	rev2^#(x,y)	(7)

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[rev^#(x₁)]	=	x₁ + 0
[rev1(x₁, x₂)]	=	x₂ + 1
[++(x₁, x₂)]	=	x₂ + 11799
[rev1^#(x₁, x₂)]	=	0
[rev2^#(x₁, x₂)]	=	x₂ + 11798
[nil]	=	1
[rev(x₁)]	=	x₁ + 0
[rev2(x₁, x₂)]	=	x₂ + 0

together with the usable rules

rev(nil)	→	nil	(1)
rev2(x,nil)	→	nil	(5)
rev2(x,++(y,z))	→	rev(++(x,rev(rev2(y,z))))	(6)
rev(++(x,y))	→	++(rev1(x,y),rev2(x,y))	(2)

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

rev2^#(x,++(y,z))	→	rev^#(rev2(y,z))	(12)
rev2^#(x,++(y,z))	→	rev^#(++(x,rev(rev2(y,z))))	(10)
rev2^#(x,++(y,z))	→	rev2^#(y,z)	(9)
rev^#(++(x,y))	→	rev2^#(x,y)	(7)

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,++(y,z))

→

rev1^#(y,z)

(8)

1.1.2 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[rev^#(x₁)]	=	x₁ + 0
[rev1(x₁, x₂)]	=	x₂ + 1
[++(x₁, x₂)]	=	x₂ + 32286
[rev1^#(x₁, x₂)]	=	x₂ + 0
[rev2^#(x₁, x₂)]	=	11798
[nil]	=	1
[rev(x₁)]	=	x₁ + 0
[rev2(x₁, x₂)]	=	x₂ + 0

together with the usable rules

rev(nil)	→	nil	(1)
rev2(x,nil)	→	nil	(5)
rev2(x,++(y,z))	→	rev(++(x,rev(rev2(y,z))))	(6)
rev(++(x,y))	→	++(rev1(x,y),rev2(x,y))	(2)

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

rev1^#(x,++(y,z))

→

rev1^#(y,z)

(8)

could be deleted.

1.1.2.1 Dependency Graph Processor

The dependency pairs are split into 0 components.