Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/PEANO_nosorts-noand_FR)

The rewrite relation of the following TRS is considered.

U11(tt,M,N)	→	U12(tt,activate(M),activate(N))	(1)
U12(tt,M,N)	→	s(plus(activate(N),activate(M)))	(2)
plus(N,0)	→	N	(3)
plus(N,s(M))	→	U11(tt,M,N)	(4)
activate(X)	→	X	(5)

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.

U12^#(tt,M,N)	→	activate^#(N)	(6)
U11^#(tt,M,N)	→	activate^#(N)	(7)
plus^#(N,s(M))	→	U11^#(tt,M,N)	(8)
U11^#(tt,M,N)	→	activate^#(M)	(9)
U12^#(tt,M,N)	→	activate^#(M)	(10)
U12^#(tt,M,N)	→	plus^#(activate(N),activate(M))	(11)
U11^#(tt,M,N)	→	U12^#(tt,activate(M),activate(N))	(12)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

U11^#(tt,M,N)	→	U12^#(tt,activate(M),activate(N))	(12)
U12^#(tt,M,N)	→	plus^#(activate(N),activate(M))	(11)
plus^#(N,s(M))	→	U11^#(tt,M,N)	(8)

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[U11(x₁, x₂, x₃)]	=	0
[s(x₁)]	=	x₁ + 5
[activate(x₁)]	=	x₁ + 1
[plus^#(x₁, x₂)]	=	x₂ + 0
[activate^#(x₁)]	=	0
[U12(x₁, x₂, x₃)]	=	0
[U12^#(x₁, x₂, x₃)]	=	x₂ + 2
[0]	=	0
[plus(x₁, x₂)]	=	0
[U11^#(x₁, x₂, x₃)]	=	x₁ + x₂ + 1
[tt]	=	3

together with the usable rule

activate(X)

→

X

(5)

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

U11^#(tt,M,N)	→	U12^#(tt,activate(M),activate(N))	(12)
U12^#(tt,M,N)	→	plus^#(activate(N),activate(M))	(11)
plus^#(N,s(M))	→	U11^#(tt,M,N)	(8)

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.