Certification Problem

Input (TPDB TRS_Standard/Mixed_TRS/jones4)

The rewrite relation of the following TRS is considered.

p(m,n,s(r))	→	p(m,r,n)	(1)
p(m,s(n),0)	→	p(0,n,m)	(2)
p(m,0,0)	→	m	(3)

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.

p^#(m,n,s(r))	→	p^#(m,r,n)	(4)
p^#(m,s(n),0)	→	p^#(0,n,m)	(5)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

p^#(m,s(n),0) → p^#(0,n,m) (5)

p^#(m,n,s(r)) → p^#(m,r,n) (4)

1.1.1 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation

[s(x₁)] = x₁ + 1

[p^#(x₁, x₂, x₃)] = x₁ + x₂ + x₃ + 0

[p(x₁, x₂, x₃)] = 0

[0] = 1

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

p^#(m,s(n),0) → p^#(0,n,m) (5)

p^#(m,n,s(r)) → p^#(m,r,n) (4)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.