Certification Problem

Input (TPDB TRS_Standard/Various_04/23)

The rewrite relation of the following TRS is considered.

g(0,f(x,x))	→	x	(1)
g(x,s(y))	→	g(f(x,y),0)	(2)
g(s(x),y)	→	g(f(x,y),0)	(3)
g(f(x,y),0)	→	f(g(x,0),g(y,0))	(4)

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.

g^#(s(x),y)	→	g^#(f(x,y),0)	(5)
g^#(f(x,y),0)	→	g^#(x,0)	(6)
g^#(x,s(y))	→	g^#(f(x,y),0)	(7)
g^#(f(x,y),0)	→	g^#(y,0)	(8)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

g^#(f(x,y),0) → g^#(y,0) (8)

g^#(f(x,y),0) → g^#(x,0) (6)

g^#(s(x),y) → g^#(f(x,y),0) (5)

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

[s(x₁)] = x₁ + 23678

[f(x₁, x₂)] = x₁ + x₂ + 1

[0] = 23676

[g^#(x₁, x₂)] = x₁ + x₂ + 0

[g(x₁, x₂)] = 0

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

g^#(f(x,y),0) → g^#(y,0) (8)

g^#(f(x,y),0) → g^#(x,0) (6)

g^#(s(x),y) → g^#(f(x,y),0) (5)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.