Certification Problem

Input (TPDB TRS_Standard/SK90/4.38)

The rewrite relation of the following TRS is considered.

g(f(x,y),z)	→	f(x,g(y,z))	(1)
g(h(x,y),z)	→	g(x,f(y,z))	(2)
g(x,h(y,z))	→	h(g(x,y),z)	(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.

g^#(x,h(y,z))	→	g^#(x,y)	(4)
g^#(f(x,y),z)	→	g^#(y,z)	(5)
g^#(h(x,y),z)	→	g^#(x,f(y,z))	(6)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

g^#(h(x,y),z) → g^#(x,f(y,z)) (6)

g^#(f(x,y),z) → g^#(y,z) (5)

g^#(x,h(y,z)) → g^#(x,y) (4)

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

[h(x₁, x₂)] = x₁ + x₂ + 2

[f(x₁, x₂)] = x₂ + 1

[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^#(h(x,y),z) → g^#(x,f(y,z)) (6)

g^#(f(x,y),z) → g^#(y,z) (5)

g^#(x,h(y,z)) → g^#(x,y) (4)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.