Certification Problem

Input (TPDB TRS_Standard/AotoYamada_05/019)

The rewrite relation of the following TRS is considered.

app(app(app(comp,f),g),x)	→	app(f,app(g,x))	(1)
app(twice,f)	→	app(app(comp,f),f)	(2)

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.

app^#(twice,f)	→	app^#(comp,f)	(3)
app^#(app(app(comp,f),g),x)	→	app^#(f,app(g,x))	(4)
app^#(twice,f)	→	app^#(app(comp,f),f)	(5)
app^#(app(app(comp,f),g),x)	→	app^#(g,x)	(6)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

app^#(app(app(comp,f),g),x) → app^#(g,x) (6)

app^#(app(app(comp,f),g),x) → app^#(f,app(g,x)) (4)

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

[twice] = 31455

[app^#(x₁, x₂)] = x₁ + 0

[comp] = 10803

[app(x₁, x₂)] = x₁ + x₂ + 20653

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

app^#(app(app(comp,f),g),x) → app^#(g,x) (6)

app^#(app(app(comp,f),g),x) → app^#(f,app(g,x)) (4)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.