Certification Problem

Input (TPDB TRS_Standard/Applicative_05/Ex2_6_1Composition)

The rewrite relation of the following TRS is considered.

app(app(app(compose,f),g),x)

→

app(f,app(g,x))

(1)

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^#(app(app(compose,f),g),x)	→	app^#(g,x)	(2)
app^#(app(app(compose,f),g),x)	→	app^#(f,app(g,x))	(3)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

app^#(app(app(compose,f),g),x) → app^#(f,app(g,x)) (3)

app^#(app(app(compose,f),g),x) → app^#(g,x) (2)

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

[compose] = 1

[app^#(x₁, x₂)] = x₁ + x₂ + 0

[app(x₁, x₂)] = x₁ + x₂ + 23676

together with the usable rule
app(app(app(compose,f),g),x) → app(f,app(g,x)) (1)
(w.r.t. the implicit argument filter of the reduction pair), the pairs

app^#(app(app(compose,f),g),x) → app^#(f,app(g,x)) (3)

app^#(app(app(compose,f),g),x) → app^#(g,x) (2)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.