Certification Problem

Input (TPDB TRS_Standard/AotoYamada_05/017)

The rewrite relation of the following TRS is considered.

app(app(app(uncurry,f),x),y)

→

app(app(f,x),y)

(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(uncurry,f),x),y)	→	app^#(f,x)	(2)
app^#(app(app(uncurry,f),x),y)	→	app^#(app(f,x),y)	(3)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

app^#(app(app(uncurry,f),x),y) → app^#(app(f,x),y) (3)

app^#(app(app(uncurry,f),x),y) → app^#(f,x) (2)

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

[uncurry] = 1

[app^#(x₁, x₂)] = x₁ + 0

[app(x₁, x₂)] = x₁ + x₂ + 7758

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

app^#(app(app(uncurry,f),x),y) → app^#(app(f,x),y) (3)

app^#(app(app(uncurry,f),x),y) → app^#(f,x) (2)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.