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 ttt2 @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

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

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

1.1.1 Size-Change Termination

Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.

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

2 ≥ 2

1 > 1

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

1 > 1

As there is no critical graph in the transitive closure, there are no infinite chains.