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 AProVE @ termCOMP 2023)

1 Uncurrying

We uncurry the binary symbol app in combination with the following symbol map which also determines the applicative arities of these symbols.

comp	is mapped to	comp,	comp1(x₁),	comp2(x₁, x₂),	comp3(x₁, x₂, x₃)
twice	is mapped to	twice,	twice1(x₁)

There are no uncurry rules.
No rules have to be added for the eta-expansion.

Uncurrying the rules and adding the uncurrying rules yields the new set of rules

comp3(f,g,x)	→	app(f,app(g,x))	(7)
twice1(f)	→	comp2(f,f)	(8)
app(comp,y1)	→	comp1(y1)	(3)
app(comp1(x0),y1)	→	comp2(x0,y1)	(4)
app(comp2(x0,x1),y1)	→	comp3(x0,x1,y1)	(5)
app(twice,y1)	→	twice1(y1)	(6)

1.1 Switch to Innermost Termination

The TRS is overlay and locally confluent:

10

Hence, it suffices to show innermost termination in the following.

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

comp3^#(f,g,x)	→	app^#(f,app(g,x))	(9)
comp3^#(f,g,x)	→	app^#(g,x)	(10)
app^#(comp2(x0,x1),y1)	→	comp3^#(x0,x1,y1)	(11)
app^#(twice,y1)	→	twice1^#(y1)	(12)

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

app^#(comp2(x0,x1),y1) → comp3^#(x0,x1,y1) (11)

comp3^#(f,g,x) → app^#(f,app(g,x)) (9)

comp3^#(f,g,x) → app^#(g,x) (10)

1.1.1.1.1 Size-Change Termination

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

app^#(comp2(x0,x1),y1) → comp3^#(x0,x1,y1) (11)

1 > 1

1 > 2

2 ≥ 3

comp3^#(f,g,x) → app^#(f,app(g,x)) (9)

1 ≥ 1

comp3^#(f,g,x) → app^#(g,x) (10)

2 ≥ 1

3 ≥ 2

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