Certification Problem

Input (TPDB TRS_Standard/AotoYamada_05/002)

The rewrite relation of the following TRS is considered.

app(app(filter,f),nil)	→	nil	(1)
app(app(filter,f),app(app(cons,y),ys))	→	app(app(app(filtersub,app(f,y)),f),app(app(cons,y),ys))	(2)
app(app(app(filtersub,true),f),app(app(cons,y),ys))	→	app(app(cons,y),app(app(filter,f),ys))	(3)
app(app(app(filtersub,false),f),app(app(cons,y),ys))	→	app(app(filter,f),ys)	(4)

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.

filter	is mapped to	filter,	filter1(x₁),	filter2(x₁, x₂)
nil	is mapped to	nil
cons	is mapped to	cons,	cons1(x₁),	cons2(x₁, x₂)
filtersub	is mapped to	filtersub,	filtersub1(x₁),	filtersub2(x₁, x₂),	filtersub3(x₁, x₂, x₃)
true	is mapped to	true
false	is mapped to	false

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

filter2(f,nil)	→	nil	(12)
filter2(f,cons2(y,ys))	→	filtersub3(app(f,y),f,cons2(y,ys))	(13)
filtersub3(true,f,cons2(y,ys))	→	cons2(y,filter2(f,ys))	(14)
filtersub3(false,f,cons2(y,ys))	→	filter2(f,ys)	(15)
app(filter,y1)	→	filter1(y1)	(5)
app(filter1(x0),y1)	→	filter2(x0,y1)	(6)
app(cons,y1)	→	cons1(y1)	(7)
app(cons1(x0),y1)	→	cons2(x0,y1)	(8)
app(filtersub,y1)	→	filtersub1(y1)	(9)
app(filtersub1(x0),y1)	→	filtersub2(x0,y1)	(10)
app(filtersub2(x0,x1),y1)	→	filtersub3(x0,x1,y1)	(11)

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.

filter2^#(f,cons2(y,ys))	→	filtersub3^#(app(f,y),f,cons2(y,ys))	(16)
filter2^#(f,cons2(y,ys))	→	app^#(f,y)	(17)
filtersub3^#(true,f,cons2(y,ys))	→	filter2^#(f,ys)	(18)
filtersub3^#(false,f,cons2(y,ys))	→	filter2^#(f,ys)	(19)
app^#(filter1(x0),y1)	→	filter2^#(x0,y1)	(20)
app^#(filtersub2(x0,x1),y1)	→	filtersub3^#(x0,x1,y1)	(21)

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

filter2^#(f,cons2(y,ys))	→	app^#(f,y)	(17)
app^#(filter1(x0),y1)	→	filter2^#(x0,y1)	(20)
app^#(filtersub2(x0,x1),y1)	→	filtersub3^#(x0,x1,y1)	(21)
filtersub3^#(true,f,cons2(y,ys))	→	filter2^#(f,ys)	(18)
filtersub3^#(false,f,cons2(y,ys))	→	filter2^#(f,ys)	(19)

1.1.1.1.1 Usable Rules Processor

We restrict the rewrite rules to the following usable rules of the DP problem.

There are no rules.

1.1.1.1.1.1 Innermost Lhss Removal Processor

We restrict the innermost strategy to the following left hand sides.

There are no lhss.

1.1.1.1.1.1.1 Size-Change Termination

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

app^#(filter1(x0),y1)	→	filter2^#(x0,y1)	(20)

1	>	1
2	≥	2
app^#(filtersub2(x0,x1),y1)	→	filtersub3^#(x0,x1,y1)	(21)

1	>	1
1	>	2
2	≥	3
filter2^#(f,cons2(y,ys))	→	app^#(f,y)	(17)

1	≥	1
2	>	2
filtersub3^#(true,f,cons2(y,ys))	→	filter2^#(f,ys)	(18)

2	≥	1
3	>	2
filtersub3^#(false,f,cons2(y,ys))	→	filter2^#(f,ys)	(19)

2	≥	1
3	>	2

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