Certification Problem

Input (TPDB TRS_Standard/Der95/32)

The rewrite relation of the following TRS is considered.

sort(nil)	→	nil	(1)
sort(cons(x,y))	→	insert(x,sort(y))	(2)
insert(x,nil)	→	cons(x,nil)	(3)
insert(x,cons(v,w))	→	choose(x,cons(v,w),x,v)	(4)
choose(x,cons(v,w),y,0)	→	cons(x,cons(v,w))	(5)
choose(x,cons(v,w),0,s(z))	→	cons(v,insert(x,w))	(6)
choose(x,cons(v,w),s(y),s(z))	→	choose(x,cons(v,w),y,z)	(7)

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.

sort^#(cons(x,y))	→	sort^#(y)	(8)
sort^#(cons(x,y))	→	insert^#(x,sort(y))	(9)
insert^#(x,cons(v,w))	→	choose^#(x,cons(v,w),x,v)	(10)
choose^#(x,cons(v,w),0,s(z))	→	insert^#(x,w)	(11)
choose^#(x,cons(v,w),s(y),s(z))	→	choose^#(x,cons(v,w),y,z)	(12)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair
sort^#(cons(x,y)) → sort^#(y) (8)

1.1.1 Size-Change Termination

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

sort^#(cons(x,y)) → sort^#(y) (8)

1 > 1

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

The 2^nd component contains the pair

choose^#(x,cons(v,w),s(y),s(z))	→	choose^#(x,cons(v,w),y,z)	(12)
choose^#(x,cons(v,w),0,s(z))	→	insert^#(x,w)	(11)
insert^#(x,cons(v,w))	→	choose^#(x,cons(v,w),x,v)	(10)

1.1.2 Size-Change Termination

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

choose^#(x,cons(v,w),s(y),s(z))	→	choose^#(x,cons(v,w),y,z)	(12)

4	>	4
3	>	3
2	≥	2
1	≥	1
choose^#(x,cons(v,w),0,s(z))	→	insert^#(x,w)	(11)

2	>	2
1	≥	1
insert^#(x,cons(v,w))	→	choose^#(x,cons(v,w),x,v)	(10)

2	>	4
2	≥	2
1	≥	3
1	≥	1

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