Certification Problem

Input (TPDB TRS_Standard/Rubio_04/revlist)

The rewrite relation of the following TRS is considered.

rev1(0,nil)	→	0	(1)
rev1(s(X),nil)	→	s(X)	(2)
rev1(X,cons(Y,L))	→	rev1(Y,L)	(3)
rev(nil)	→	nil	(4)
rev(cons(X,L))	→	cons(rev1(X,L),rev2(X,L))	(5)
rev2(X,nil)	→	nil	(6)
rev2(X,cons(Y,L))	→	rev(cons(X,rev(rev2(Y,L))))	(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.

rev1^#(X,cons(Y,L))	→	rev1^#(Y,L)	(8)
rev^#(cons(X,L))	→	rev2^#(X,L)	(9)
rev^#(cons(X,L))	→	rev1^#(X,L)	(10)
rev2^#(X,cons(Y,L))	→	rev2^#(Y,L)	(11)
rev2^#(X,cons(Y,L))	→	rev^#(rev2(Y,L))	(12)
rev2^#(X,cons(Y,L))	→	rev^#(cons(X,rev(rev2(Y,L))))	(13)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair

rev2^#(X,cons(Y,L))	→	rev2^#(Y,L)	(11)
rev2^#(X,cons(Y,L))	→	rev^#(rev2(Y,L))	(12)
rev^#(cons(X,L))	→	rev2^#(X,L)	(9)
rev2^#(X,cons(Y,L))	→	rev^#(cons(X,rev(rev2(Y,L))))	(13)

1.1.1 Subterm Criterion Processor

We use the projection to multisets

π(rev2^#)	=	{ 2, 2 }
π(rev^#)	=	{ 1, 1 }
π(rev2)	=	{ 2 }
π(rev)	=	{ 1 }
π(cons)	=	{ 2, 2 }

to remove the pairs:

rev2^#(X,cons(Y,L))	→	rev2^#(Y,L)	(11)
rev2^#(X,cons(Y,L))	→	rev^#(rev2(Y,L))	(12)
rev^#(cons(X,L))	→	rev2^#(X,L)	(9)

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.

The 2^nd component contains the pair
rev1^#(X,cons(Y,L)) → rev1^#(Y,L) (8)

1.1.2 Size-Change Termination

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

rev1^#(X,cons(Y,L)) → rev1^#(Y,L) (8)

2 > 2

2 > 1

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