Certification Problem

Input (TPDB TRS_Standard/AG01/#3.38)

The rewrite relation of the following TRS is considered.

rev(nil)	→	nil	(1)
rev(cons(x,l))	→	cons(rev1(x,l),rev2(x,l))	(2)
rev1(0,nil)	→	0	(3)
rev1(s(x),nil)	→	s(x)	(4)
rev1(x,cons(y,l))	→	rev1(y,l)	(5)
rev2(x,nil)	→	nil	(6)
rev2(x,cons(y,l))	→	rev(cons(x,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.

rev^#(cons(x,l))	→	rev2^#(x,l)	(8)
rev^#(cons(x,l))	→	rev1^#(x,l)	(9)
rev1^#(x,cons(y,l))	→	rev1^#(y,l)	(10)
rev2^#(x,cons(y,l))	→	rev2^#(y,l)	(11)
rev2^#(x,cons(y,l))	→	rev^#(cons(x,rev2(y,l)))	(12)

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^#(cons(x,rev2(y,l)))	(12)
rev^#(cons(x,l))	→	rev2^#(x,l)	(8)

1.1.1 Subterm Criterion Processor

We use the projection to multisets

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

to remove the pairs:

rev2^#(x,cons(y,l))	→	rev2^#(y,l)	(11)
rev2^#(x,cons(y,l))	→	rev^#(cons(x,rev2(y,l)))	(12)

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) (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.

rev1^#(x,cons(y,l)) → rev1^#(y,l) (10)

2 > 2

2 > 1

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