Certification Problem

Input (TPDB TRS_Standard/CiME_04/append)

The rewrite relation of the following TRS is considered.

is_empty(nil) → true (1)
is_empty(cons(x,l)) → false (2)
hd(cons(x,l)) → x (3)
tl(cons(x,l)) → l (4)
append(l1,l2) → ifappend(l1,l2,l1) (5)
ifappend(l1,l2,nil) → l2 (6)
ifappend(l1,l2,cons(x,l)) → cons(x,append(l,l2)) (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.
append#(l1,l2) → ifappend#(l1,l2,l1) (8)
ifappend#(l1,l2,cons(x,l)) → append#(l,l2) (9)

1.1 Subterm Criterion Processor

We use the projection

π(ifappend#) = 3
π(append#) = 1

and remove the pairs:
ifappend#(l1,l2,cons(x,l)) → append#(l,l2) (9)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.