Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/ExProp7_Luc06_GM)

The rewrite relation of the following TRS is considered.

a__f(0) cons(0,f(s(0))) (1)
a__f(s(0)) a__f(a__p(s(0))) (2)
a__p(s(X)) mark(X) (3)
mark(f(X)) a__f(mark(X)) (4)
mark(p(X)) a__p(mark(X)) (5)
mark(0) 0 (6)
mark(cons(X1,X2)) cons(mark(X1),X2) (7)
mark(s(X)) s(mark(X)) (8)
a__f(X) f(X) (9)
a__p(X) p(X) (10)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by NaTT @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a__f#(s(0)) a__p#(s(0)) (11)
a__p#(s(X)) mark#(X) (12)
mark#(f(X)) mark#(X) (13)
mark#(s(X)) mark#(X) (14)
mark#(p(X)) mark#(X) (15)
mark#(cons(X1,X2)) mark#(X1) (16)
a__f#(s(0)) a__f#(a__p(s(0))) (17)
mark#(f(X)) a__f#(mark(X)) (18)
mark#(p(X)) a__p#(mark(X)) (19)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.