Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/PALINDROME_nosorts_noand_GM)

The rewrite relation of the following TRS is considered.

a____(__(X,Y),Z) a____(mark(X),a____(mark(Y),mark(Z))) (1)
a____(X,nil) mark(X) (2)
a____(nil,X) mark(X) (3)
a__U11(tt) a__U12(tt) (4)
a__U12(tt) tt (5)
a__isNePal(__(I,__(P,I))) a__U11(tt) (6)
mark(__(X1,X2)) a____(mark(X1),mark(X2)) (7)
mark(U11(X)) a__U11(mark(X)) (8)
mark(U12(X)) a__U12(mark(X)) (9)
mark(isNePal(X)) a__isNePal(mark(X)) (10)
mark(nil) nil (11)
mark(tt) tt (12)
a____(X1,X2) __(X1,X2) (13)
a__U11(X) U11(X) (14)
a__U12(X) U12(X) (15)
a__isNePal(X) isNePal(X) (16)

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__isNePal#(__(I,__(P,I))) a__U11#(tt) (17)
mark#(__(X1,X2)) mark#(X1) (18)
a____#(__(X,Y),Z) a____#(mark(Y),mark(Z)) (19)
mark#(isNePal(X)) mark#(X) (20)
a____#(__(X,Y),Z) a____#(mark(X),a____(mark(Y),mark(Z))) (21)
a____#(__(X,Y),Z) mark#(X) (22)
a____#(__(X,Y),Z) mark#(Y) (23)
mark#(__(X1,X2)) mark#(X2) (24)
a____#(nil,X) mark#(X) (25)
mark#(__(X1,X2)) a____#(mark(X1),mark(X2)) (26)
mark#(U11(X)) mark#(X) (27)
mark#(U11(X)) a__U11#(mark(X)) (28)
mark#(U12(X)) a__U12#(mark(X)) (29)
a____#(X,nil) mark#(X) (30)
mark#(isNePal(X)) a__isNePal#(mark(X)) (31)
a____#(__(X,Y),Z) mark#(Z) (32)
mark#(U12(X)) mark#(X) (33)
a__U11#(tt) a__U12#(tt) (34)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.