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 AProVE @ termCOMP 2023)

1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(nil) = 7 weight(nil) = 1
prec(tt) = 0 weight(tt) = 4
prec(mark) = 10 weight(mark) = 0
prec(a__U11) = 5 weight(a__U11) = 1
prec(a__U12) = 2 weight(a__U12) = 1
prec(a__isNePal) = 6 weight(a__isNePal) = 2
prec(U11) = 3 weight(U11) = 1
prec(U12) = 1 weight(U12) = 1
prec(isNePal) = 4 weight(isNePal) = 2
prec(__) = 8 weight(__) = 0
prec(a____) = 9 weight(a____) = 0
all of the following rules can be deleted.
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)

1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.