Certification Problem

Input (TPDB TRS_Innermost/Transformed_CSR_innermost_04/PALINDROME_nosorts_noand_iGM)

The rewrite relation of the following TRS is considered.

active(__(__(X,Y),Z)) mark(__(X,__(Y,Z))) (1)
active(__(X,nil)) mark(X) (2)
active(__(nil,X)) mark(X) (3)
active(U11(tt)) mark(U12(tt)) (4)
active(U12(tt)) mark(tt) (5)
active(isNePal(__(I,__(P,I)))) mark(U11(tt)) (6)
mark(__(X1,X2)) active(__(mark(X1),mark(X2))) (7)
mark(nil) active(nil) (8)
mark(U11(X)) active(U11(mark(X))) (9)
mark(tt) active(tt) (10)
mark(U12(X)) active(U12(mark(X))) (11)
mark(isNePal(X)) active(isNePal(mark(X))) (12)
__(mark(X1),X2) __(X1,X2) (13)
__(X1,mark(X2)) __(X1,X2) (14)
__(active(X1),X2) __(X1,X2) (15)
__(X1,active(X2)) __(X1,X2) (16)
U11(mark(X)) U11(X) (17)
U11(active(X)) U11(X) (18)
U12(mark(X)) U12(X) (19)
U12(active(X)) U12(X) (20)
isNePal(mark(X)) isNePal(X) (21)
isNePal(active(X)) isNePal(X) (22)
The evaluation strategy is innermost.

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over the naturals
[active(x1)] = 1 · x1
[__(x1, x2)] = 2 + 1 · x1 + 1 · x2
[mark(x1)] = 1 · x1
[nil] = 0
[U11(x1)] = 2 + 1 · x1
[tt] = 1
[U12(x1)] = 2 · x1
[isNePal(x1)] = 1 · x1
all of the following rules can be deleted.
active(__(X,nil)) mark(X) (2)
active(__(nil,X)) mark(X) (3)
active(U11(tt)) mark(U12(tt)) (4)
active(U12(tt)) mark(tt) (5)
active(isNePal(__(I,__(P,I)))) mark(U11(tt)) (6)

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[active(x1)] = 1 · x1
[__(x1, x2)] = 2 + 2 · x1 + 1 · x2
[mark(x1)] = 1 · x1
[nil] = 0
[U11(x1)] = 2 · x1
[tt] = 0
[U12(x1)] = 1 · x1
[isNePal(x1)] = 1 · x1
all of the following rules can be deleted.
active(__(__(X,Y),Z)) mark(__(X,__(Y,Z))) (1)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[mark(x1)] = 2 · x1
[__(x1, x2)] = 1 + 2 · x1 + 1 · x2
[active(x1)] = 1 · x1
[nil] = 1
[U11(x1)] = 1 + 2 · x1
[tt] = 1
[U12(x1)] = 1 + 2 · x1
[isNePal(x1)] = 2 · x1
all of the following rules can be deleted.
mark(__(X1,X2)) active(__(mark(X1),mark(X2))) (7)
mark(nil) active(nil) (8)
mark(U11(X)) active(U11(mark(X))) (9)
mark(tt) active(tt) (10)
mark(U12(X)) active(U12(mark(X))) (11)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[mark(x1)] = 2 + 2 · x1
[isNePal(x1)] = 2 + 2 · x1
[active(x1)] = 1 · x1
[__(x1, x2)] = 1 · x1 + 1 · x2
[U11(x1)] = 2 · x1
[U12(x1)] = 2 · x1
all of the following rules can be deleted.
__(mark(X1),X2) __(X1,X2) (13)
__(X1,mark(X2)) __(X1,X2) (14)
U11(mark(X)) U11(X) (17)
U12(mark(X)) U12(X) (19)
isNePal(mark(X)) isNePal(X) (21)

1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[mark(x1)] = 2 · x1
[isNePal(x1)] = 2 + 2 · x1
[active(x1)] = 1 · x1
[__(x1, x2)] = 1 · x1 + 2 · x2
[U11(x1)] = 2 · x1
[U12(x1)] = 2 · x1
all of the following rules can be deleted.
mark(isNePal(X)) active(isNePal(mark(X))) (12)

1.1.1.1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(active) = 4 weight(active) = 0
prec(U11) = 1 weight(U11) = 1
prec(U12) = 2 weight(U12) = 1
prec(isNePal) = 3 weight(isNePal) = 1
prec(__) = 0 weight(__) = 0
all of the following rules can be deleted.
__(active(X1),X2) __(X1,X2) (15)
__(X1,active(X2)) __(X1,X2) (16)
U11(active(X)) U11(X) (18)
U12(active(X)) U12(X) (20)
isNePal(active(X)) isNePal(X) (22)

1.1.1.1.1.1.1 R is empty

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