Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/PALINDROME_nosorts_C)

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(and(tt,X)) mark(X) (4)
active(isNePal(__(I,__(P,I)))) mark(tt) (5)
active(__(X1,X2)) __(active(X1),X2) (6)
active(__(X1,X2)) __(X1,active(X2)) (7)
active(and(X1,X2)) and(active(X1),X2) (8)
active(isNePal(X)) isNePal(active(X)) (9)
__(mark(X1),X2) mark(__(X1,X2)) (10)
__(X1,mark(X2)) mark(__(X1,X2)) (11)
and(mark(X1),X2) mark(and(X1,X2)) (12)
isNePal(mark(X)) mark(isNePal(X)) (13)
proper(__(X1,X2)) __(proper(X1),proper(X2)) (14)
proper(nil) ok(nil) (15)
proper(and(X1,X2)) and(proper(X1),proper(X2)) (16)
proper(tt) ok(tt) (17)
proper(isNePal(X)) isNePal(proper(X)) (18)
__(ok(X1),ok(X2)) ok(__(X1,X2)) (19)
and(ok(X1),ok(X2)) ok(and(X1,X2)) (20)
isNePal(ok(X)) ok(isNePal(X)) (21)
top(mark(X)) top(proper(X)) (22)
top(ok(X)) top(active(X)) (23)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[tt] =
0 0 0
0 0 0
0 0 0
[ok(x1)] =
1 0 0
0 0 1
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[active(x1)] =
1 0 0
0 0 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[nil] =
0 0 0
0 0 0
0 0 0
[and(x1, x2)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 1 1
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[__(x1, x2)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[proper(x1)] =
1 0 0
0 0 0
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[top(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[isNePal(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
1 0 0
0 0 0
0 0 0
[mark(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
active(isNePal(__(I,__(P,I)))) mark(tt) (5)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[tt] =
1 0 0
0 0 0
0 0 0
[ok(x1)] =
1 0 0
0 0 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[active(x1)] =
1 0 0
1 1 0
0 0 0
· x1 +
0 0 0
1 0 0
1 0 0
[nil] =
0 0 0
0 0 0
0 0 0
[and(x1, x2)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 1
0 0 0
· x2 +
0 0 0
0 0 0
1 0 0
[__(x1, x2)] =
1 0 0
1 0 0
0 0 0
· x1 +
1 0 0
1 0 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[proper(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[top(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[isNePal(x1)] =
1 0 0
0 0 0
0 0 1
· x1 +
0 0 0
1 0 0
0 0 0
[mark(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
active(and(tt,X)) mark(X) (4)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[tt] =
1 0 0
0 0 0
0 0 0
[ok(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[active(x1)] =
1 0 0
0 1 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[nil] =
0 0 0
1 0 0
0 0 0
[and(x1, x2)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[__(x1, x2)] =
1 0 0
1 0 0
0 0 0
· x1 +
1 0 0
0 1 0
0 0 0
· x2 +
1 0 0
0 0 0
0 0 0
[proper(x1)] =
1 0 0
1 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[top(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
[isNePal(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
[mark(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
active(__(X,nil)) mark(X) (2)
active(__(nil,X)) mark(X) (3)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[tt] =
0 0 0
0 0 0
1 0 0
[ok(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[active(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[nil] =
0 0 0
0 0 0
0 0 0
[and(x1, x2)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 1 1
· x2 +
0 0 0
0 0 0
0 0 0
[__(x1, x2)] =
1 1 1
0 1 0
0 0 0
· x1 +
1 0 1
0 1 0
0 0 0
· x2 +
0 0 0
1 0 0
1 0 0
[proper(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[top(x1)] =
1 0 0
0 0 1
1 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[isNePal(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[mark(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
active(__(__(X,Y),Z)) mark(__(X,__(Y,Z))) (1)

1.1.1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
active#(__(X1,X2)) active#(X1) (24)
active#(__(X1,X2)) __#(active(X1),X2) (25)
active#(__(X1,X2)) active#(X2) (26)
active#(__(X1,X2)) __#(X1,active(X2)) (27)
active#(and(X1,X2)) active#(X1) (28)
active#(and(X1,X2)) and#(active(X1),X2) (29)
active#(isNePal(X)) active#(X) (30)
active#(isNePal(X)) isNePal#(active(X)) (31)
__#(mark(X1),X2) __#(X1,X2) (32)
__#(X1,mark(X2)) __#(X1,X2) (33)
and#(mark(X1),X2) and#(X1,X2) (34)
isNePal#(mark(X)) isNePal#(X) (35)
proper#(__(X1,X2)) proper#(X2) (36)
proper#(__(X1,X2)) proper#(X1) (37)
proper#(__(X1,X2)) __#(proper(X1),proper(X2)) (38)
proper#(and(X1,X2)) proper#(X2) (39)
proper#(and(X1,X2)) proper#(X1) (40)
proper#(and(X1,X2)) and#(proper(X1),proper(X2)) (41)
proper#(isNePal(X)) proper#(X) (42)
proper#(isNePal(X)) isNePal#(proper(X)) (43)
__#(ok(X1),ok(X2)) __#(X1,X2) (44)
and#(ok(X1),ok(X2)) and#(X1,X2) (45)
isNePal#(ok(X)) isNePal#(X) (46)
top#(mark(X)) proper#(X) (47)
top#(mark(X)) top#(proper(X)) (48)
top#(ok(X)) active#(X) (49)
top#(ok(X)) top#(active(X)) (50)

1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 6 components.