Certification Problem

Input (TPDB TRS_Relative/Relative_05/rtL-me3)

The relative rewrite relation R/S is considered where R is the following TRS

topB(i,N1(x),y) → topA(1,T1(x),y) (1)
topA(i,x,N2(y)) → topB(0,x,T2(y)) (2)
topB(i,S1(x),y) → topA(i,N1(x),y) (3)
topA(i,x,S2(y)) → topB(i,x,N2(y)) (4)
topA(i,N1(x),T2(y)) → topB(i,N1(x),S2(y)) (5)
topA(1,T1(x),T2(y)) → topB(1,T1(x),S2(y)) (6)

and S is the following TRS.

topA(i,N1(x),y) → topA(1,T1(x),y) (7)
topB(i,x,N2(y)) → topB(0,x,T2(y)) (8)
topA(i,S1(x),y) → topA(i,N1(x),y) (9)
topB(i,x,S2(y)) → topB(i,x,N2(y)) (10)
topB(i,N1(x),T2(y)) → topB(i,N1(x),S2(y)) (11)
topB(1,T1(x),T2(y)) → topB(1,T1(x),S2(y)) (12)
topA(i,N1(x),y) → topA(i,N1(C(x)),y) (13)
topB(i,x,N2(y)) → topB(i,x,N2(C(y))) (14)
topA(i,T1(x),y) → topA(i,T1(x),y) (15)
topB(i,x,T2(y)) → topB(i,x,T2(y)) (16)
topB(i,x,S2(y)) → topB(i,x,S2(D(y))) (17)

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
[topA(x1, x2, x3)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
1 0 0
0 0 0
0 0 0
· x3 +
0 0 0
0 0 0
0 0 0
[S1(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
[topB(x1, x2, x3)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
1 0 0
0 0 0
0 0 0
· x3 +
0 0 0
0 0 0
0 0 0
[T1(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[N2(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[C(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
1 0 0
[N1(x1)] =
1 0 0
0 1 1
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[T2(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[D(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
1 0 0
[S2(x1)] =
1 0 0
1 1 1
1 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[0] =
0 0 0
0 0 0
0 0 0
[1] =
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
topB(i,S1(x),y) → topA(i,N1(x),y) (3)
topA(i,S1(x),y) → topA(i,N1(x),y) (9)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[topA(x1, x2, x3)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 1 0
0 0 0
1 0 0
· x2 +
1 0 0
0 0 1
1 0 1
· x3 +
0 0 0
0 0 0
0 0 0
[topB(x1, x2, x3)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
1 0 0
· x2 +
1 0 0
0 0 1
1 0 1
· x3 +
0 0 0
0 0 0
0 0 0
[T1(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[N2(x1)] =
1 0 0
0 0 0
1 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[C(x1)] =
1 0 0
0 0 0
1 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[N1(x1)] =
1 0 0
1 0 0
1 0 1
· x1 +
0 0 0
1 0 0
0 0 0
[T2(x1)] =
1 0 0
0 0 0
1 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[D(x1)] =
1 0 0
0 0 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[S2(x1)] =
1 0 0
0 0 1
1 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[0] =
0 0 0
0 0 0
0 0 0
[1] =
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
topA(i,N1(x),T2(y)) → topB(i,N1(x),S2(y)) (5)
topA(i,N1(x),y) → topA(1,T1(x),y) (7)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[topA(x1, x2, x3)] =
1 1 0
0 0 0
0 0 0
· x1 +
1 1 0
0 0 0
0 0 0
· x2 +
1 0 0
0 0 0
0 0 0
· x3 +
0 0 0
0 0 0
0 0 0
[topB(x1, x2, x3)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 1 0
0 0 0
0 0 0
· x2 +
1 0 0
0 0 0
0 0 0
· x3 +
0 0 0
0 0 0
0 0 0
[T1(x1)] =
1 0 0
1 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[N2(x1)] =
1 0 0
0 1 1
0 1 1
· x1 +
0 0 0
0 0 0
0 0 0
[C(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
1 0 0
[N1(x1)] =
1 0 0
1 0 0
1 1 1
· x1 +
0 0 0
1 0 0
0 0 0
[T2(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[D(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[S2(x1)] =
1 0 0
0 1 1
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[0] =
0 0 0
0 0 0
0 0 0
[1] =
0 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
topA(1,T1(x),T2(y)) → topB(1,T1(x),S2(y)) (6)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[topA(x1, x2, x3)] =
1 0 0
1 0 0
0 0 0
· x1 +
1 1 0
0 0 0
0 0 0
· x2 +
1 0 0
0 1 1
0 0 0
· x3 +
0 0 0
0 0 0
0 0 0
[topB(x1, x2, x3)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 1 0
0 0 0
0 0 0
· x2 +
1 0 0
0 1 1
0 0 0
· x3 +
0 0 0
0 0 0
0 0 0
[T1(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
[N2(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
[C(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[N1(x1)] =
1 0 0
1 0 0
0 0 0
· x1 +
1 0 0
1 0 0
0 0 0
[T2(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
[D(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[S2(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
1 0 0
[0] =
0 0 0
0 0 0
0 0 0
[1] =
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
topB(i,N1(x),y) → topA(1,T1(x),y) (1)

1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[topA(x1, x2, x3)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 1 0
1 0 1
· x2 +
1 1 1
0 0 0
1 1 0
· x3 +
0 0 0
1 0 0
0 0 0
[topB(x1, x2, x3)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
1 0 0
0 0 0
0 0 0
· x3 +
0 0 0
0 0 0
0 0 0
[T1(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
[N2(x1)] =
1 0 0
1 0 0
0 1 1
· x1 +
0 0 0
0 0 0
0 0 0
[C(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
1 0 0
[N1(x1)] =
1 0 0
1 0 0
0 0 0
· x1 +
1 0 0
0 0 0
1 0 0
[T2(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[D(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[S2(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
1 0 0
[0] =
0 0 0
0 0 0
0 0 0
[1] =
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
topA(i,x,S2(y)) → topB(i,x,N2(y)) (4)

1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[topA(x1, x2, x3)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
1 1 1
0 1 0
0 0 0
· x3 +
0 0 0
0 0 0
0 0 0
[topB(x1, x2, x3)] =
1 0 0
0 0 1
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
1 0 0
0 0 0
0 0 0
· x3 +
0 0 0
1 0 0
0 0 0
[T1(x1)] =
1 0 1
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[N2(x1)] =
1 0 1
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
1 0 0
[C(x1)] =
1 0 1
1 0 1
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[N1(x1)] =
1 0 1
1 1 0
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[T2(x1)] =
1 0 1
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[D(x1)] =
1 0 0
1 0 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[S2(x1)] =
1 0 1
0 1 0
0 0 0
· x1 +
0 0 0
0 0 0
1 0 0
[0] =
0 0 0
0 0 0
0 0 0
[1] =
0 0 0
0 0 0
1 0 0
all of the following rules can be deleted.
topA(i,x,N2(y)) → topB(0,x,T2(y)) (2)

1.1.1.1.1.1.1 R is empty

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