Certification Problem

Input (TPDB TRS_Relative/Relative_05/rtL-me2)

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

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

and S is the following TRS.

topA(i,N1,y) → topA(1,T1,y) (7)
topB(i,x,N2) → topB(0,x,T2) (8)
topA(i,S1,y) → topA(i,N1,y) (9)
topB(i,x,S2) → topB(i,x,N2) (10)
topB(i,N1,T2) → topB(i,N1,S2) (11)
topB(1,T1,T2) → topB(1,T1,S2) (12)

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 1
1 0 1
1 1 0
· x2 +
1 0 0
0 0 0
0 0 0
· x3 +
0 0 0
0 0 0
0 0 0
[S1] =
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
1 0 0
1 0 0
· x2 +
1 0 0
1 1 1
1 0 1
· x3 +
0 0 0
0 0 0
0 0 0
[T1] =
0 0 0
0 0 0
0 0 0
[N2] =
0 0 0
0 0 0
0 0 0
[N1] =
1 0 0
0 0 0
0 0 0
[T2] =
0 0 0
0 0 0
0 0 0
[S2] =
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,N1,y) → topA(1,T1,y) (1)
topA(i,N1,y) → topA(1,T1,y) (7)

1.1 Rule Removal

Using the linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1 over the naturals
[topA(x1, x2, x3)] =
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 1 0
· x1 +
1 0 1 0 0
0 0 0 0 0
0 1 0 0 0
0 0 1 0 1
0 0 0 1 0
· x2 +
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x3 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[S1] =
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[topB(x1, x2, x3)] =
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 1 0
· x1 +
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 1 0 1
0 0 0 0 0
· x2 +
1 1 1 0 0
0 0 1 1 1
0 0 0 0 1
1 0 1 0 0
0 0 1 1 0
· x3 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[T1] =
1 0 0 0 0
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[N2] =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[N1] =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
[T2] =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[S2] =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[0] =
0 0 0 0 0
0 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 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,S1,y) → topA(i,N1,y) (9)

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 1
1 1 0
1 1 1
· x2 +
1 0 0
0 0 0
0 0 0
· x3 +
0 0 0
0 0 0
1 0 0
[S1] =
1 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
1 1 0
· x2 +
1 0 0
1 1 0
1 1 1
· x3 +
0 0 0
0 0 0
1 0 0
[T1] =
1 0 0
0 0 0
0 0 0
[N2] =
0 0 0
0 0 0
0 0 0
[N1] =
0 0 0
0 0 0
1 0 0
[T2] =
0 0 0
0 0 0
0 0 0
[S2] =
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,T2) → topB(i,N1,S2) (5)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1 over the naturals
[topA(x1, x2, x3)] =
1 0 0 0 0
0 1 0 0 0
0 1 0 0 0
0 0 0 0 0
0 0 0 0 0
· x1 +
1 0 1 1 1
0 0 1 0 1
0 1 0 0 0
1 0 1 1 1
0 1 1 0 0
· x2 +
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x3 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
[S1] =
1 0 0 0 0
1 0 0 0 0
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[topB(x1, x2, x3)] =
1 0 0 0 0
0 1 0 0 0
0 1 0 0 0
0 0 0 0 0
0 0 0 0 0
· x1 +
1 0 0 0 0
0 0 1 0 0
0 1 0 0 0
0 0 0 0 0
0 0 1 0 0
· x2 +
1 0 0 0 0
1 0 0 0 0
0 0 0 0 1
1 0 0 0 1
0 1 1 1 0
· x3 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[T1] =
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[N2] =
0 0 0 0 0
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[N1] =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[T2] =
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[S2] =
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[0] =
0 0 0 0 0
0 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
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
all of the following rules can be deleted.
topB(i,S1,y) → topA(i,N1,y) (3)

1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1 over the naturals
[topA(x1, x2, x3)] =
1 0 0 0 1
0 1 0 0 1
0 0 0 0 1
0 0 1 0 0
1 0 0 0 0
· x1 +
1 0 0 0 0
0 1 1 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x2 +
1 0 0 0 1
0 1 0 0 1
0 0 0 0 0
0 0 0 0 0
0 0 1 1 0
· x3 +
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
[topB(x1, x2, x3)] =
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x1 +
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x2 +
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 1 1 0
0 0 0 0 0
· x3 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
[T1] =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[N2] =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
1 0 0 0 0
[N1] =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[T2] =
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
[S2] =
0 0 0 0 0
1 0 0 0 0
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[0] =
0 0 0 0 0
0 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 0
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
all of the following rules can be deleted.
topA(i,x,N2) → topB(0,x,T2) (2)
topA(1,T1,T2) → topB(1,T1,S2) (6)

1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1 over the naturals
[topA(x1, x2, x3)] =
1 0 0 1 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x1 +
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x2 +
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x3 +
1 0 0 0 0
0 0 0 0 0
1 0 0 0 0
1 0 0 0 0
1 0 0 0 0
[topB(x1, x2, x3)] =
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x1 +
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x2 +
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 1 0 1
0 0 0 0 0
· x3 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[T1] =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[N2] =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
[N1] =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[T2] =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
[S2] =
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[0] =
0 0 0 0 0
0 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 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
all of the following rules can be deleted.
topA(i,x,S2) → topB(i,x,N2) (4)

1.1.1.1.1.1.1 R is empty

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