Certification Problem

Input (TPDB TRS_Relative/Relative_05/rt2-3)

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

R(x,B2) → B2 (1)
W(x,B2) → B2 (2)

and S is the following TRS.

B1 → W(T,B1) (3)
B1 → R(T,B1) (4)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[R(x1, x2)] =
0
0
+
1 0
1 0
· x1 +
1 1
0 1
· x2
[B2] =
1
1
[W(x1, x2)] =
0
0
+
1 0
1 0
· x1 +
1 0
0 1
· x2
[B1] =
1
0
[T] =
0
1
all of the following rules can be deleted.
R(x,B2) → B2 (1)

1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[W(x1, x2)] =
0
0
+
1 0
1 0
· x1 +
1 1
1 0
· x2
[B2] =
1
1
[B1] =
0
0
[T] =
0
1
[R(x1, x2)] =
0
0
+
1 0
1 0
· x1 +
1 1
1 1
· x2
all of the following rules can be deleted.
W(x,B2) → B2 (2)

1.1.1 R is empty

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