Certification Problem

Input (TPDB TRS_Relative/Mixed_relative_TRS/ijcar2006)

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

f(a,g(y),z) → f(a,y,g(y)) (1)
f(b,g(y),z) → f(a,y,z) (2)
a → b (3)

and S is the following TRS.

f(x,y,z) → f(x,y,g(z)) (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
[f(x1, x2, x3)] =
0
0
+
1 1
1 0
· x1 +
1 1
0 1
· x2 +
1 0
0 0
· x3
[a] =
0
1
[g(x1)] =
0
1
+
1 0
1 1
· x1
[b] =
0
0
all of the following rules can be deleted.
f(a,g(y),z) → f(a,y,g(y)) (1)

1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[f(x1, x2, x3)] =
0
1
+
1 1
1 1
· x1 +
1 1
1 1
· x2 +
1 0
1 0
· x3
[b] =
0
1
[g(x1)] =
0
1
+
1 0
1 1
· x1
[a] =
1
1
all of the following rules can be deleted.
a → b (3)

1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[f(x1, x2, x3)] =
1
1
+
1 0
1 1
· x1 +
1 1
1 1
· x2 +
1 0
1 0
· x3
[b] =
0
1
[g(x1)] =
0
1
+
1 0
1 1
· x1
[a] =
0
1
all of the following rules can be deleted.
f(b,g(y),z) → f(a,y,z) (2)

1.1.1.1 R is empty

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