Certification Problem

Input (TPDB TRS_Relative/Relative_05/rt3-9)

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

l(m(x)) → m(l(x)) (1)
m(r(x)) → r(m(x)) (2)
f(m(x),y) → f(x,m(y)) (3)

and S is the following TRS.

f(x,y) → f(x,r(y)) (4)
b → l(b) (5)

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
[l(x1)] =
0
0
+
1 1
1 1
· x1
[m(x1)] =
0
1
+
1 0
0 1
· x1
[r(x1)] =
0
0
+
1 0
0 1
· x1
[f(x1, x2)] =
1
1
+
1 1
1 1
· x1 +
1 1
1 1
· x2
[b] =
0
0
all of the following rules can be deleted.
l(m(x)) → m(l(x)) (1)

1.1 Rule Removal

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

1.1.1 Rule Removal

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

1.1.1.1 R is empty

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