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.

b → l(b) (4)
f(x,y) → f(x,r(y)) (5)

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

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[b] =
0 0 0
0 0 0
0 0 0
[l(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[f(x1, x2)] =
1 1 1
0 0 1
0 0 1
· x1 +
1 1 0
0 1 0
0 0 0
· x2 +
0 0 0
1 0 0
0 0 0
[m(x1)] =
1 0 0
0 0 0
0 1 1
· x1 +
0 0 0
1 0 0
1 0 0
[r(x1)] =
1 0 0
0 1 0
0 0 0
· x1 +
0 0 0
0 0 0
1 0 0
all of the following rules can be deleted.
f(m(x),y) → f(x,m(y)) (3)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[b] =
1 0 0
0 0 0
0 0 0
[l(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[f(x1, x2)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 1 0
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[m(x1)] =
1 0 1
0 0 1
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[r(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
0 0 0
0 0 0
1 0 0
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.