Certification Problem

Input (TPDB SRS_Relative/Mixed_relative_SRS/zr10)

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

a(b(a(x1))) → a(b(b(a(x1)))) (1)
b(a(b(x1))) → b(a(a(b(x1)))) (2)

and S is the following TRS.

a(x1) → a(a(a(x1))) (3)
b(x1) → b(b(b(x1))) (4)

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(x1)] =
1 0 1
0 0 1
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
[a(x1)] =
1 0 0
0 0 0
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
b(a(b(x1))) → b(a(a(b(x1)))) (2)

1.1 Rule Removal

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

1.1.1 R is empty

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