Certification Problem

Input (TPDB SRS_Standard/Bouchare_06/14)

The rewrite relation of the following TRS is considered.

a(a(a(x1))) → b(x1) (1)
b(b(x1)) → b(a(b(x1))) (2)
b(b(x1)) → a(a(x1)) (3)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the arctic semiring over the integers
[b(x1)] =
0 2
0 2
· x1 +
-∞ -∞
-∞ -∞
[a(x1)] =
1 2
0 0
· x1 +
-∞ -∞
-∞ -∞
all of the following rules can be deleted.
a(a(a(x1))) → b(x1) (1)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[b(x1)] =
1 0 1
1 0 0
0 1 0
· x1 +
1 0 0
0 0 0
1 0 0
[a(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
b(b(x1)) → b(a(b(x1))) (2)
b(b(x1)) → a(a(x1)) (3)

1.1.1 R is empty

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