Certification Problem

Input (TPDB SRS_Standard/Bouchare_06/01)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
b(b(b(x1))) → a(x1) (1)
a(a(x1)) → a(b(a(x1))) (2)
a(a(a(x1))) → a(a(b(x1))) (4)

1.1 Rule Removal

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

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[a(x1)] =
1 1 0
1 0 1
0 1 1
· x1 +
0 0 0
1 0 0
0 0 0
[b(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.
a(a(x1)) → a(b(a(x1))) (2)
a(a(a(x1))) → a(a(b(x1))) (4)

1.1.1.1 R is empty

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