Certification Problem

Input (TPDB SRS_Standard/Zantema_06/16)

The rewrite relation of the following TRS is considered.

a(x1) → b(x1) (1)
a(a(x1)) → a(b(a(x1))) (2)
a(b(x1)) → b(b(b(x1))) (3)
a(a(a(x1))) → a(a(b(a(a(x1))))) (4)
a(a(b(x1))) → a(b(b(a(b(x1))))) (5)
a(b(a(x1))) → b(a(b(b(a(x1))))) (6)
a(b(b(x1))) → b(b(b(b(b(x1))))) (7)
b(a(x1)) → b(b(b(x1))) (8)
a(b(a(x1))) → a(b(b(a(b(x1))))) (9)
b(a(a(x1))) → b(a(b(b(a(x1))))) (10)
b(b(a(x1))) → b(b(b(b(b(x1))))) (11)

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 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[a(x1)] =
1 0 1
0 0 1
0 1 0
· x1 +
0 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
a(a(a(x1))) → a(a(b(a(a(x1))))) (4)

1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[b(x1)] = 0 · x1 + -∞
[a(x1)] = 2 · x1 + -∞
all of the following rules can be deleted.
a(x1) → b(x1) (1)
a(b(x1)) → b(b(b(x1))) (3)
a(b(b(x1))) → b(b(b(b(b(x1))))) (7)
b(a(x1)) → b(b(b(x1))) (8)
b(b(a(x1))) → b(b(b(b(b(x1))))) (11)

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 0
0 1 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[a(x1)] =
1 0 1
0 0 1
1 1 0
· x1 +
0 0 0
0 0 0
1 0 0
all of the following rules can be deleted.
a(a(x1)) → a(b(a(x1))) (2)
a(a(b(x1))) → a(b(b(a(b(x1))))) (5)
b(a(a(x1))) → b(a(b(b(a(x1))))) (10)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[b(x1)] =
1 0 0
0 0 1
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[a(x1)] =
1 1 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.
a(b(a(x1))) → b(a(b(b(a(x1))))) (6)
a(b(a(x1))) → a(b(b(a(b(x1))))) (9)

1.1.1.1.1 R is empty

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