Certification Problem

Input (TPDB SRS_Standard/Bouchare_06/15)

The rewrite relation of the following TRS is considered.

a(b(a(x1))) b(x1) (1)
b(a(b(x1))) b(b(b(x1))) (2)
b(b(x1)) a(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)] =
1 2
0 1
· x1 +
-∞ -∞
-∞ -∞
[a(x1)] =
0 0
0 0
· x1 +
-∞ -∞
-∞ -∞
all of the following rules can be deleted.
b(b(x1)) a(a(a(x1))) (3)

1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(b) = 0 weight(b) = 2
prec(a) = 1 weight(a) = 2
all of the following rules can be deleted.
a(b(a(x1))) b(x1) (1)
b(a(b(x1))) b(b(b(x1))) (2)

1.1.1 R is empty

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