Certification Problem

Input (TPDB SRS_Standard/Secret_06_SRS/7)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a#(b(b(x1))) → a#(x1) (3)
a#(b(b(x1))) → a#(a(x1)) (4)
a#(b(b(x1))) → a#(a(a(x1))) (5)

1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[a(x1)] =
0 1 0
0 0 1
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[b(x1)] =
0 0 1
1 0 0
1 0 0
· x1 +
0 0 0
0 0 0
1 0 0
[a#(x1)] =
0 0 1
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
together with the usable rules
a(b(b(x1))) → b(a(a(a(x1)))) (1)
a(a(b(x1))) → b(a(b(x1))) (2)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
a#(b(b(x1))) → a#(x1) (3)
a#(b(b(x1))) → a#(a(x1)) (4)
a#(b(b(x1))) → a#(a(a(x1))) (5)
could be deleted.

1.1.1 P is empty

There are no pairs anymore.