Certification Problem

Input (TPDB SRS_Relative/Waldmann_06_relative/r10)

The relative rewrite relation R/S is considered where R is the following TRS

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

and S is the following TRS.

b(x1) → b(b(b(x1))) (3)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Rule Removal

Using the matrix interpretations of dimension 6 with strict dimension 1 over the naturals
[b(x1)] =
1 0 2 0 0 0
0 0 0 1 0 0
0 0 0 0 0 0
0 0 1 0 0 0
0 1 1 3 0 1
0 2 2 0 0 0
· x1 +
0
0
0
0
1
0
[a(x1)] =
1 0 0 0 0 1
0 0 0 0 1 1
0 1 0 0 3 0
0 1 1 0 1 0
0 0 0 0 0 1
0 0 0 0 0 0
· x1 +
0
0
0
1
0
0
all of the following rules can be deleted.
b(a(a(b(x1)))) → a(a(a(a(x1)))) (1)
a(b(b(a(x1)))) → b(b(b(b(x1)))) (2)

1.1 R is empty

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