Certification Problem

Input (TPDB SRS_Relative/Mixed_relative_SRS/zr06)

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

a(b(a(x1))) → a(b(b(a(x1)))) (1)

and S is the following TRS.

b(x1) → b(b(b(x1))) (2)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Semantic Labeling

Root-labeling is applied.

We obtain the labeled TRS
ab(ba(aa(x1))) → ab(bb(ba(aa(x1)))) (3)
ab(ba(ab(x1))) → ab(bb(ba(ab(x1)))) (4)
ba(x1) → bb(bb(ba(x1))) (5)
bb(x1) → bb(bb(bb(x1))) (6)

1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[ab(x1)] =
0
0
+
1 2
0 0
· x1
[ba(x1)] =
0
0
+
1 0
0 2
· x1
[aa(x1)] =
0
2
+
1 0
0 0
· x1
[bb(x1)] =
0
0
+
1 0
0 0
· x1
all of the following rules can be deleted.
ab(ba(aa(x1))) → ab(bb(ba(aa(x1)))) (3)

1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[ab(x1)] =
0
2
+
1 2
0 0
· x1
[ba(x1)] =
0
0
+
2 0
0 2
· x1
[bb(x1)] =
0
0
+
1 0
0 0
· x1
all of the following rules can be deleted.
ab(ba(ab(x1))) → ab(bb(ba(ab(x1)))) (4)

1.1.1.1 R is empty

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