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 AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

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

1.1 Reduction Pair Processor

Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the naturals
[a#(x1)] =
0
-∞
-∞
+
0 0 -∞
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[b(x1)] =
0
1
-∞
+
-∞ -∞ -∞
-∞ -∞ 0
0 1 -∞
· x1
[a(x1)] =
0
-∞
0
+
-∞ 0 -∞
-∞ 0 -∞
1 0 -∞
· x1
the pairs
a#(b(b(x1))) a#(a(a(x1))) (3)
a#(b(b(x1))) a#(a(x1)) (4)
could be deleted.

1.1.1 Monotonic Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals
[b(x1)] = 1 · x1
[a#(x1)] = 1 · x1
having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the rule could be deleted.

1.1.1.1 Size-Change Termination

Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.

a#(b(b(x1))) a#(x1) (5)
1 > 1

As there is no critical graph in the transitive closure, there are no infinite chains.