Certification Problem

Input (TPDB SRS_Standard/Secret_06_SRS/multum3)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
b(a(a(x1))) → a(a(b(b(x1)))) (3)
b(a(b(x1))) → a(a(a(a(x1)))) (2)

1.1 Dependency Pair Transformation

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

1.1.1 Reduction Pair Processor

Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the naturals
[b#(x1)] =
0
-∞
-∞
+
0 -∞ 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[a(x1)] =
1
-∞
-∞
+
-∞ 0 1
0 0 0
0 -∞ -∞
· x1
[b(x1)] =
0
-∞
1
+
0 -∞ -∞
1 0 0
1 0 0
· x1
the pair
b#(a(a(x1))) → b#(x1) (5)
could be deleted.

1.1.1.1 Reduction Pair Processor

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

1.1.1.1.1 P is empty

There are no pairs anymore.