Certification Problem

Input (TPDB SRS_Standard/Mixed_SRS/07)

The rewrite relation of the following TRS is considered.

a(a(b(b(x1)))) → a(b(a(a(x1)))) (1)
a(x1) → b(b(b(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(b(a(a(x1)))) → a(a(b(a(x1)))) (3)
a(x1) → b(b(b(x1))) (2)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
b#(b(a(a(x1)))) → a#(a(b(a(x1)))) (4)
b#(b(a(a(x1)))) → a#(b(a(x1))) (5)
b#(b(a(a(x1)))) → b#(a(x1)) (6)
a#(x1) → b#(b(b(x1))) (7)
a#(x1) → b#(b(x1)) (8)
a#(x1) → b#(x1) (9)

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 -∞ -∞
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[b(x1)] =
-∞
-∞
-∞
+
0 -∞ -∞
0 -∞ 0
0 0 -∞
· x1
[a(x1)] =
-∞
-∞
0
+
0 -∞ 0
0 -∞ 0
1 0 1
· x1
[a#(x1)] =
0
-∞
-∞
+
0 0 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1
the pair
b#(b(a(a(x1)))) → b#(a(x1)) (6)
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 0 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[b(x1)] =
0
0
0
+
-∞ -∞ 0
-∞ -∞ 0
0 0 -∞
· x1
[a(x1)] =
0
1
0
+
0 1 0
1 0 0
0 0 -∞
· x1
[a#(x1)] =
0
-∞
-∞
+
0 0 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1
the pairs
b#(b(a(a(x1)))) → a#(a(b(a(x1)))) (4)
b#(b(a(a(x1)))) → a#(b(a(x1))) (5)
could be deleted.

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.