Certification Problem

Input (TPDB SRS_Standard/Bouchare_06/11)

The rewrite relation of the following TRS is considered.

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

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

1.1 Dependency Pair Transformation

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

1.1.1 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
[a(x1)] =
-∞
-∞
0
+
1 0 0
0 -∞ -∞
1 0 -∞
· x1
[b#(x1)] =
-∞
-∞
-∞
+
0 0 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[b(x1)] =
-∞
0
0
+
0 1 -∞
1 0 0
0 1 0
· x1
the pair
a#(a(a(x1))) → b#(x1) (4)
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
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[b(x1)] =
1
1
-∞
+
0 1 1
1 -∞ 0
-∞ -∞ -∞
· x1
[a#(x1)] =
0
-∞
-∞
+
-∞ 0 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[a(x1)] =
0
1
-∞
+
-∞ 0 0
0 1 1
-∞ -∞ -∞
· x1
the pair
b#(b(x1)) → a#(x1) (6)
could be deleted.

1.1.1.1.1 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 -∞
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[b(x1)] =
0
1
0
+
0 1 0
1 0 0
-∞ 0 -∞
· x1
[a#(x1)] =
-∞
-∞
-∞
+
0 -∞ -∞
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[a(x1)] =
1
0
0
+
1 0 0
0 -∞ -∞
0 -∞ 0
· x1
the pair
a#(a(x1)) → b#(a(x1)) (8)
could be deleted.

1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.