Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-77)

The rewrite relation of the following TRS is considered.

b(a(b(b(x1)))) → a(a(b(a(x1)))) (1)
a(a(a(a(x1)))) → b(a(b(b(x1)))) (2)
b(b(b(a(x1)))) → b(b(a(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
b(b(a(b(x1)))) → a(b(a(a(x1)))) (4)
a(a(a(a(x1)))) → b(b(a(b(x1)))) (5)
a(b(b(b(x1)))) → a(a(b(b(x1)))) (6)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
b#(b(a(b(x1)))) → a#(b(a(a(x1)))) (7)
b#(b(a(b(x1)))) → b#(a(a(x1))) (8)
b#(b(a(b(x1)))) → a#(a(x1)) (9)
b#(b(a(b(x1)))) → a#(x1) (10)
a#(a(a(a(x1)))) → b#(b(a(b(x1)))) (11)
a#(a(a(a(x1)))) → b#(a(b(x1))) (12)
a#(a(a(a(x1)))) → a#(b(x1)) (13)
a#(a(a(a(x1)))) → b#(x1) (14)
a#(b(b(b(x1)))) → a#(a(b(b(x1)))) (15)
a#(b(b(b(x1)))) → a#(b(b(x1))) (16)

1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[b#(x1)] = 1 + 1 · x1
[b(x1)] = 1 + 1 · x1
[a(x1)] = 1 + 1 · x1
[a#(x1)] = 1 + 1 · x1
the pairs
b#(b(a(b(x1)))) → b#(a(a(x1))) (8)
b#(b(a(b(x1)))) → a#(a(x1)) (9)
b#(b(a(b(x1)))) → a#(x1) (10)
a#(a(a(a(x1)))) → b#(a(b(x1))) (12)
a#(a(a(a(x1)))) → a#(b(x1)) (13)
a#(a(a(a(x1)))) → b#(x1) (14)
a#(b(b(b(x1)))) → a#(b(b(x1))) (16)
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)] =
0
-∞
-∞
+
-∞ 1 0
-∞ -∞ 0
0 0 1
· x1
[a(x1)] =
0
0
0
+
-∞ -∞ 0
-∞ 1 0
0 0 -∞
· x1
[a#(x1)] =
0
-∞
-∞
+
-∞ 0 -∞
-∞ -∞ -∞
-∞ -∞ -∞
· x1
the pair
b#(b(a(b(x1)))) → a#(b(a(a(x1)))) (7)
could be deleted.

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.