Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-55)

The rewrite relation of the following TRS is considered.

b(b(b(a(x1)))) → b(a(a(a(x1)))) (1)
b(b(b(b(x1)))) → b(a(a(b(x1)))) (2)
a(a(b(a(x1)))) → b(a(b(a(x1)))) (3)
b(a(a(b(x1)))) → a(a(a(b(x1)))) (4)

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(b(b(b(x1)))) → a(a(a(b(x1)))) (5)
b(b(b(b(x1)))) → b(a(a(b(x1)))) (2)
a(b(a(a(x1)))) → a(b(a(b(x1)))) (6)
b(a(a(b(x1)))) → b(a(a(a(x1)))) (7)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a#(b(b(b(x1)))) → a#(a(a(b(x1)))) (8)
a#(b(b(b(x1)))) → a#(a(b(x1))) (9)
a#(b(b(b(x1)))) → a#(b(x1)) (10)
b#(b(b(b(x1)))) → b#(a(a(b(x1)))) (11)
b#(b(b(b(x1)))) → a#(a(b(x1))) (12)
b#(b(b(b(x1)))) → a#(b(x1)) (13)
a#(b(a(a(x1)))) → a#(b(a(b(x1)))) (14)
a#(b(a(a(x1)))) → b#(a(b(x1))) (15)
a#(b(a(a(x1)))) → a#(b(x1)) (16)
a#(b(a(a(x1)))) → b#(x1) (17)
b#(a(a(b(x1)))) → b#(a(a(a(x1)))) (18)
b#(a(a(b(x1)))) → a#(a(a(x1))) (19)
b#(a(a(b(x1)))) → a#(a(x1)) (20)
b#(a(a(b(x1)))) → a#(x1) (21)

1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[a#(x1)] = 1 · x1
[b(x1)] = 1 + 1 · x1
[a(x1)] = 1 + 1 · x1
[b#(x1)] = 1 · x1
the pairs
a#(b(b(b(x1)))) → a#(a(b(x1))) (9)
a#(b(b(b(x1)))) → a#(b(x1)) (10)
b#(b(b(b(x1)))) → a#(a(b(x1))) (12)
b#(b(b(b(x1)))) → a#(b(x1)) (13)
a#(b(a(a(x1)))) → b#(a(b(x1))) (15)
a#(b(a(a(x1)))) → a#(b(x1)) (16)
a#(b(a(a(x1)))) → b#(x1) (17)
b#(a(a(b(x1)))) → a#(a(a(x1))) (19)
b#(a(a(b(x1)))) → a#(a(x1)) (20)
b#(a(a(b(x1)))) → a#(x1) (21)
could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.