Certification Problem

Input (TPDB SRS_Standard/Bouchare_06/10)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
b(b(b(x1))) → b(a(a(x1))) (4)
a(a(b(x1))) → b(a(a(x1))) (5)
a(a(x1)) → b(a(b(x1))) (3)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
b#(b(b(x1))) → b#(a(a(x1))) (6)
b#(b(b(x1))) → a#(x1) (7)
b#(b(b(x1))) → a#(a(x1)) (8)
a#(a(x1)) → b#(x1) (9)
a#(a(x1)) → b#(a(b(x1))) (10)
a#(a(x1)) → a#(b(x1)) (11)
a#(a(b(x1))) → b#(a(a(x1))) (12)
a#(a(b(x1))) → a#(x1) (13)
a#(a(b(x1))) → a#(a(x1)) (14)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.