Certification Problem

Input (TPDB SRS_Standard/Trafo_06/un14)

The rewrite relation of the following TRS is considered.

b(a(b(x1))) → b(a(a(a(b(x1))))) (1)
b(a(a(a(b(a(a(b(x1)))))))) → b(a(a(b(a(a(b(a(a(a(b(b(x1)))))))))))) (2)
b(a(a(a(b(a(a(a(b(x1))))))))) → b(x1) (3)
b(a(a(a(b(b(b(x1))))))) → b(b(b(a(a(a(b(x1))))))) (4)
b(a(a(b(b(x1))))) → b(x1) (5)
b(b(a(a(b(x1))))) → b(x1) (6)
b(a(a(a(b(a(b(x1))))))) → b(x1) (7)
b(a(b(a(a(a(b(x1))))))) → b(x1) (8)

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(a(b(x1))) → b(a(a(a(b(x1))))) (1)
b(a(a(b(a(a(a(b(x1)))))))) → b(b(a(a(a(b(a(a(b(a(a(b(x1)))))))))))) (9)
b(a(a(a(b(a(a(a(b(x1))))))))) → b(x1) (3)
b(b(b(a(a(a(b(x1))))))) → b(a(a(a(b(b(b(x1))))))) (10)
b(b(a(a(b(x1))))) → b(x1) (6)
b(a(a(b(b(x1))))) → b(x1) (5)
b(a(b(a(a(a(b(x1))))))) → b(x1) (8)
b(a(a(a(b(a(b(x1))))))) → b(x1) (7)

1.1 Dependency Pair Transformation

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

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.