Certification Problem

Input (TPDB SRS_Standard/Secret_06_SRS/multum4)

The rewrite relation of the following TRS is considered.

b(b(a(b(b(a(b(b(b(b(b(b(a(b(x1)))))))))))))) b(b(b(b(b(a(b(b(a(b(b(a(b(b(b(b(b(b(x1)))))))))))))))))) (1)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Semantic Labeling

Root-labeling is applied.

We obtain the labeled TRS
bb(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(ab(bb(x1)))))))))))))) bb(bb(bb(bb(ba(ab(bb(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(bb(x1)))))))))))))))))) (2)
bb(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(ab(ba(x1)))))))))))))) bb(bb(bb(bb(ba(ab(bb(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(x1)))))))))))))))))) (3)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
bb#(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(ab(bb(x1)))))))))))))) bb#(bb(bb(bb(ba(ab(bb(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(bb(x1)))))))))))))))))) (4)
bb#(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(ab(bb(x1)))))))))))))) bb#(bb(bb(ba(ab(bb(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(bb(x1))))))))))))))))) (5)
bb#(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(ab(bb(x1)))))))))))))) bb#(bb(ba(ab(bb(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(bb(x1)))))))))))))))) (6)
bb#(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(ab(bb(x1)))))))))))))) bb#(ba(ab(bb(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(bb(x1))))))))))))))) (7)
bb#(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(ab(bb(x1)))))))))))))) bb#(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(bb(x1)))))))))))) (8)
bb#(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(ab(bb(x1)))))))))))))) bb#(ba(ab(bb(bb(bb(bb(bb(bb(x1))))))))) (9)
bb#(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(ab(bb(x1)))))))))))))) bb#(bb(bb(bb(bb(bb(x1)))))) (10)
bb#(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(ab(bb(x1)))))))))))))) bb#(bb(bb(bb(bb(x1))))) (11)
bb#(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(ab(bb(x1)))))))))))))) bb#(bb(bb(bb(x1)))) (12)
bb#(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(ab(bb(x1)))))))))))))) bb#(bb(bb(x1))) (13)
bb#(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(ab(bb(x1)))))))))))))) bb#(bb(x1)) (14)
bb#(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(ab(ba(x1)))))))))))))) bb#(bb(bb(bb(ba(ab(bb(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(x1)))))))))))))))))) (15)
bb#(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(ab(ba(x1)))))))))))))) bb#(bb(bb(ba(ab(bb(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(x1))))))))))))))))) (16)
bb#(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(ab(ba(x1)))))))))))))) bb#(bb(ba(ab(bb(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(x1)))))))))))))))) (17)
bb#(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(ab(ba(x1)))))))))))))) bb#(ba(ab(bb(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(x1))))))))))))))) (18)
bb#(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(ab(ba(x1)))))))))))))) bb#(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(x1)))))))))))) (19)
bb#(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(ab(ba(x1)))))))))))))) bb#(ba(ab(bb(bb(bb(bb(bb(ba(x1))))))))) (20)
bb#(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(ab(ba(x1)))))))))))))) bb#(bb(bb(bb(bb(ba(x1)))))) (21)
bb#(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(ab(ba(x1)))))))))))))) bb#(bb(bb(bb(ba(x1))))) (22)
bb#(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(ab(ba(x1)))))))))))))) bb#(bb(bb(ba(x1)))) (23)
bb#(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(ab(ba(x1)))))))))))))) bb#(bb(ba(x1))) (24)
bb#(ba(ab(bb(ba(ab(bb(bb(bb(bb(bb(ba(ab(ba(x1)))))))))))))) bb#(ba(x1)) (25)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.