Certification Problem

Input (TPDB SRS_Standard/Trafo_06/un08)

The rewrite relation of the following TRS is considered.

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

1.1 Semantic Labeling

Root-labeling is applied.

We obtain the labeled TRS
ba(ab(bb(bb(x1)))) → bb(bb(bb(ba(ab(bb(x1)))))) (7)
ba(ab(bb(ba(x1)))) → bb(bb(bb(ba(ab(ba(x1)))))) (8)
bb(ba(ab(bb(x1)))) → bb(bb(bb(x1))) (9)
bb(ba(ab(ba(x1)))) → bb(bb(ba(x1))) (10)
bb(bb(bb(x1))) → ba(ab(ba(aa(ab(bb(bb(x1))))))) (11)
bb(bb(ba(x1))) → ba(ab(ba(aa(ab(bb(ba(x1))))))) (12)

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
ba#(ab(bb(bb(x1)))) → bb#(bb(bb(ba(ab(bb(x1)))))) (13)
ba#(ab(bb(bb(x1)))) → bb#(bb(ba(ab(bb(x1))))) (14)
ba#(ab(bb(bb(x1)))) → bb#(ba(ab(bb(x1)))) (15)
ba#(ab(bb(bb(x1)))) → ba#(ab(bb(x1))) (16)
ba#(ab(bb(ba(x1)))) → bb#(bb(bb(ba(ab(ba(x1)))))) (17)
ba#(ab(bb(ba(x1)))) → bb#(bb(ba(ab(ba(x1))))) (18)
ba#(ab(bb(ba(x1)))) → bb#(ba(ab(ba(x1)))) (19)
ba#(ab(bb(ba(x1)))) → ba#(ab(ba(x1))) (20)
bb#(ba(ab(bb(x1)))) → bb#(bb(bb(x1))) (21)
bb#(ba(ab(bb(x1)))) → bb#(bb(x1)) (22)
bb#(ba(ab(ba(x1)))) → bb#(bb(ba(x1))) (23)
bb#(ba(ab(ba(x1)))) → bb#(ba(x1)) (24)
bb#(bb(bb(x1))) → ba#(ab(ba(aa(ab(bb(bb(x1))))))) (25)
bb#(bb(bb(x1))) → ba#(aa(ab(bb(bb(x1))))) (26)
bb#(bb(ba(x1))) → ba#(ab(ba(aa(ab(bb(ba(x1))))))) (27)
bb#(bb(ba(x1))) → ba#(aa(ab(bb(ba(x1))))) (28)

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.