Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-545)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Closure Under Flat Contexts

Using the flat contexts

{a(☐), b(☐)}

We obtain the transformed TRS
a(b(a(a(x1)))) → a(a(b(a(x1)))) (1)
b(a(a(b(x1)))) → b(b(b(a(x1)))) (3)
a(b(b(a(a(x1))))) → a(a(a(a(b(x1))))) (4)
b(b(b(a(a(x1))))) → b(a(a(a(b(x1))))) (5)

1.1 Semantic Labeling

Root-labeling is applied.

We obtain the labeled TRS
ab(ba(aa(aa(x1)))) → aa(ab(ba(aa(x1)))) (6)
ab(ba(aa(ab(x1)))) → aa(ab(ba(ab(x1)))) (7)
ba(aa(ab(ba(x1)))) → bb(bb(ba(aa(x1)))) (8)
ba(aa(ab(bb(x1)))) → bb(bb(ba(ab(x1)))) (9)
ab(bb(ba(aa(aa(x1))))) → aa(aa(aa(ab(ba(x1))))) (10)
ab(bb(ba(aa(ab(x1))))) → aa(aa(aa(ab(bb(x1))))) (11)
bb(bb(ba(aa(aa(x1))))) → ba(aa(aa(ab(ba(x1))))) (12)
bb(bb(ba(aa(ab(x1))))) → ba(aa(aa(ab(bb(x1))))) (13)

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
ab#(ba(aa(aa(x1)))) → ab#(ba(aa(x1))) (14)
ab#(ba(aa(aa(x1)))) → ba#(aa(x1)) (15)
ab#(ba(aa(ab(x1)))) → ab#(ba(ab(x1))) (16)
ab#(ba(aa(ab(x1)))) → ba#(ab(x1)) (17)
ba#(aa(ab(ba(x1)))) → bb#(bb(ba(aa(x1)))) (18)
ba#(aa(ab(ba(x1)))) → bb#(ba(aa(x1))) (19)
ba#(aa(ab(ba(x1)))) → ba#(aa(x1)) (20)
ba#(aa(ab(bb(x1)))) → bb#(bb(ba(ab(x1)))) (21)
ba#(aa(ab(bb(x1)))) → bb#(ba(ab(x1))) (22)
ba#(aa(ab(bb(x1)))) → ba#(ab(x1)) (23)
ba#(aa(ab(bb(x1)))) → ab#(x1) (24)
ab#(bb(ba(aa(aa(x1))))) → ab#(ba(x1)) (25)
ab#(bb(ba(aa(aa(x1))))) → ba#(x1) (26)
ab#(bb(ba(aa(ab(x1))))) → ab#(bb(x1)) (27)
ab#(bb(ba(aa(ab(x1))))) → bb#(x1) (28)
bb#(bb(ba(aa(aa(x1))))) → ba#(aa(aa(ab(ba(x1))))) (29)
bb#(bb(ba(aa(aa(x1))))) → ab#(ba(x1)) (30)
bb#(bb(ba(aa(aa(x1))))) → ba#(x1) (31)
bb#(bb(ba(aa(ab(x1))))) → ba#(aa(aa(ab(bb(x1))))) (32)
bb#(bb(ba(aa(ab(x1))))) → ab#(bb(x1)) (33)
bb#(bb(ba(aa(ab(x1))))) → bb#(x1) (34)

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.