Certification Problem

Input (TPDB SRS_Standard/Mixed_SRS/04)

The rewrite relation of the following TRS is considered.

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

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(a(a(x1))) → a(b(b(b(x1)))) (1)
a(b(b(a(b(x1))))) → a(a(a(a(x1)))) (3)
b(b(b(a(b(x1))))) → b(a(a(a(x1)))) (4)

1.1 Semantic Labeling

Root-labeling is applied.

We obtain the labeled TRS
aa(aa(aa(x1))) → ab(bb(bb(ba(x1)))) (5)
aa(aa(ab(x1))) → ab(bb(bb(bb(x1)))) (6)
ab(bb(ba(ab(ba(x1))))) → aa(aa(aa(aa(x1)))) (7)
ab(bb(ba(ab(bb(x1))))) → aa(aa(aa(ab(x1)))) (8)
bb(bb(ba(ab(ba(x1))))) → ba(aa(aa(aa(x1)))) (9)
bb(bb(ba(ab(bb(x1))))) → ba(aa(aa(ab(x1)))) (10)

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
aa#(aa(aa(x1))) → ab#(bb(bb(ba(x1)))) (11)
aa#(aa(aa(x1))) → bb#(bb(ba(x1))) (12)
aa#(aa(aa(x1))) → bb#(ba(x1)) (13)
aa#(aa(ab(x1))) → ab#(bb(bb(bb(x1)))) (14)
aa#(aa(ab(x1))) → bb#(bb(bb(x1))) (15)
aa#(aa(ab(x1))) → bb#(bb(x1)) (16)
aa#(aa(ab(x1))) → bb#(x1) (17)
ab#(bb(ba(ab(ba(x1))))) → aa#(aa(aa(aa(x1)))) (18)
ab#(bb(ba(ab(ba(x1))))) → aa#(aa(aa(x1))) (19)
ab#(bb(ba(ab(ba(x1))))) → aa#(aa(x1)) (20)
ab#(bb(ba(ab(ba(x1))))) → aa#(x1) (21)
ab#(bb(ba(ab(bb(x1))))) → aa#(aa(aa(ab(x1)))) (22)
ab#(bb(ba(ab(bb(x1))))) → aa#(aa(ab(x1))) (23)
ab#(bb(ba(ab(bb(x1))))) → aa#(ab(x1)) (24)
ab#(bb(ba(ab(bb(x1))))) → ab#(x1) (25)
bb#(bb(ba(ab(ba(x1))))) → aa#(aa(aa(x1))) (26)
bb#(bb(ba(ab(ba(x1))))) → aa#(aa(x1)) (27)
bb#(bb(ba(ab(ba(x1))))) → aa#(x1) (28)
bb#(bb(ba(ab(bb(x1))))) → aa#(aa(ab(x1))) (29)
bb#(bb(ba(ab(bb(x1))))) → aa#(ab(x1)) (30)
bb#(bb(ba(ab(bb(x1))))) → ab#(x1) (31)

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.