Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-516)

The rewrite relation of the following TRS is considered.

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

{b(☐), a(☐)}

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

1.1 Semantic Labeling

Root-labeling is applied.

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

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[bb(x1)] = 1 · x1
[ba(x1)] = 1 · x1
[ab(x1)] = 1 · x1 + 1
[aa(x1)] = 1 · x1 + 1
all of the following rules can be deleted.
ba(aa(aa(ab(x1)))) → bb(bb(ba(ab(x1)))) (8)
ba(aa(aa(aa(x1)))) → bb(bb(ba(aa(x1)))) (9)

1.1.1.1 Dependency Pair Transformation

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

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.