Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-472)

The rewrite relation of the following TRS is considered.

b(a(a(a(x1)))) → a(b(a(a(x1)))) (1)
a(b(a(a(x1)))) → a(b(b(a(x1)))) (2)
b(b(b(a(x1)))) → b(b(a(b(x1)))) (3)
b(a(b(a(x1)))) → b(a(b(b(x1)))) (4)

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
a(b(a(a(x1)))) → a(b(b(a(x1)))) (2)
b(b(b(a(x1)))) → b(b(a(b(x1)))) (3)
b(a(b(a(x1)))) → b(a(b(b(x1)))) (4)
b(b(a(a(a(x1))))) → b(a(b(a(a(x1))))) (5)
a(b(a(a(a(x1))))) → a(a(b(a(a(x1))))) (6)

1.1 Semantic Labeling

Root-labeling is applied.

We obtain the labeled TRS
ab(ba(aa(aa(x1)))) → ab(bb(ba(aa(x1)))) (7)
ab(ba(aa(ab(x1)))) → ab(bb(ba(ab(x1)))) (8)
bb(bb(ba(aa(x1)))) → bb(ba(ab(ba(x1)))) (9)
bb(bb(ba(ab(x1)))) → bb(ba(ab(bb(x1)))) (10)
ba(ab(ba(aa(x1)))) → ba(ab(bb(ba(x1)))) (11)
ba(ab(ba(ab(x1)))) → ba(ab(bb(bb(x1)))) (12)
bb(ba(aa(aa(aa(x1))))) → ba(ab(ba(aa(aa(x1))))) (13)
bb(ba(aa(aa(ab(x1))))) → ba(ab(ba(aa(ab(x1))))) (14)
ab(ba(aa(aa(aa(x1))))) → aa(ab(ba(aa(aa(x1))))) (15)
ab(ba(aa(aa(ab(x1))))) → aa(ab(ba(aa(ab(x1))))) (16)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[ab(x1)] = 1 · x1
[ba(x1)] = 1 · x1 + 1
[aa(x1)] = 1 · x1 + 1
[bb(x1)] = 1 · x1
all of the following rules can be deleted.
ab(ba(aa(aa(x1)))) → ab(bb(ba(aa(x1)))) (7)
ab(ba(aa(ab(x1)))) → ab(bb(ba(ab(x1)))) (8)
ba(ab(ba(aa(x1)))) → ba(ab(bb(ba(x1)))) (11)
ba(ab(ba(ab(x1)))) → ba(ab(bb(bb(x1)))) (12)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[bb(x1)] = 1 · x1
[ba(x1)] = 1 · x1
[aa(x1)] = 1 · x1 + 1
[ab(x1)] = 1 · x1
all of the following rules can be deleted.
bb(bb(ba(aa(x1)))) → bb(ba(ab(ba(x1)))) (9)
bb(ba(aa(aa(aa(x1))))) → ba(ab(ba(aa(aa(x1))))) (13)
bb(ba(aa(aa(ab(x1))))) → ba(ab(ba(aa(ab(x1))))) (14)

1.1.1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
bb#(bb(ba(ab(x1)))) → bb#(ba(ab(bb(x1)))) (17)
bb#(bb(ba(ab(x1)))) → ab#(bb(x1)) (18)
bb#(bb(ba(ab(x1)))) → bb#(x1) (19)
ab#(ba(aa(aa(aa(x1))))) → ab#(ba(aa(aa(x1)))) (20)
ab#(ba(aa(aa(ab(x1))))) → ab#(ba(aa(ab(x1)))) (21)

1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.