Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-320)

The rewrite relation of the following TRS is considered.

a(a(a(a(x1)))) → a(b(b(b(x1)))) (1)
a(b(b(b(x1)))) → b(a(b(a(x1)))) (2)
b(a(a(b(x1)))) → b(a(b(b(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(a(a(a(x1)))) → a(b(b(b(x1)))) (1)
b(a(a(b(x1)))) → b(a(b(b(x1)))) (3)
a(a(b(b(b(x1))))) → a(b(a(b(a(x1))))) (4)
b(a(b(b(b(x1))))) → b(b(a(b(a(x1))))) (5)

1.1 Semantic Labeling

Root-labeling is applied.

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

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[aa(x1)] = 1 · x1 + 1
[ab(x1)] = 1 · x1
[bb(x1)] = 1 · x1 + 1
[ba(x1)] = 1 · x1 + 1
all of the following rules can be deleted.
aa(aa(aa(aa(x1)))) → ab(bb(bb(ba(x1)))) (6)
aa(ab(bb(bb(ba(x1))))) → ab(ba(ab(ba(aa(x1))))) (10)
aa(ab(bb(bb(bb(x1))))) → ab(ba(ab(ba(ab(x1))))) (11)
ba(ab(bb(bb(bb(x1))))) → bb(ba(ab(ba(ab(x1))))) (13)

1.1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
ba#(aa(ab(ba(x1)))) → ba#(ab(bb(ba(x1)))) (14)
ba#(aa(ab(bb(x1)))) → ba#(ab(bb(bb(x1)))) (15)
ba#(ab(bb(bb(ba(x1))))) → ba#(ab(ba(aa(x1)))) (16)
ba#(ab(bb(bb(ba(x1))))) → ba#(aa(x1)) (17)
ba#(ab(bb(bb(ba(x1))))) → aa#(x1) (18)

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.