Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-529)

The rewrite relation of the following TRS is considered.

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

1.1 Semantic Labeling

Root-labeling is applied.

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

1.1.1 Dependency Pair Transformation

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

1.1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[bb#(x1)] = 1 · x1
[ba(x1)] = 1 + 1 · x1
[aa(x1)] = 1 + 1 · x1
[ab#(x1)] = 1 · x1
[ab(x1)] = 1 + 1 · x1
[bb(x1)] = 1 + 1 · x1
the pairs
bb#(ba(aa(aa(x1)))) ab#(ba(aa(x1))) (14)
bb#(ba(aa(ab(x1)))) ab#(ba(ab(x1))) (15)
ab#(ba(aa(aa(aa(x1))))) ab#(bb(ba(x1))) (16)
ab#(ba(aa(aa(aa(x1))))) bb#(ba(x1)) (17)
ab#(ba(aa(aa(ab(x1))))) ab#(bb(bb(x1))) (18)
ab#(ba(aa(aa(ab(x1))))) bb#(bb(x1)) (19)
ab#(ba(aa(aa(ab(x1))))) bb#(x1) (20)
bb#(ba(aa(aa(aa(x1))))) ab#(bb(ba(x1))) (21)
bb#(ba(aa(aa(aa(x1))))) bb#(ba(x1)) (22)
bb#(ba(aa(aa(ab(x1))))) ab#(bb(bb(x1))) (23)
bb#(ba(aa(aa(ab(x1))))) bb#(bb(x1)) (24)
bb#(ba(aa(aa(ab(x1))))) bb#(x1) (25)
could be deleted.

1.1.1.1.1 P is empty

There are no pairs anymore.