Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-466)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
a(b(a(a(x1)))) a(b(b(a(x1)))) (4)
b(b(a(b(x1)))) b(a(b(b(x1)))) (5)
a(b(a(b(x1)))) a(a(b(b(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(ba(ab(ba(x1)))) ba(ab(bb(ba(x1)))) (9)
bb(ba(ab(bb(x1)))) ba(ab(bb(bb(x1)))) (10)
ab(ba(ab(ba(x1)))) aa(ab(bb(ba(x1)))) (11)
ab(ba(ab(bb(x1)))) aa(ab(bb(bb(x1)))) (12)

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)

1.1.1.1 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
all of the following rules can be deleted.
ab(ba(ab(ba(x1)))) aa(ab(bb(ba(x1)))) (11)
ab(ba(ab(bb(x1)))) aa(ab(bb(bb(x1)))) (12)

1.1.1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
bb#(ba(ab(ba(x1)))) bb#(ba(x1)) (13)
bb#(ba(ab(bb(x1)))) bb#(bb(x1)) (14)

1.1.1.1.1.1 Size-Change Termination

Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.

bb#(ba(ab(ba(x1)))) bb#(ba(x1)) (13)
1 > 1
bb#(ba(ab(bb(x1)))) bb#(bb(x1)) (14)
1 > 1

As there is no critical graph in the transitive closure, there are no infinite chains.