Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-268)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a#(b(b(b(x1)))) b#(a(b(a(x1)))) (4)
a#(b(b(b(x1)))) a#(b(a(x1))) (5)
a#(b(b(b(x1)))) b#(a(x1)) (6)
a#(b(b(b(x1)))) a#(x1) (7)
b#(a(b(a(x1)))) b#(b(b(b(x1)))) (8)
b#(a(b(a(x1)))) b#(b(b(x1))) (9)
b#(a(b(a(x1)))) b#(b(x1)) (10)
b#(a(b(a(x1)))) b#(x1) (11)
a#(a(a(b(x1)))) a#(b(a(a(x1)))) (12)
a#(a(a(b(x1)))) b#(a(a(x1))) (13)
a#(a(a(b(x1)))) a#(a(x1)) (14)
a#(a(a(b(x1)))) a#(x1) (15)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.