Certification Problem

Input (TPDB SRS_Standard/Zantema_06/16)

The rewrite relation of the following TRS is considered.

a(x1) → b(x1) (1)
a(a(x1)) → a(b(a(x1))) (2)
a(b(x1)) → b(b(b(x1))) (3)
a(a(a(x1))) → a(a(b(a(a(x1))))) (4)
a(a(b(x1))) → a(b(b(a(b(x1))))) (5)
a(b(a(x1))) → b(a(b(b(a(x1))))) (6)
a(b(b(x1))) → b(b(b(b(b(x1))))) (7)
b(a(x1)) → b(b(b(x1))) (8)
a(b(a(x1))) → a(b(b(a(b(x1))))) (9)
b(a(a(x1))) → b(a(b(b(a(x1))))) (10)
b(b(a(x1))) → b(b(b(b(b(x1))))) (11)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by NaTT @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a#(a(x1)) → b#(a(x1)) (12)
b#(a(x1)) → b#(b(b(x1))) (13)
a#(b(a(x1))) → a#(b(b(a(b(x1))))) (14)
b#(b(a(x1))) → b#(b(b(x1))) (15)
a#(b(a(x1))) → a#(b(x1)) (16)
b#(b(a(x1))) → b#(b(b(b(b(x1))))) (17)
a#(a(b(x1))) → b#(b(a(b(x1)))) (18)
a#(a(b(x1))) → a#(b(b(a(b(x1))))) (19)
b#(b(a(x1))) → b#(b(b(b(x1)))) (20)
b#(a(x1)) → b#(x1) (21)
b#(b(a(x1))) → b#(b(x1)) (22)
b#(a(x1)) → b#(b(x1)) (23)
a#(a(b(x1))) → b#(a(b(x1))) (24)
b#(a(a(x1))) → a#(b(b(a(x1)))) (25)
a#(b(a(x1))) → b#(b(a(b(x1)))) (26)
a#(b(a(x1))) → b#(x1) (27)
a#(a(a(x1))) → a#(b(a(a(x1)))) (28)
a#(b(a(x1))) → b#(b(a(x1))) (29)
a#(x1) → b#(x1) (30)
b#(a(a(x1))) → b#(a(x1)) (31)
a#(a(a(x1))) → b#(a(a(x1))) (32)
a#(b(x1)) → b#(b(x1)) (33)
a#(b(b(x1))) → b#(b(b(b(x1)))) (34)
a#(a(a(x1))) → a#(a(b(a(a(x1))))) (35)
b#(a(a(x1))) → b#(a(b(b(a(x1))))) (36)
a#(b(x1)) → b#(b(b(x1))) (37)
b#(a(a(x1))) → b#(b(a(x1))) (38)
a#(b(b(x1))) → b#(b(b(b(b(x1))))) (39)
a#(b(a(x1))) → b#(a(b(b(a(x1))))) (40)
a#(a(x1)) → a#(b(a(x1))) (41)
a#(b(a(x1))) → b#(a(b(x1))) (42)
b#(b(a(x1))) → b#(x1) (43)
a#(b(a(x1))) → a#(b(b(a(x1)))) (44)
a#(b(b(x1))) → b#(b(b(x1))) (45)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.