Certification Problem

Input (TPDB SRS_Standard/Secret_07_SRS/x09)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
b(b(a(a(a(x1))))) → a(a(a(b(b(a(b(b(x1)))))))) (5)
a(b(b(x1))) → x1 (6)
a(x1) → b(b(b(x1))) (3)
a(x1) → b(x1) (4)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
b#(b(a(a(a(x1))))) → b#(x1) (7)
b#(b(a(a(a(x1))))) → b#(b(x1)) (8)
b#(b(a(a(a(x1))))) → b#(b(a(b(b(x1))))) (9)
b#(b(a(a(a(x1))))) → b#(a(b(b(x1)))) (10)
b#(b(a(a(a(x1))))) → a#(b(b(x1))) (11)
b#(b(a(a(a(x1))))) → a#(b(b(a(b(b(x1)))))) (12)
b#(b(a(a(a(x1))))) → a#(a(b(b(a(b(b(x1))))))) (13)
b#(b(a(a(a(x1))))) → a#(a(a(b(b(a(b(b(x1)))))))) (14)
a#(x1) → b#(x1) (15)
a#(x1) → b#(b(x1)) (16)
a#(x1) → b#(b(b(x1))) (17)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.