Certification Problem

Input (TPDB SRS_Standard/Secret_06_SRS/multum2)

The rewrite relation of the following TRS is considered.

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

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

1.1 Dependency Pair Transformation

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

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.