Certification Problem

Input (TPDB SRS_Standard/Secret_05_SRS/jambox4)

The rewrite relation of the following TRS is considered.

c(c(x1)) → a(b(x1)) (1)
b(x1) → a(a(x1)) (2)
b(b(b(x1))) → a(c(b(x1))) (3)
a(c(a(x1))) → a(c(c(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
c(c(x1)) → b(a(x1)) (5)
b(x1) → a(a(x1)) (2)
b(b(b(x1))) → b(c(a(x1))) (6)
a(c(a(x1))) → c(c(a(x1))) (7)

1.1 Dependency Pair Transformation

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

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.