Certification Problem

Input (TPDB SRS_Standard/Secret_05_SRS/jambox1)

The rewrite relation of the following TRS is considered.

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

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(a(x1)) → a(x1) (4)
c(b(c(a(x1)))) → c(c(b(c(x1)))) (5)
c(x1) → a(a(b(x1))) (6)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
c#(b(c(a(x1)))) → c#(x1) (7)
c#(b(c(a(x1)))) → c#(c(b(c(x1)))) (8)
c#(b(c(a(x1)))) → c#(b(c(x1))) (9)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.