Certification Problem

Input (TPDB SRS_Standard/Secret_05_SRS/jambox3)

The rewrite relation of the following TRS is considered.

a(a(c(x1))) → b(c(b(a(x1)))) (1)
b(x1) → d(a(x1)) (2)
b(a(c(d(x1)))) → a(a(a(x1))) (3)
c(x1) → x1 (4)
b(x1) → c(d(x1)) (5)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
c(a(a(x1))) → a(b(c(b(x1)))) (6)
b(x1) → a(d(x1)) (7)
d(c(a(b(x1)))) → a(a(a(x1))) (8)
c(x1) → x1 (4)
b(x1) → d(c(x1)) (9)

1.1 Dependency Pair Transformation

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

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.