Certification Problem

Input (TPDB SRS_Standard/Secret_06_SRS/secr10)

The rewrite relation of the following TRS is considered.

a(a(x1)) c(b(a(b(a(x1))))) (1)
b(a(b(x1))) b(x1) (2)
a(a(a(x1))) c(c(a(x1))) (3)
c(c(x1)) a(b(c(b(a(x1))))) (4)
a(c(a(x1))) c(c(a(x1))) (5)
c(a(c(x1))) a(a(c(x1))) (6)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a#(a(x1)) c#(b(a(b(a(x1))))) (7)
a#(a(x1)) b#(a(b(a(x1)))) (8)
a#(a(x1)) a#(b(a(x1))) (9)
a#(a(x1)) b#(a(x1)) (10)
a#(a(a(x1))) c#(c(a(x1))) (11)
a#(a(a(x1))) c#(a(x1)) (12)
c#(c(x1)) a#(b(c(b(a(x1))))) (13)
c#(c(x1)) b#(c(b(a(x1)))) (14)
c#(c(x1)) c#(b(a(x1))) (15)
c#(c(x1)) b#(a(x1)) (16)
c#(c(x1)) a#(x1) (17)
a#(c(a(x1))) c#(c(a(x1))) (18)
c#(a(c(x1))) a#(a(c(x1))) (19)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.