YES

Problem 1: 

(VAR v_NonEmpty:S x1:S)
(RULES
a(a(x1:S)) -> a(b(c(a(x1:S))))
a(x1:S) -> x1:S
c(b(x1:S)) -> a(b(a(x1:S)))
)

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 A(a(x1:S)) -> A(b(c(a(x1:S))))
 A(a(x1:S)) -> C(a(x1:S))
 C(b(x1:S)) -> A(b(a(x1:S)))
 C(b(x1:S)) -> A(x1:S)
-> Rules:
 a(a(x1:S)) -> a(b(c(a(x1:S))))
 a(x1:S) -> x1:S
 c(b(x1:S)) -> a(b(a(x1:S)))

Problem 1: 

SCC Processor:
-> Pairs:
 A(a(x1:S)) -> A(b(c(a(x1:S))))
 A(a(x1:S)) -> C(a(x1:S))
 C(b(x1:S)) -> A(b(a(x1:S)))
 C(b(x1:S)) -> A(x1:S)
-> Rules:
 a(a(x1:S)) -> a(b(c(a(x1:S))))
 a(x1:S) -> x1:S
 c(b(x1:S)) -> a(b(a(x1:S)))
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 A(a(x1:S)) -> C(a(x1:S))
 C(b(x1:S)) -> A(x1:S)
->->-> Rules:
 a(a(x1:S)) -> a(b(c(a(x1:S))))
 a(x1:S) -> x1:S
 c(b(x1:S)) -> a(b(a(x1:S)))

Problem 1: 

Subterm Processor:
-> Pairs:
 A(a(x1:S)) -> C(a(x1:S))
 C(b(x1:S)) -> A(x1:S)
-> Rules:
 a(a(x1:S)) -> a(b(c(a(x1:S))))
 a(x1:S) -> x1:S
 c(b(x1:S)) -> a(b(a(x1:S)))
->Projection:
 pi(A) = 1
 pi(C) = 1

Problem 1: 

SCC Processor:
-> Pairs:
 A(a(x1:S)) -> C(a(x1:S))
-> Rules:
 a(a(x1:S)) -> a(b(c(a(x1:S))))
 a(x1:S) -> x1:S
 c(b(x1:S)) -> a(b(a(x1:S)))
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.