YES

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

Problem 1: 

Reduction Pair Processor:
-> Pairs:
 A(c(x1:S)) -> A(a(x1:S))
 A(c(x1:S)) -> A(x1:S)
-> Rules:
 a(a(b(x1:S))) -> c(x1:S)
 a(c(x1:S)) -> b(c(a(a(x1:S))))
 b(c(x1:S)) -> x1:S
-> Usable rules:
 a(a(b(x1:S))) -> c(x1:S)
 a(c(x1:S)) -> b(c(a(a(x1:S))))
 b(c(x1:S)) -> x1:S
->Interpretation type:
 Linear
->Coefficients:
 All rationals
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[a](X) = 2.X
[b](X) = 1/2.X + 1/2
[c](X) = 2.X + 1
[A](X) = X

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

The problem is finite.