YES

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

Problem 1: 

Reduction Pair Processor:
-> Pairs:
 A(a(x1:S)) -> A(b(a(x1:S)))
 A(a(x1:S)) -> B(a(x1:S))
 B(a(b(x1:S))) -> A(x1:S)
-> Rules:
 a(a(x1:S)) -> a(b(a(x1:S)))
 b(a(b(x1:S))) -> a(c(a(x1:S)))
-> Usable rules:
 a(a(x1:S)) -> a(b(a(x1:S)))
 b(a(b(x1:S))) -> a(c(a(x1:S)))
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[a](X) = 2.X + 2
[b](X) = X
[c](X) = 0
[A](X) = 2.X + 2
[B](X) = 2.X

Problem 1: 

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

Problem 1: 

Reduction Pair Processor:
-> Pairs:
 A(a(x1:S)) -> A(b(a(x1:S)))
-> Rules:
 a(a(x1:S)) -> a(b(a(x1:S)))
 b(a(b(x1:S))) -> a(c(a(x1:S)))
-> Usable rules:
 a(a(x1:S)) -> a(b(a(x1:S)))
 b(a(b(x1:S))) -> a(c(a(x1:S)))
->Interpretation type:
 Linear
->Coefficients:
 All rationals
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[a](X) = 2.X + 2
[b](X) = 1/2.X + 1/2
[c](X) = 0
[A](X) = 2.X

Problem 1: 

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

The problem is finite.