YES

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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


The problem is decomposed in 2 subproblems.

Problem 1.1: 

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

Problem 1.1: 

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

The problem is finite.

Problem 1.2: 

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

Problem 1.2: 

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

The problem is finite.