YES

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

The problem is finite.