YES

Problem 1: 

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

Problem 1: 

Innermost Equivalent Processor:
-> Rules:
 a(b(a(a(b(b(x1:S)))))) -> a(a(b(a(b(b(a(x1:S)))))))
-> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination.
 

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

The problem is finite.