YES

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

The problem is finite.