YES

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

The problem is finite.