YES

Problem 1: 

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

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 F(a,f(b,f(a,x:S))) -> F(a,f(b,f(b,f(a,x:S))))
 F(a,f(b,f(a,x:S))) -> F(b,f(b,f(a,x:S)))
-> Rules:
 f(a,f(b,f(a,x:S))) -> f(a,f(b,f(b,f(a,x:S))))
 f(b,f(b,f(b,x:S))) -> f(b,f(b,x:S))

Problem 1: 

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

Problem 1: 

Reduction Pair Processor:
-> Pairs:
 F(a,f(b,f(a,x:S))) -> F(a,f(b,f(b,f(a,x:S))))
-> Rules:
 f(a,f(b,f(a,x:S))) -> f(a,f(b,f(b,f(a,x:S))))
 f(b,f(b,f(b,x:S))) -> f(b,f(b,x:S))
-> Usable rules:
 f(a,f(b,f(a,x:S))) -> f(a,f(b,f(b,f(a,x:S))))
 f(b,f(b,f(b,x:S))) -> f(b,f(b,x:S))
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 2
->Bound:
 1
->Interpretation:
 
[f](X1,X2) = [0 1;1 1].X1 + [0 1;0 0].X2
[a] = [1;1]
[b] = 0
[F](X1,X2) = [1 0;1 1].X2

Problem 1: 

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

The problem is finite.