YES

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

Problem 1: 

Subterm Processor:
-> Pairs:
 F(a,f(f(a,x:S),a)) -> F(a,f(a,x:S))
-> Rules:
 f(a,f(f(a,x:S),a)) -> f(f(a,f(a,x:S)),a)
->Projection:
 pi(F) = 2

Problem 1: 

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

The problem is finite.