YES

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

Problem 1: 

Subterm Processor:
-> Pairs:
 F(f(a,f(x:S,a)),a) -> F(f(x:S,a),a)
-> Rules:
 f(f(a,f(x:S,a)),a) -> f(a,f(f(x:S,a),a))
->Projection:
 pi(F) = 1

Problem 1: 

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

The problem is finite.