YES

Problem 1: 

(VAR v_NonEmpty:S X:S)
(RULES
a__f(f(a)) -> a__f(g(f(a)))
a__f(X:S) -> f(X:S)
mark(a) -> a
mark(f(X:S)) -> a__f(mark(X:S))
mark(g(X:S)) -> g(X:S)
) 
(STRATEGY INNERMOST)

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 A__F(f(a)) -> A__F(g(f(a)))
 MARK(f(X:S)) -> A__F(mark(X:S))
 MARK(f(X:S)) -> MARK(X:S)
-> Rules:
 a__f(f(a)) -> a__f(g(f(a)))
 a__f(X:S) -> f(X:S)
 mark(a) -> a
 mark(f(X:S)) -> a__f(mark(X:S))
 mark(g(X:S)) -> g(X:S)

Problem 1: 

SCC Processor:
-> Pairs:
 A__F(f(a)) -> A__F(g(f(a)))
 MARK(f(X:S)) -> A__F(mark(X:S))
 MARK(f(X:S)) -> MARK(X:S)
-> Rules:
 a__f(f(a)) -> a__f(g(f(a)))
 a__f(X:S) -> f(X:S)
 mark(a) -> a
 mark(f(X:S)) -> a__f(mark(X:S))
 mark(g(X:S)) -> g(X:S)
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 MARK(f(X:S)) -> MARK(X:S)
->->-> Rules:
 a__f(f(a)) -> a__f(g(f(a)))
 a__f(X:S) -> f(X:S)
 mark(a) -> a
 mark(f(X:S)) -> a__f(mark(X:S))
 mark(g(X:S)) -> g(X:S)

Problem 1: 

Subterm Processor:
-> Pairs:
 MARK(f(X:S)) -> MARK(X:S)
-> Rules:
 a__f(f(a)) -> a__f(g(f(a)))
 a__f(X:S) -> f(X:S)
 mark(a) -> a
 mark(f(X:S)) -> a__f(mark(X:S))
 mark(g(X:S)) -> g(X:S)
->Projection:
 pi(MARK) = 1

Problem 1: 

SCC Processor:
-> Pairs:
 Empty
-> Rules:
 a__f(f(a)) -> a__f(g(f(a)))
 a__f(X:S) -> f(X:S)
 mark(a) -> a
 mark(f(X:S)) -> a__f(mark(X:S))
 mark(g(X:S)) -> g(X:S)
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.