YES

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

The problem is finite.