YES

Problem 1: 

(VAR v_NonEmpty:S X:S Y:S)
(RULES
f(0,1,X:S) -> f(g(X:S,X:S),X:S,X:S)
g(X:S,Y:S) -> X:S
g(X:S,Y:S) -> Y:S
) 
(STRATEGY INNERMOST)

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 F(0,1,X:S) -> F(g(X:S,X:S),X:S,X:S)
 F(0,1,X:S) -> G(X:S,X:S)
-> Rules:
 f(0,1,X:S) -> f(g(X:S,X:S),X:S,X:S)
 g(X:S,Y:S) -> X:S
 g(X:S,Y:S) -> Y:S

Problem 1: 

SCC Processor:
-> Pairs:
 F(0,1,X:S) -> F(g(X:S,X:S),X:S,X:S)
 F(0,1,X:S) -> G(X:S,X:S)
-> Rules:
 f(0,1,X:S) -> f(g(X:S,X:S),X:S,X:S)
 g(X:S,Y:S) -> X:S
 g(X:S,Y:S) -> Y:S
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 F(0,1,X:S) -> F(g(X:S,X:S),X:S,X:S)
->->-> Rules:
 f(0,1,X:S) -> f(g(X:S,X:S),X:S,X:S)
 g(X:S,Y:S) -> X:S
 g(X:S,Y:S) -> Y:S

Problem 1: 

Instantiation Processor:
-> Pairs:
 F(0,1,X:S) -> F(g(X:S,X:S),X:S,X:S)
-> Rules:
 f(0,1,X:S) -> f(g(X:S,X:S),X:S,X:S)
 g(X:S,Y:S) -> X:S
 g(X:S,Y:S) -> Y:S
->Instantiated Pairs:
->->Original Pair:
 F(0,1,X:S) -> F(g(X:S,X:S),X:S,X:S)
->-> Instantiated pairs:
 F(0,1,1) -> F(g(1,1),1,1)

Problem 1: 

SCC Processor:
-> Pairs:
 F(0,1,1) -> F(g(1,1),1,1)
-> Rules:
 f(0,1,X:S) -> f(g(X:S,X:S),X:S,X:S)
 g(X:S,Y:S) -> X:S
 g(X:S,Y:S) -> Y:S
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 F(0,1,1) -> F(g(1,1),1,1)
->->-> Rules:
 f(0,1,X:S) -> f(g(X:S,X:S),X:S,X:S)
 g(X:S,Y:S) -> X:S
 g(X:S,Y:S) -> Y:S

Problem 1: 

Reduction Pairs Processor:
-> Pairs:
 F(0,1,1) -> F(g(1,1),1,1)
-> Rules:
 f(0,1,X:S) -> f(g(X:S,X:S),X:S,X:S)
 g(X:S,Y:S) -> X:S
 g(X:S,Y:S) -> Y:S
-> Usable rules:
 g(X:S,Y:S) -> X:S
 g(X:S,Y:S) -> Y:S
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[f](X1,X2,X3) = 0
[g](X1,X2) = X1 + 2.X2 + 1
[0] = 2
[1] = 0
[fSNonEmpty] = 0
[F](X1,X2,X3) = 2.X1
[G](X1,X2) = 0

Problem 1: 

SCC Processor:
-> Pairs:
 Empty
-> Rules:
 f(0,1,X:S) -> f(g(X:S,X:S),X:S,X:S)
 g(X:S,Y:S) -> X:S
 g(X:S,Y:S) -> Y:S
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.