YES

Problem 1: 

(VAR v_NonEmpty:S x:S y:S z:S)
(RULES
0 -> 2
1 -> 2
f(0,1,x:S) -> f(x:S,x:S,x:S)
f(x:S,y:S,z:S) -> 2
)

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 F(0,1,x:S) -> F(x:S,x:S,x:S)
-> Rules:
 0 -> 2
 1 -> 2
 f(0,1,x:S) -> f(x:S,x:S,x:S)
 f(x:S,y:S,z:S) -> 2

Problem 1: 

SCC Processor:
-> Pairs:
 F(0,1,x:S) -> F(x:S,x:S,x:S)
-> Rules:
 0 -> 2
 1 -> 2
 f(0,1,x:S) -> f(x:S,x:S,x:S)
 f(x:S,y:S,z:S) -> 2
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 F(0,1,x:S) -> F(x:S,x:S,x:S)
->->-> Rules:
 0 -> 2
 1 -> 2
 f(0,1,x:S) -> f(x:S,x:S,x:S)
 f(x:S,y:S,z:S) -> 2

Problem 1: 

Forward Instantiation Processor:
-> Pairs:
 F(0,1,x:S) -> F(x:S,x:S,x:S)
-> Rules:
 0 -> 2
 1 -> 2
 f(0,1,x:S) -> f(x:S,x:S,x:S)
 f(x:S,y:S,z:S) -> 2
->Instantiated Pairs:
->->Original Pair:
 F(0,1,x:S) -> F(x:S,x:S,x:S)
->-> Instantiated pairs:
 Empty

Problem 1: 

SCC Processor:
-> Pairs:
 Empty
-> Rules:
 0 -> 2
 1 -> 2
 f(0,1,x:S) -> f(x:S,x:S,x:S)
 f(x:S,y:S,z:S) -> 2
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.