YES

Problem:
 f(f(x,y,z),u,f(x,y,v)) -> f(x,y,f(z,u,v))
 f(x,y,y) -> y
 f(x,y,g(y)) -> x
 f(x,x,y) -> x
 f(g(x),x,y) -> y

Proof:
 DP Processor:
  DPs:
   f#(f(x,y,z),u,f(x,y,v)) -> f#(z,u,v)
   f#(f(x,y,z),u,f(x,y,v)) -> f#(x,y,f(z,u,v))
  TRS:
   f(f(x,y,z),u,f(x,y,v)) -> f(x,y,f(z,u,v))
   f(x,y,y) -> y
   f(x,y,g(y)) -> x
   f(x,x,y) -> x
   f(g(x),x,y) -> y
  CDG Processor:
   DPs:
    f#(f(x,y,z),u,f(x,y,v)) -> f#(z,u,v)
    f#(f(x,y,z),u,f(x,y,v)) -> f#(x,y,f(z,u,v))
   TRS:
    f(f(x,y,z),u,f(x,y,v)) -> f(x,y,f(z,u,v))
    f(x,y,y) -> y
    f(x,y,g(y)) -> x
    f(x,x,y) -> x
    f(g(x),x,y) -> y
   graph:
    
   Qed