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
  KBO Processor:
   argument filtering:
    pi(f) = [0,1,2]
    pi(g) = []
    pi(f#) = [0,1,2]
   weight function:
    w0 = 1
    w(f#) = w(g) = w(f) = 1
   precedence:
    f# ~ g ~ f
   problem:
    DPs:
     
    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
   Qed