YES

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

Proof:
 DP Processor:
  DPs:
   g#(f(x,y),z) -> g#(y,z)
   g#(h(x,y),z) -> g#(x,f(y,z))
   g#(x,h(y,z)) -> g#(x,y)
  TRS:
   g(f(x,y),z) -> f(x,g(y,z))
   g(h(x,y),z) -> g(x,f(y,z))
   g(x,h(y,z)) -> h(g(x,y),z)
  Subterm Criterion Processor:
   simple projection:
    pi(g#) = 0
   problem:
    DPs:
     g#(x,h(y,z)) -> g#(x,y)
    TRS:
     g(f(x,y),z) -> f(x,g(y,z))
     g(h(x,y),z) -> g(x,f(y,z))
     g(x,h(y,z)) -> h(g(x,y),z)
   Subterm Criterion Processor:
    simple projection:
     pi(g#) = 1
    problem:
     DPs:
      
     TRS:
      g(f(x,y),z) -> f(x,g(y,z))
      g(h(x,y),z) -> g(x,f(y,z))
      g(x,h(y,z)) -> h(g(x,y),z)
    Qed