YES

Problem:
 strict:
  f(g(x),y,z) -> f(x,y,g(z))
 weak:
  f(x,y,g(z)) -> f(x,g(y),z)
  f(x,a(),z) -> f(x,g(a()),z)
  f(x,y,z) -> f(x,y,g(z))

Proof:
 Bounds Processor:
  bound: 1
  enrichment: match-rt
  automaton:
   final states: {4}
   transitions:
    f1(4,4,6) -> 4*
    f1(4,6,4) -> 4*
    f1(4,6,6) -> 4*
    a1() -> 4*
    g1(6) -> 6*
    g1(4) -> 6*
    f0(4,4,4) -> 4*
    g0(4) -> 4*
    a0() -> 4*
  problem:
   strict:
    
   weak:
    f(x,y,g(z)) -> f(x,g(y),z)
    f(x,a(),z) -> f(x,g(a()),z)
    f(x,y,z) -> f(x,y,g(z))
  Qed