YES

1 decompositions
 #0 -----------
   1: f(f(x,y),z) -> f(x,f(y,z))
   2: f(1(),x) -> x

@Jouannaud and Kirchner's criterion
--- R
   1: f(f(x,y),z) -> f(x,f(y,z))
   2: f(1(),x) -> x

--- S
   1: f(f(x,y),z) -> f(x,f(y,z))
   2: f(1(),x) -> x