YES

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

@Strongly Commuting
--- R
   1: f(f(x,y),z) -> f(x,f(y,z))
   2: f(x,y) -> f(y,x)

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