YES

1 decompositions
 #1 -----------
   1: f(f(x1,x2),x3) -> f(x1,f(x2,x3))
   2: f(1(),x4) -> x4

@Knuth and Bendix' criterion
--- R
   1: f(f(x1,x2),x3) -> f(x1,f(x2,x3))
   2: f(1(),x4) -> x4

--- S
   1: f(f(x1,x2),x3) -> f(x1,f(x2,x3))
   2: f(1(),x4) -> x4