YES

1 decompositions
 #0 -----------
   1: f(a(),f(x,a())) -> f(a(),f(f(a(),a()),a()))

@Mutually Orthogonal
--- R
   1: f(a(),f(x,a())) -> f(a(),f(f(a(),a()),a()))

--- S
   1: f(a(),f(x,a())) -> f(a(),f(f(a(),a()),a()))