YES

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

@Strongly Commuting
--- R
   1: f(0(),0()) -> f(0(),1())
   2: f(1(),0()) -> f(0(),0())
   3: f(x,y) -> f(y,x)

--- S
   1: f(0(),0()) -> f(0(),1())
   2: f(1(),0()) -> f(0(),0())
   3: f(x,y) -> f(y,x)