YES

1 decompositions
 #1 -----------
   1: +(x1,0()) -> x1
   2: +(x2,s(x3)) -> s(+(x2,x3))
   3: d(0()) -> 0()
   4: d(s(x4)) -> s(s(d(x4)))
   5: f(0()) -> 0()
   6: f(s(x5)) -> +(+(s(x5),s(x5)),s(x5))
   7: f(g(0())) -> +(+(g(0()),g(0())),g(0()))
   8: g(x6) -> s(d(x6))

@Knuth and Bendix' criterion
--- R
   1: +(x1,0()) -> x1
   2: +(x2,s(x3)) -> s(+(x2,x3))
   3: d(0()) -> 0()
   4: d(s(x4)) -> s(s(d(x4)))
   5: f(0()) -> 0()
   6: f(s(x5)) -> +(+(s(x5),s(x5)),s(x5))
   7: f(g(0())) -> +(+(g(0()),g(0())),g(0()))
   8: g(x6) -> s(d(x6))

--- S
   1: +(x1,0()) -> x1
   2: +(x2,s(x3)) -> s(+(x2,x3))
   3: d(0()) -> 0()
   4: d(s(x4)) -> s(s(d(x4)))
   5: f(0()) -> 0()
   6: f(s(x5)) -> +(+(s(x5),s(x5)),s(x5))
   7: f(g(0())) -> +(+(g(0()),g(0())),g(0()))
   8: g(x6) -> s(d(x6))