YES

Problem:
 +(x,0()) -> x
 +(x,s(y)) -> s(+(x,y))
 d(0()) -> 0()
 d(s(x)) -> s(s(d(x)))
 f(0()) -> 0()
 f(s(x)) -> +(+(s(x),s(x)),s(x))
 f(g(0())) -> +(+(g(0()),g(0())),g(0()))
 g(x) -> s(d(x))

Proof:
 Church Rosser Transformation Processor:
  f(s(x104)) -> +(+(s(x104),s(x104)),s(x104))
  g(0()) -> s(d(0()))
  critical peaks: joinable
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [f](x0) = 7x0,
    
    [+](x0, x1) = 2x0 + x1,
    
    [d](x0) = 2x0,
    
    [g](x0) = 6x0 + 3,
    
    [0] = 2,
    
    [s](x0) = 2x0
   orientation:
    f(s(x104)) = 14x104 >= 14x104 = +(+(s(x104),s(x104)),s(x104))
    
    g(0()) = 15 >= 8 = s(d(0()))
   problem:
    f(s(x104)) -> +(+(s(x104),s(x104)),s(x104))
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [f](x0) = 7x0 + 7,
     
     [+](x0, x1) = x0 + 2x1 + 5,
     
     [s](x0) = 2x0 + 3
    orientation:
     f(s(x104)) = 14x104 + 28 >= 10x104 + 25 = +(+(s(x104),s(x104)),s(x104))
    problem:
     
    Qed