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:
 Matrix Interpretation Processor:
  dimension: 1
  interpretation:
   [g](x0) = 2x0 + 4,
   
   [f](x0) = 4x0,
   
   [d](x0) = 2x0,
   
   [s](x0) = x0 + 4,
   
   [+](x0, x1) = x0 + x1 + 1,
   
   [0] = 0
  orientation:
   +(x,0()) = x + 1 >= x = x
   
   +(x,s(y)) = x + y + 5 >= x + y + 5 = s(+(x,y))
   
   d(0()) = 0 >= 0 = 0()
   
   d(s(x)) = 2x + 8 >= 2x + 8 = s(s(d(x)))
   
   f(0()) = 0 >= 0 = 0()
   
   f(s(x)) = 4x + 16 >= 3x + 14 = +(+(s(x),s(x)),s(x))
   
   f(g(0())) = 16 >= 14 = +(+(g(0()),g(0())),g(0()))
   
   g(x) = 2x + 4 >= 2x + 4 = s(d(x))
  problem:
   +(x,s(y)) -> s(+(x,y))
   d(0()) -> 0()
   d(s(x)) -> s(s(d(x)))
   f(0()) -> 0()
   g(x) -> s(d(x))
  DP Processor:
   DPs:
    +#(x,s(y)) -> +#(x,y)
    d#(s(x)) -> d#(x)
    g#(x) -> d#(x)
   TRS:
    +(x,s(y)) -> s(+(x,y))
    d(0()) -> 0()
    d(s(x)) -> s(s(d(x)))
    f(0()) -> 0()
    g(x) -> s(d(x))
   TDG Processor:
    DPs:
     +#(x,s(y)) -> +#(x,y)
     d#(s(x)) -> d#(x)
     g#(x) -> d#(x)
    TRS:
     +(x,s(y)) -> s(+(x,y))
     d(0()) -> 0()
     d(s(x)) -> s(s(d(x)))
     f(0()) -> 0()
     g(x) -> s(d(x))
    graph:
     g#(x) -> d#(x) -> d#(s(x)) -> d#(x)
     d#(s(x)) -> d#(x) -> d#(s(x)) -> d#(x)
     +#(x,s(y)) -> +#(x,y) -> +#(x,s(y)) -> +#(x,y)
    SCC Processor:
     #sccs: 2
     #rules: 2
     #arcs: 3/9
     DPs:
      +#(x,s(y)) -> +#(x,y)
     TRS:
      +(x,s(y)) -> s(+(x,y))
      d(0()) -> 0()
      d(s(x)) -> s(s(d(x)))
      f(0()) -> 0()
      g(x) -> s(d(x))
     Subterm Criterion Processor:
      simple projection:
       pi(+#) = 1
      problem:
       DPs:
        
       TRS:
        +(x,s(y)) -> s(+(x,y))
        d(0()) -> 0()
        d(s(x)) -> s(s(d(x)))
        f(0()) -> 0()
        g(x) -> s(d(x))
      Qed
     
     DPs:
      d#(s(x)) -> d#(x)
     TRS:
      +(x,s(y)) -> s(+(x,y))
      d(0()) -> 0()
      d(s(x)) -> s(s(d(x)))
      f(0()) -> 0()
      g(x) -> s(d(x))
     Subterm Criterion Processor:
      simple projection:
       pi(d#) = 0
      problem:
       DPs:
        
       TRS:
        +(x,s(y)) -> s(+(x,y))
        d(0()) -> 0()
        d(s(x)) -> s(s(d(x)))
        f(0()) -> 0()
        g(x) -> s(d(x))
      Qed