YES

Problem:
 din(der(plus(X,Y))) -> u21(din(der(X)),X,Y)
 u21(dout(DX),X,Y) -> u22(din(der(Y)),X,Y,DX)
 u22(dout(DY),X,Y,DX) -> dout(plus(DX,DY))
 din(der(times(X,Y))) -> u31(din(der(X)),X,Y)
 u31(dout(DX),X,Y) -> u32(din(der(Y)),X,Y,DX)
 u32(dout(DY),X,Y,DX) -> dout(plus(times(X,DY),times(Y,DX)))
 din(der(der(X))) -> u41(din(der(X)),X)
 u41(dout(DX),X) -> u42(din(der(DX)),X,DX)
 u42(dout(DDX),X,DX) -> dout(DDX)

Proof:
 Bounds Processor:
  bound: 2
  enrichment: roof
  automaton:
   final states: {1}
   transitions:
    dout1(7) -> 1*
    dout1(1) -> 1*
    u421(14,1,7) -> 1*
    u421(4,1,1) -> 1*
    din1(13) -> 14*
    din1(3) -> 4*
    der1(7) -> 13*
    der1(1) -> 3*
    u411(4,1) -> 18,4,1
    plus1(10,9) -> 7*
    plus1(1,1) -> 7*
    plus1(1,7) -> 1*
    times1(1,1) -> 10,9
    times1(1,7) -> 10*
    u321(4,1,1,7) -> 1*
    u321(4,1,1,1) -> 1*
    u311(4,1,1) -> 18,4,1
    u221(4,1,1,7) -> 1*
    u221(4,1,1,1) -> 1*
    u211(4,1,7) -> 1*
    u211(4,1,1) -> 18,4,1
    u312(18,1,7) -> 20*
    u312(18,1,1) -> 20*
    din0(1) -> 1*
    din2(17) -> 18*
    din2(19) -> 20*
    der0(1) -> 1*
    der2(10) -> 19*
    der2(1) -> 17*
    plus0(1,1) -> 1*
    u212(18,1,7) -> 18,4
    u212(18,1,1) -> 14*
    u212(20,10,9) -> 14*
    u210(1,1,1) -> 1*
    dout0(1) -> 1*
    u220(1,1,1,1) -> 1*
    times0(1,1) -> 1*
    u310(1,1,1) -> 1*
    u320(1,1,1,1) -> 1*
    u410(1,1) -> 1*
    u420(1,1,1) -> 1*
  problem:
   
  Qed