YES

We show the termination of the TRS R:

  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)

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: din#(der(plus(X,Y))) -> u21#(din(der(X)),X,Y)
p2: din#(der(plus(X,Y))) -> din#(der(X))
p3: u21#(dout(DX),X,Y) -> u22#(din(der(Y)),X,Y,DX)
p4: u21#(dout(DX),X,Y) -> din#(der(Y))
p5: din#(der(times(X,Y))) -> u31#(din(der(X)),X,Y)
p6: din#(der(times(X,Y))) -> din#(der(X))
p7: u31#(dout(DX),X,Y) -> u32#(din(der(Y)),X,Y,DX)
p8: u31#(dout(DX),X,Y) -> din#(der(Y))
p9: din#(der(der(X))) -> u41#(din(der(X)),X)
p10: din#(der(der(X))) -> din#(der(X))
p11: u41#(dout(DX),X) -> u42#(din(der(DX)),X,DX)
p12: u41#(dout(DX),X) -> din#(der(DX))

and R consists of:

r1: din(der(plus(X,Y))) -> u21(din(der(X)),X,Y)
r2: u21(dout(DX),X,Y) -> u22(din(der(Y)),X,Y,DX)
r3: u22(dout(DY),X,Y,DX) -> dout(plus(DX,DY))
r4: din(der(times(X,Y))) -> u31(din(der(X)),X,Y)
r5: u31(dout(DX),X,Y) -> u32(din(der(Y)),X,Y,DX)
r6: u32(dout(DY),X,Y,DX) -> dout(plus(times(X,DY),times(Y,DX)))
r7: din(der(der(X))) -> u41(din(der(X)),X)
r8: u41(dout(DX),X) -> u42(din(der(DX)),X,DX)
r9: u42(dout(DDX),X,DX) -> dout(DDX)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p4, p5, p6, p8, p9, p10, p12}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: din#(der(plus(X,Y))) -> u21#(din(der(X)),X,Y)
p2: u21#(dout(DX),X,Y) -> din#(der(Y))
p3: din#(der(der(X))) -> din#(der(X))
p4: din#(der(der(X))) -> u41#(din(der(X)),X)
p5: u41#(dout(DX),X) -> din#(der(DX))
p6: din#(der(times(X,Y))) -> din#(der(X))
p7: din#(der(times(X,Y))) -> u31#(din(der(X)),X,Y)
p8: u31#(dout(DX),X,Y) -> din#(der(Y))
p9: din#(der(plus(X,Y))) -> din#(der(X))

and R consists of:

r1: din(der(plus(X,Y))) -> u21(din(der(X)),X,Y)
r2: u21(dout(DX),X,Y) -> u22(din(der(Y)),X,Y,DX)
r3: u22(dout(DY),X,Y,DX) -> dout(plus(DX,DY))
r4: din(der(times(X,Y))) -> u31(din(der(X)),X,Y)
r5: u31(dout(DX),X,Y) -> u32(din(der(Y)),X,Y,DX)
r6: u32(dout(DY),X,Y,DX) -> dout(plus(times(X,DY),times(Y,DX)))
r7: din(der(der(X))) -> u41(din(der(X)),X)
r8: u41(dout(DX),X) -> u42(din(der(DX)),X,DX)
r9: u42(dout(DDX),X,DX) -> dout(DDX)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          din#_A(x1) = x1 + 9
          der_A(x1) = max{275, x1 + 213}
          plus_A(x1,x2) = max{69, x1, x2 + 12}
          u21#_A(x1,x2,x3) = max{276, x1 - 43, x2 + 212, x3 + 222}
          din_A(x1) = max{282, x1 + 51}
          dout_A(x1) = max{511, x1 + 228}
          u41#_A(x1,x2) = max{x1 - 6, x2 + 213}
          times_A(x1,x2) = max{100, x1 + 34, x2 + 10}
          u31#_A(x1,x2,x3) = max{x1 - 227, x3 + 222}
          u22_A(x1,x2,x3,x4) = max{x1 + 12, x2 + 263, x3 + 276, x4 + 228}
          u32_A(x1,x2,x3,x4) = max{x1 + 10, x2 + 298, x3 + 274, x4 + 250}
          u42_A(x1,x2,x3) = max{x1 + 46, x2 + 1, x3 + 327}
          u21_A(x1,x2,x3) = max{x1, x2 + 263, x3 + 276}
          u31_A(x1,x2,x3) = max{x1 + 34, x2 + 298, x3 + 274}
          u41_A(x1,x2) = max{x1 + 99, x2 + 477}
    
      precedence:
      
        u21# > der = u41# > din = u31# > din# = times = u31 > u32 = u41 > plus = u42 = u21 > dout = u22
      
      partial status:
    
        pi(din#) = [1]
        pi(der) = [1]
        pi(plus) = [1, 2]
        pi(u21#) = [3]
        pi(din) = [1]
        pi(dout) = []
        pi(u41#) = [2]
        pi(times) = [1]
        pi(u31#) = [3]
        pi(u22) = [4]
        pi(u32) = [4]
        pi(u42) = [1, 3]
        pi(u21) = [1, 3]
        pi(u31) = [1, 2]
        pi(u41) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          din#_A(x1) = max{23, x1 - 1}
          der_A(x1) = x1 + 49
          plus_A(x1,x2) = max{x1 + 66, x2 + 150}
          u21#_A(x1,x2,x3) = max{22, x3}
          din_A(x1) = max{40, x1 + 30}
          dout_A(x1) = 10
          u41#_A(x1,x2) = max{1, x2}
          times_A(x1,x2) = x1 + 90
          u31#_A(x1,x2,x3) = 22
          u22_A(x1,x2,x3,x4) = 229
          u32_A(x1,x2,x3,x4) = max{170, x4 + 1}
          u42_A(x1,x2,x3) = max{130, x1, x3 + 1}
          u21_A(x1,x2,x3) = max{229, x3 + 1}
          u31_A(x1,x2,x3) = max{169, x1 + 14, x2 + 10}
          u41_A(x1,x2) = 129
    
      precedence:
      
        din# = der = u21# = din = dout = u41# = times = u31# = u32 = u42 > u22 = u21 = u31 = u41 > plus
      
      partial status:
    
        pi(din#) = []
        pi(der) = [1]
        pi(plus) = []
        pi(u21#) = [3]
        pi(din) = []
        pi(dout) = []
        pi(u41#) = [2]
        pi(times) = [1]
        pi(u31#) = []
        pi(u22) = []
        pi(u32) = [4]
        pi(u42) = [3]
        pi(u21) = []
        pi(u31) = [1]
        pi(u41) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6, p7, p8, p9

We remove them from the problem.  Then no dependency pair remains.