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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      din#_A(x1) = ((0,0,0,0),(1,0,0,0),(0,1,0,0),(0,0,0,0)) x1 + (2,1,13,11)
      der_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,0,0),(1,0,0,0)) x1 + (1,0,1,1)
      plus_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,0,0,0),(0,0,0,0)) x1 + (5,0,0,0)
      u21#_A(x1,x2,x3) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,1,1)) x1 + ((0,0,0,0),(1,0,0,0),(1,1,0,0),(0,0,1,0)) x2 + ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x3 + (0,3,7,2)
      din_A(x1) = ((0,0,0,0),(1,0,0,0),(1,1,0,0),(1,0,0,0)) x1 + (1,4,6,0)
      dout_A(x1) = ((0,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x1 + (3,0,0,10)
      u41#_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,0,1,1)) x1 + ((0,0,0,0),(1,0,0,0),(1,0,0,0),(0,0,0,0)) x2 + (0,3,0,0)
      times_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,0,0),(1,1,0,0)) x1 + ((1,0,0,0),(0,1,0,0),(0,0,0,0),(1,0,0,0)) x2
      u31#_A(x1,x2,x3) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1 + ((0,0,0,0),(0,0,0,0),(1,0,0,0),(1,1,0,0)) x2 + ((0,0,0,0),(1,0,0,0),(1,1,0,0),(1,1,0,0)) x3 + (2,0,14,12)
      u22_A(x1,x2,x3,x4) = x1 + (1,1,4,0)
      u32_A(x1,x2,x3,x4) = ((1,0,0,0),(1,0,0,0),(0,1,0,0),(1,1,1,0)) x1 + ((0,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x2 + (0,0,4,6)
      u42_A(x1,x2,x3) = ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + (3,1,1,0)
      u21_A(x1,x2,x3) = ((1,0,0,0),(0,1,0,0),(1,1,1,0),(1,1,0,1)) x1 + ((0,0,0,0),(0,0,0,0),(1,0,0,0),(0,1,0,0)) x2 + (0,4,0,0)
      u31_A(x1,x2,x3) = ((0,0,0,0),(1,0,0,0),(1,1,0,0),(1,0,1,0)) x1 + ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x3 + (1,0,0,0)
      u41_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(1,1,0,0),(0,1,1,0)) x1 + ((0,0,0,0),(1,0,0,0),(0,1,0,0),(0,0,1,0)) x2 + (0,2,3,2)

The next rules are strictly ordered:

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

We remove them from the problem.

-- SCC decomposition.

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

p1: din#(der(times(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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: din#(der(times(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

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      din#_A(x1) = ((1,0,0,0),(1,1,0,0),(1,0,0,0),(1,0,0,0)) x1
      der_A(x1) = ((1,0,0,0),(0,0,0,0),(0,1,0,0),(1,1,1,0)) x1 + (1,1,1,1)
      times_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x2 + (1,1,1,1)

The next rules are strictly ordered:

  p1

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