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. matrix interpretations:
    
      carrier: N^2
      order: lexicographic order
      interpretations:
        din#_A(x1) = x1 + (2,2)
        der_A(x1) = (1,1)
        plus_A(x1,x2) = ((0,0),(1,0)) x1 + ((0,0),(1,0)) x2 + (1,1)
        u21#_A(x1,x2,x3) = x1
        din_A(x1) = x1 + (0,5)
        dout_A(x1) = ((0,0),(1,0)) x1 + (4,1)
        u41#_A(x1,x2) = ((0,0),(1,0)) x1 + (3,0)
        times_A(x1,x2) = ((0,0),(1,0)) x1 + ((0,0),(1,0)) x2 + (1,0)
        u31#_A(x1,x2,x3) = x1 + (0,3)
        u22_A(x1,x2,x3,x4) = x1 + (1,3)
        u32_A(x1,x2,x3,x4) = (4,3)
        u42_A(x1,x2,x3) = x1 + (1,1)
        u21_A(x1,x2,x3) = x1 + (0,4)
        u31_A(x1,x2,x3) = x1 + (0,4)
        u41_A(x1,x2) = ((1,0),(1,0)) x1 + (0,3)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        plus > din > u21 > u22 > dout > u31 > u32 > times > der > u41 > u42 > u31# > din# > u41# > u21#
      
      argument filter:
    
        pi(din#) = []
        pi(der) = []
        pi(plus) = []
        pi(u21#) = []
        pi(din) = []
        pi(dout) = []
        pi(u41#) = []
        pi(times) = []
        pi(u31#) = []
        pi(u22) = []
        pi(u32) = []
        pi(u42) = [1]
        pi(u21) = []
        pi(u31) = []
        pi(u41) = []
    

The next rules are strictly ordered:

  p1, p2, p4, p5, p7, p8

We remove them from the problem.

-- SCC decomposition.

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

p1: din#(der(der(X))) -> din#(der(X))
p2: din#(der(times(X,Y))) -> din#(der(X))
p3: 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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: din#(der(der(X))) -> din#(der(X))
p2: din#(der(plus(X,Y))) -> din#(der(X))
p3: 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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: lexicographic order
      interpretations:
        din#_A(x1) = ((1,0),(1,0)) x1
        der_A(x1) = ((1,0),(1,0)) x1 + (1,1)
        plus_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (1,1)
        times_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        der > din# > times > plus
      
      argument filter:
    
        pi(din#) = []
        pi(der) = []
        pi(plus) = [1, 2]
        pi(times) = []
    

The next rules are strictly ordered:

  p1, p2, p3

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