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:

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

The next rules are strictly ordered:

  p8

We remove them from the problem.

-- 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: 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: 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, p4, p5, p6, p8}


-- 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(plus(X,Y))) -> din#(der(X))
p4: din#(der(times(X,Y))) -> din#(der(X))
p5: din#(der(der(X))) -> u41#(din(der(X)),X)
p6: u41#(dout(DX),X) -> din#(der(DX))
p7: din#(der(der(X))) -> 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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          din#_A(x1) = (1,2)
          der_A(x1) = ((0,0),(1,1)) x1 + (3,3)
          plus_A(x1,x2) = ((1,1),(0,1)) x2
          u21#_A(x1,x2,x3) = ((0,1),(0,0)) x1 + (0,2)
          din_A(x1) = ((0,1),(0,0)) x1 + (1,0)
          dout_A(x1) = (6,2)
          times_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,1),(0,0)) x2
          u41#_A(x1,x2) = (1,2)
          u22_A(x1,x2,x3,x4) = ((1,0),(0,0)) x1 + (0,2)
          u32_A(x1,x2,x3,x4) = x1 + (1,0)
          u42_A(x1,x2,x3) = ((0,0),(0,1)) x1 + (6,0)
          u21_A(x1,x2,x3) = ((0,0),(0,1)) x1 + ((1,1),(0,0)) x3 + (4,0)
          u31_A(x1,x2,x3) = ((1,1),(0,0)) x1 + ((1,1),(0,0)) x3
          u41_A(x1,x2) = ((1,0),(0,0)) x1 + (1,0)
  
    precedence:
    
      din > u32 = u21 = u31 > times > din# = der = plus = u21# = dout = u41# = u22 = u42 = u41
    
    partial status:
  
      pi(din#) = []
      pi(der) = []
      pi(plus) = [2]
      pi(u21#) = []
      pi(din) = []
      pi(dout) = []
      pi(times) = []
      pi(u41#) = []
      pi(u22) = []
      pi(u32) = []
      pi(u42) = []
      pi(u21) = []
      pi(u31) = []
      pi(u41) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- 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: din#(der(times(X,Y))) -> din#(der(X))
p4: din#(der(der(X))) -> u41#(din(der(X)),X)
p5: u41#(dout(DX),X) -> din#(der(DX))
p6: din#(der(der(X))) -> 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:

  {p2, p3, p4, p5, p6}


-- Reduction pair.

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

p1: din#(der(plus(X,Y))) -> din#(der(X))
p2: din#(der(der(X))) -> din#(der(X))
p3: din#(der(der(X))) -> u41#(din(der(X)),X)
p4: u41#(dout(DX),X) -> din#(der(DX))
p5: 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

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          din#_A(x1) = (2,7)
          der_A(x1) = ((0,1),(1,0)) x1 + (5,8)
          plus_A(x1,x2) = ((0,0),(0,1)) x1
          u41#_A(x1,x2) = ((0,1),(0,0)) x1 + (0,7)
          din_A(x1) = ((1,1),(0,0)) x1 + (3,1)
          dout_A(x1) = ((0,0),(1,0)) x1 + (3,6)
          times_A(x1,x2) = ((1,1),(0,0)) x1 + ((0,0),(1,0)) x2
          u22_A(x1,x2,x3,x4) = ((0,1),(0,0)) x1 + ((0,0),(1,0)) x4 + (3,6)
          u32_A(x1,x2,x3,x4) = ((0,0),(0,1)) x1 + ((1,1),(0,0)) x2 + (4,0)
          u42_A(x1,x2,x3) = ((0,0),(0,1)) x1 + (4,5)
          u21_A(x1,x2,x3) = ((0,0),(0,1)) x1 + (5,0)
          u31_A(x1,x2,x3) = ((1,1),(0,0)) x2 + ((1,0),(0,0)) x3 + (6,1)
          u41_A(x1,x2) = x1 + (13,0)
  
    precedence:
    
      din = u21 > u31 > times = u32 > din# = der = u41# > plus > u22 > u42 = u41 > dout
    
    partial status:
  
      pi(din#) = []
      pi(der) = []
      pi(plus) = []
      pi(u41#) = []
      pi(din) = []
      pi(dout) = []
      pi(times) = []
      pi(u22) = []
      pi(u32) = []
      pi(u42) = []
      pi(u21) = []
      pi(u31) = []
      pi(u41) = []

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

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

p1: din#(der(plus(X,Y))) -> din#(der(X))
p2: din#(der(der(X))) -> din#(der(X))
p3: din#(der(der(X))) -> u41#(din(der(X)),X)
p4: 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, p2, p4}


-- Reduction pair.

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

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

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          din#_A(x1) = ((0,1),(0,0)) x1 + (1,2)
          der_A(x1) = x1 + (1,3)
          plus_A(x1,x2) = ((1,1),(0,1)) x1 + ((1,1),(1,1)) x2 + (5,2)
          times_A(x1,x2) = ((0,0),(0,1)) x1 + ((1,1),(1,1)) x2 + (0,1)
  
    precedence:
    
      din# = der = plus = times
    
    partial status:
  
      pi(din#) = []
      pi(der) = []
      pi(plus) = [1]
      pi(times) = [2]

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

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

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


-- Reduction pair.

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

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

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          din#_A(x1) = ((0,1),(0,0)) x1 + (1,5)
          der_A(x1) = ((0,0),(0,1)) x1 + (4,1)
          plus_A(x1,x2) = ((0,0),(1,1)) x1 + ((1,1),(1,1)) x2 + (5,3)
          times_A(x1,x2) = ((0,0),(0,1)) x1 + ((1,1),(0,1)) x2 + (5,1)
  
    precedence:
    
      der = plus = times > din#
    
    partial status:
  
      pi(din#) = []
      pi(der) = []
      pi(plus) = [2]
      pi(times) = []

The next rules are strictly ordered:

  p1

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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          din#_A(x1) = ((0,1),(0,1)) x1 + (1,5)
          der_A(x1) = ((0,0),(1,0)) x1 + (3,1)
          times_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,1),(1,1)) x2 + (4,1)
  
    precedence:
    
      din# = der = times
    
    partial status:
  
      pi(din#) = []
      pi(der) = []
      pi(times) = [1]

The next rules are strictly ordered:

  p1

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