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:
  
      max/plus interpretations on natural numbers:
      
        din#_A(x1) = max{0, x1 - 2}
        der_A(x1) = x1 + 89
        plus_A(x1,x2) = max{28, x1, x2 + 16}
        u21#_A(x1,x2,x3) = max{x1 - 115, x2 + 48, x3 + 89}
        din_A(x1) = max{114, x1 + 88}
        dout_A(x1) = max{129, x1 + 117}
        u41#_A(x1,x2) = max{x1 - 24, x2 + 175}
        times_A(x1,x2) = max{x1 + 90, x2 + 48}
        u31#_A(x1,x2,x3) = max{x1 - 23, x2 + 176, x3 + 88}
        u22_A(x1,x2,x3,x4) = max{x1 + 16, x2 + 133, x3 + 144, x4 + 117}
        u32_A(x1,x2,x3,x4) = max{x1 + 48, x2 + 208, x3 + 224, x4 + 182}
        u42_A(x1,x2,x3) = max{204, x1 + 13, x2 + 130, x3 - 1}
        u21_A(x1,x2,x3) = max{x1, x2 + 134, x3 + 193}
        u31_A(x1,x2,x3) = max{x1 + 67, x2 + 267, x3 + 225}
        u41_A(x1,x2) = max{x1 + 74, x2 + 265}
  
    precedence:
    
      din# = der = plus = u21# = din = dout = u41# = times = u31# = u22 = u32 = u42 = u21 = u31 = u41
    
    partial status:
  
      pi(din#) = []
      pi(der) = [1]
      pi(plus) = [1, 2]
      pi(u21#) = []
      pi(din) = [1]
      pi(dout) = []
      pi(u41#) = []
      pi(times) = [2]
      pi(u31#) = []
      pi(u22) = []
      pi(u32) = []
      pi(u42) = [2]
      pi(u21) = []
      pi(u31) = [2]
      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:
  
      max/plus interpretations on natural numbers:
      
        din#_A(x1) = 58
        der_A(x1) = max{17, x1 + 7}
        plus_A(x1,x2) = max{38, x1, x2 + 21}
        u21#_A(x1,x2,x3) = x1 + 3
        din_A(x1) = 15
        dout_A(x1) = x1 + 58
        times_A(x1,x2) = 15
        u41#_A(x1,x2) = max{57, x1}
        u22_A(x1,x2,x3,x4) = max{x1 + 38, x4 + 58}
        u32_A(x1,x2,x3,x4) = max{x1 + 41, x4 + 39}
        u42_A(x1,x2,x3) = x1 + 1
        u21_A(x1,x2,x3) = max{15, x1}
        u31_A(x1,x2,x3) = max{15, x1}
        u41_A(x1,x2) = max{13, x1}
  
    precedence:
    
      der > plus > din = u41# = u32 = u31 > din# = u21# = dout = times = u22 = u42 = u21 = u41
    
    partial status:
  
      pi(din#) = []
      pi(der) = []
      pi(plus) = []
      pi(u21#) = []
      pi(din) = []
      pi(dout) = []
      pi(times) = []
      pi(u41#) = [1]
      pi(u22) = []
      pi(u32) = [4]
      pi(u42) = []
      pi(u21) = []
      pi(u31) = []
      pi(u41) = []

The next rules are strictly ordered:

  p5

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(plus(X,Y))) -> din#(der(X))
p4: din#(der(times(X,Y))) -> din#(der(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:

  {p1, p2, p3, p4, p6}


-- 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(times(X,Y))) -> din#(der(X))
p5: 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:
  
      max/plus interpretations on natural numbers:
      
        din#_A(x1) = max{0, x1 - 8}
        der_A(x1) = max{8, x1 + 3}
        plus_A(x1,x2) = max{x1 + 5, x2 + 10}
        u21#_A(x1,x2,x3) = max{x1 - 63, x2, x3}
        din_A(x1) = x1 + 60
        dout_A(x1) = 63
        times_A(x1,x2) = max{24, x1 + 23, x2 + 17}
        u22_A(x1,x2,x3,x4) = max{x1 + 5, x3 + 73}
        u32_A(x1,x2,x3,x4) = max{x1 + 10, x2 + 86, x3 + 79}
        u42_A(x1,x2,x3) = 64
        u21_A(x1,x2,x3) = max{x1 + 5, x3 + 73}
        u31_A(x1,x2,x3) = max{x1 + 17, x2 + 86, x3 + 79}
        u41_A(x1,x2) = max{71, x1 + 2}
  
    precedence:
    
      der > din > din# = plus = u21# = dout = times = u22 = u32 = u42 = u21 = u31 = u41
    
    partial status:
  
      pi(din#) = []
      pi(der) = []
      pi(plus) = []
      pi(u21#) = []
      pi(din) = [1]
      pi(dout) = []
      pi(times) = [2]
      pi(u22) = []
      pi(u32) = [3]
      pi(u42) = []
      pi(u21) = []
      pi(u31) = [3]
      pi(u41) = [1]

The next rules are strictly ordered:

  p5

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(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, p3, p4}


-- 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(times(X,Y))) -> din#(der(X))
p4: 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:
  
      max/plus interpretations on natural numbers:
      
        din#_A(x1) = max{19, x1 - 154}
        der_A(x1) = x1 + 131
        plus_A(x1,x2) = max{x1 + 146, x2 + 49}
        u21#_A(x1,x2,x3) = max{x1 - 131, x2 + 19, x3 + 18}
        din_A(x1) = max{145, x1 + 10}
        dout_A(x1) = 149
        times_A(x1,x2) = max{x1 + 132, x2 + 68}
        u22_A(x1,x2,x3,x4) = max{x1 + 39, x2 + 277, x3 + 179}
        u32_A(x1,x2,x3,x4) = max{x1 + 68, x2 + 272, x3 + 180}
        u42_A(x1,x2,x3) = x2 + 272
        u21_A(x1,x2,x3) = max{x1 + 129, x2 + 277, x3 + 180}
        u31_A(x1,x2,x3) = max{x1 + 119, x2 + 273, x3 + 209}
        u41_A(x1,x2) = max{x1 + 127, x2 + 272}
  
    precedence:
    
      din# = der = plus = u21# = din = dout = times = u22 = u32 = u42 = u21 = u31 = u41
    
    partial status:
  
      pi(din#) = []
      pi(der) = [1]
      pi(plus) = []
      pi(u21#) = [3]
      pi(din) = [1]
      pi(dout) = []
      pi(times) = [2]
      pi(u22) = []
      pi(u32) = []
      pi(u42) = []
      pi(u21) = [1, 2, 3]
      pi(u31) = []
      pi(u41) = []

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: u21#(dout(DX),X,Y) -> din#(der(Y))
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 estimated dependency graph contains the following SCCs:

  {p2, p3}


-- Reduction pair.

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

p1: din#(der(times(X,Y))) -> din#(der(X))
p2: 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 monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        din#_A(x1) = max{11, x1 + 7}
        der_A(x1) = x1 + 4
        times_A(x1,x2) = max{x1 + 10, x2 + 10}
  
    precedence:
    
      din# = der = times
    
    partial status:
  
      pi(din#) = [1]
      pi(der) = [1]
      pi(times) = [1, 2]

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(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:

  {p1}


-- Reduction pair.

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

p1: 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:
  
      max/plus interpretations on natural numbers:
      
        din#_A(x1) = max{1, x1 - 5}
        der_A(x1) = max{7, x1 + 4}
  
    precedence:
    
      din# = der
    
    partial status:
  
      pi(din#) = []
      pi(der) = [1]

The next rules are strictly ordered:

  p1

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