YES
We show the termination of the TRS R:

  D(t()) -> |1|()
  D(constant()) -> |0|()
  D(+(x,y)) -> +(D(x),D(y))
  D(*(x,y)) -> +(*(y,D(x)),*(x,D(y)))
  D(-(x,y)) -> -(D(x),D(y))
  D(minus(x)) -> minus(D(x))
  D(div(x,y)) -> -(div(D(x),y),div(*(x,D(y)),pow(y,|2|())))
  D(ln(x)) -> div(D(x),x)
  D(pow(x,y)) -> +(*(*(y,pow(x,-(y,|1|()))),D(x)),*(*(pow(x,y),ln(x)),D(y)))


-- SCC decomposition.

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

p1: D#(+(x,y)) -> D#(x)
p2: D#(+(x,y)) -> D#(y)
p3: D#(*(x,y)) -> D#(x)
p4: D#(*(x,y)) -> D#(y)
p5: D#(-(x,y)) -> D#(x)
p6: D#(-(x,y)) -> D#(y)
p7: D#(minus(x)) -> D#(x)
p8: D#(div(x,y)) -> D#(x)
p9: D#(div(x,y)) -> D#(y)
p10: D#(ln(x)) -> D#(x)
p11: D#(pow(x,y)) -> D#(x)
p12: D#(pow(x,y)) -> D#(y)

and R consists of:

r1: D(t()) -> |1|()
r2: D(constant()) -> |0|()
r3: D(+(x,y)) -> +(D(x),D(y))
r4: D(*(x,y)) -> +(*(y,D(x)),*(x,D(y)))
r5: D(-(x,y)) -> -(D(x),D(y))
r6: D(minus(x)) -> minus(D(x))
r7: D(div(x,y)) -> -(div(D(x),y),div(*(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> +(*(*(y,pow(x,-(y,|1|()))),D(x)),*(*(pow(x,y),ln(x)),D(y)))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12}


-- Reduction pair.

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

p1: D#(+(x,y)) -> D#(x)
p2: D#(pow(x,y)) -> D#(y)
p3: D#(pow(x,y)) -> D#(x)
p4: D#(ln(x)) -> D#(x)
p5: D#(div(x,y)) -> D#(y)
p6: D#(div(x,y)) -> D#(x)
p7: D#(minus(x)) -> D#(x)
p8: D#(-(x,y)) -> D#(y)
p9: D#(-(x,y)) -> D#(x)
p10: D#(*(x,y)) -> D#(y)
p11: D#(*(x,y)) -> D#(x)
p12: D#(+(x,y)) -> D#(y)

and R consists of:

r1: D(t()) -> |1|()
r2: D(constant()) -> |0|()
r3: D(+(x,y)) -> +(D(x),D(y))
r4: D(*(x,y)) -> +(*(y,D(x)),*(x,D(y)))
r5: D(-(x,y)) -> -(D(x),D(y))
r6: D(minus(x)) -> minus(D(x))
r7: D(div(x,y)) -> -(div(D(x),y),div(*(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> +(*(*(y,pow(x,-(y,|1|()))),D(x)),*(*(pow(x,y),ln(x)),D(y)))

The set of usable rules consists of

  r1

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          D#_A(x1) = 1
          +_A(x1,x2) = 2
          pow_A(x1,x2) = x1 + 3
          ln_A(x1) = 0
          div_A(x1,x2) = x1 + x2 + 5
          minus_A(x1) = 2
          -_A(x1,x2) = 4
          *_A(x1,x2) = x1 + x2
          D_A(x1) = 2
          t_A() = 2
          |1|_A() = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          D#_A(x1) = max{0, x1 - 122}
          +_A(x1,x2) = max{231, x1 + 110, x2 + 110}
          pow_A(x1,x2) = max{x1 + 18, x2 + 36}
          ln_A(x1) = max{127, x1 + 37}
          div_A(x1,x2) = max{x1, x2 + 12}
          minus_A(x1) = max{231, x1 + 73}
          -_A(x1,x2) = max{x1 + 73, x2 + 73}
          *_A(x1,x2) = max{x1 + 121, x2 + 12}
          D_A(x1) = max{230, x1 - 31}
          t_A = 15
          |1|_A = 0
          D#!_A(x1) = max{55, x1 + 19}
          +!_A(x1,x2) = max{56, x2 - 383}
          pow!_A(x1,x2) = max{x1 + 57, x2 + 20}
          ln!_A(x1) = 57
          div!_A(x1,x2) = max{x1 + 18, x2 - 21}
          minus!_A(x1) = max{57, x1 - 231}
          -!_A(x1,x2) = max{21, x1 - 233, x2 - 464}
          *!_A(x1,x2) = max{2, x1 - 140}
          D!_A(x1) = max{58, x1 - 14}
          t!_A = 58
          |1|!_A = 57
  
    partial status:
  
      pi(D#) = []
      pi(+) = []
      pi(pow) = [1]
      pi(ln) = []
      pi(div) = [2]
      pi(minus) = []
      pi(-) = []
      pi(*) = []
      pi(D) = []
      pi(t) = []
      pi(|1|) = []

The following rules are removed:

  p1, p4, p7, p8, p9, p10, p11, p12



-- SCC decomposition.

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

p1: D#(pow(x,y)) -> D#(y)
p2: D#(pow(x,y)) -> D#(x)
p3: D#(div(x,y)) -> D#(y)
p4: D#(div(x,y)) -> D#(x)

and R consists of:

r1: D(t()) -> |1|()
r2: D(constant()) -> |0|()
r3: D(+(x,y)) -> +(D(x),D(y))
r4: D(*(x,y)) -> +(*(y,D(x)),*(x,D(y)))
r5: D(-(x,y)) -> -(D(x),D(y))
r6: D(minus(x)) -> minus(D(x))
r7: D(div(x,y)) -> -(div(D(x),y),div(*(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> +(*(*(y,pow(x,-(y,|1|()))),D(x)),*(*(pow(x,y),ln(x)),D(y)))

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: D#(pow(x,y)) -> D#(y)
p2: D#(div(x,y)) -> D#(x)
p3: D#(div(x,y)) -> D#(y)
p4: D#(pow(x,y)) -> D#(x)

and R consists of:

r1: D(t()) -> |1|()
r2: D(constant()) -> |0|()
r3: D(+(x,y)) -> +(D(x),D(y))
r4: D(*(x,y)) -> +(*(y,D(x)),*(x,D(y)))
r5: D(-(x,y)) -> -(D(x),D(y))
r6: D(minus(x)) -> minus(D(x))
r7: D(div(x,y)) -> -(div(D(x),y),div(*(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> +(*(*(y,pow(x,-(y,|1|()))),D(x)),*(*(pow(x,y),ln(x)),D(y)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          D#_A(x1) = x1
          pow_A(x1,x2) = x1 + x2 + 3
          div_A(x1,x2) = x1 + x2 + 5
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          D#_A(x1) = max{0, x1 - 269}
          pow_A(x1,x2) = max{82, x1 + 76, x2}
          div_A(x1,x2) = max{431, x1 + 233, x2 + 268}
          D#!_A(x1) = max{1, x1 - 432}
          pow!_A(x1,x2) = max{4, x2 - 134}
          div!_A(x1,x2) = max{5, x1 - 228, x2 - 120}
  
    partial status:
  
      pi(D#) = [1]
      pi(pow) = []
      pi(div) = []

The following rules are removed:

  p1, p2, p3, p4

Then no dependency pair remains.