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

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            D#_A(x1) = max{0, x1 - 104}
            +_A(x1,x2) = max{103, x1 + 21, x2 + 2}
            pow_A(x1,x2) = max{x1 + 23, x2 + 1}
            ln_A(x1) = max{19, x1}
            div_A(x1,x2) = max{103, x1 + 74, x2 + 78}
            minus_A(x1) = x1 + 103
            -_A(x1,x2) = max{63, x1 + 47, x2 + 2}
            *_A(x1,x2) = max{x1 + 1, x2 + 3}
            D#!_A(x1) = max{2, x1 - 104}
            +!_A(x1,x2) = max{4, x1 - 90, x2 - 110}
            pow!_A(x1,x2) = max{5, x1 - 229, x2 - 278}
            ln!_A(x1) = max{3, x1 - 228}
            div!_A(x1,x2) = max{4, x1 - 105, x2 - 179}
            minus!_A(x1) = max{5, x1 - 8}
            -!_A(x1,x2) = max{1, x1 - 279, x2 - 281}
            *!_A(x1,x2) = max{1, x1 - 231, x2 - 58}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            D#_A(x1) = max{0, x1 - 218}
            +_A(x1,x2) = max{245, x1 + 217, x2 + 107}
            pow_A(x1,x2) = max{x1 + 19, x2 + 134}
            ln_A(x1) = x1 + 86
            div_A(x1,x2) = max{130, x1 + 87, x2 + 1}
            minus_A(x1) = max{245, x1 + 1}
            -_A(x1,x2) = max{38, x1 + 1, x2 + 11}
            *_A(x1,x2) = max{x1 + 11, x2 + 28}
            D#!_A(x1) = max{4, x1 - 240}
            +!_A(x1,x2) = max{6, x1 - 212, x2 - 239}
            pow!_A(x1,x2) = max{5, x1 - 445, x2 - 460}
            ln!_A(x1) = max{2, x1 - 304}
            div!_A(x1,x2) = max{6, x1 - 57, x2 - 135}
            minus!_A(x1) = max{114, x1 - 131}
            -!_A(x1,x2) = max{1, x1 - 461, x2 - 156}
            *!_A(x1,x2) = max{3, x1 - 459, x2 - 198}
        
  
    partial status:
  
      pi(D#) = []
      pi(+) = []
      pi(pow) = [1, 2]
      pi(ln) = []
      pi(div) = [1]
      pi(minus) = [1]
      pi(-) = []
      pi(*) = []

The following rules are removed:

  p1, p7, 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#(ln(x)) -> D#(x)
p4: D#(div(x,y)) -> D#(y)
p5: D#(div(x,y)) -> D#(x)
p6: D#(-(x,y)) -> D#(y)
p7: D#(-(x,y)) -> D#(x)
p8: D#(*(x,y)) -> D#(y)
p9: D#(*(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, p5, p6, p7, p8, p9}


-- Reduction pair.

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

p1: D#(pow(x,y)) -> D#(y)
p2: D#(*(x,y)) -> D#(x)
p3: D#(*(x,y)) -> D#(y)
p4: D#(-(x,y)) -> D#(x)
p5: D#(-(x,y)) -> D#(y)
p6: D#(div(x,y)) -> D#(x)
p7: D#(div(x,y)) -> D#(y)
p8: D#(ln(x)) -> D#(x)
p9: 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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            D#_A(x1) = max{0, x1 - 16}
            pow_A(x1,x2) = max{x1 + 15, x2 + 2}
            *_A(x1,x2) = max{x1 + 6, x2 + 1}
            -_A(x1,x2) = max{x1 + 3, x2 + 3}
            div_A(x1,x2) = max{13, x1 + 8, x2 + 6}
            ln_A(x1) = max{13, x1 + 1}
            D#!_A(x1) = max{5, x1 - 104}
            pow!_A(x1,x2) = max{2, x1 - 105, x2 - 109}
            *!_A(x1,x2) = max{4, x1 - 71, x2 - 36}
            -!_A(x1,x2) = max{4, x1 - 48, x2 - 22}
            div!_A(x1,x2) = max{7, x1 - 24, x2 - 28}
            ln!_A(x1) = max{6, x1 - 67}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            D#_A(x1) = max{0, x1 - 64}
            pow_A(x1,x2) = max{x1 + 47, x2 + 1}
            *_A(x1,x2) = max{73, x1 + 23, x2 + 15}
            -_A(x1,x2) = max{54, x1 + 14, x2 + 1}
            div_A(x1,x2) = max{117, x1 + 63, x2 + 47}
            ln_A(x1) = max{55, x1}
            D#!_A(x1) = max{7, x1 - 264}
            pow!_A(x1,x2) = max{1, x1 - 403, x2 - 463}
            *!_A(x1,x2) = max{4, x1 - 71, x2 - 63}
            -!_A(x1,x2) = max{6, x1 - 316, x2 - 263}
            div!_A(x1,x2) = max{5, x1 - 266, x2 - 200}
            ln!_A(x1) = max{8, x1 - 213}
        
  
    partial status:
  
      pi(D#) = [1]
      pi(pow) = [1, 2]
      pi(*) = [1, 2]
      pi(-) = [1, 2]
      pi(div) = [1, 2]
      pi(ln) = [1]

The following rules are removed:

  p1, p2, p3, p4, p5, p6, p7, p8, p9

Then no dependency pair remains.