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:
  
      max/plus interpretation on natural numbers:
      
        D#_A(x1) = 0
        +_A(x1,x2) = 7
        pow_A(x1,x2) = max{13, x1 + 6, x2 + 6}
        ln_A(x1) = max{7, x1 - 15}
        div_A(x1,x2) = max{17, x1 - 15, x2 - 15}
        minus_A(x1) = max{0, x1 - 15}
        -_A(x1,x2) = max{2, x1 - 18, x2 - 19}
        *_A(x1,x2) = max{8, x2 - 15}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          D#_A(x1) = max{0, x1 - 162}
          +_A(x1,x2) = max{x1 + 60, x2 + 155}
          pow_A(x1,x2) = max{x1 + 161, x2 + 1}
          ln_A(x1) = x1 + 152
          div_A(x1,x2) = max{155, x1 + 153, x2 + 55}
          minus_A(x1) = max{347, x1 + 161}
          -_A(x1,x2) = max{x1 + 161, x2 + 155}
          *_A(x1,x2) = max{151, x1 + 1, x2 + 39}
          D#!_A(x1) = max{4, x1 - 156}
          +!_A(x1,x2) = max{2, x1 - 207, x2 - 755}
          pow!_A(x1,x2) = max{7, x1 - 577, x2 - 579}
          ln!_A(x1) = max{1, x1 - 753}
          div!_A(x1,x2) = max{6, x1 - 250, x2 - 578}
          minus!_A(x1) = max{3, x1 - 242}
          -!_A(x1,x2) = max{2, x1 - 752, x2 - 1000}
          *!_A(x1,x2) = max{5, x1 - 781, x2 - 742}
  
    partial status:
  
      pi(D#) = []
      pi(+) = []
      pi(pow) = [2]
      pi(ln) = []
      pi(div) = []
      pi(minus) = []
      pi(-) = []
      pi(*) = []

The following rules are removed:

  p2, p3, p7, p8, p9



-- SCC decomposition.

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

p1: D#(+(x,y)) -> D#(x)
p2: D#(ln(x)) -> D#(x)
p3: D#(div(x,y)) -> D#(y)
p4: D#(div(x,y)) -> D#(x)
p5: D#(*(x,y)) -> D#(y)
p6: D#(*(x,y)) -> D#(x)
p7: 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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7}


-- Reduction pair.

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#(div(x,y)) -> D#(x)
p6: D#(div(x,y)) -> D#(y)
p7: D#(ln(x)) -> 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:
  
      max/plus interpretation on natural numbers:
      
        D#_A(x1) = 0
        +_A(x1,x2) = max{2, x1 - 17, x2 - 31}
        *_A(x1,x2) = max{16, x1 - 17, x2 - 31}
        div_A(x1,x2) = max{4, x1 - 4, x2 - 4}
        ln_A(x1) = max{17, x1 - 3}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          D#_A(x1) = max{0, x1 - 400}
          +_A(x1,x2) = max{886, x1 + 34, x2 + 34}
          *_A(x1,x2) = max{75, x1 + 21, x2 + 8}
          div_A(x1,x2) = max{617, x1 + 399, x2 + 3}
          ln_A(x1) = max{869, x1}
          D#!_A(x1) = max{15, x1 - 870}
          +!_A(x1,x2) = max{11, x1 - 927, x2 - 940}
          *!_A(x1,x2) = max{14, x1 - 896, x2 - 993}
          div!_A(x1,x2) = max{16, x1 - 127, x2 - 1386}
          ln!_A(x1) = max{17, x1 - 20}
  
    partial status:
  
      pi(D#) = []
      pi(+) = []
      pi(*) = []
      pi(div) = []
      pi(ln) = []

The following rules are removed:

  p1, p2



-- 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#(div(x,y)) -> D#(x)
p4: D#(div(x,y)) -> D#(y)
p5: D#(ln(x)) -> 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}


-- Reduction pair.

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

p1: D#(*(x,y)) -> D#(x)
p2: D#(ln(x)) -> D#(x)
p3: D#(div(x,y)) -> D#(y)
p4: D#(div(x,y)) -> D#(x)
p5: 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:
  
      max/plus interpretation on natural numbers:
      
        D#_A(x1) = 0
        *_A(x1,x2) = max{0, x1 - 60, x2 - 60}
        ln_A(x1) = max{59, x1 - 60}
        div_A(x1,x2) = max{7, x1 - 24, x2 - 39}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          D#_A(x1) = max{0, x1 - 23}
          *_A(x1,x2) = max{x1 + 22, x2 + 18}
          ln_A(x1) = max{8, x1 + 6}
          div_A(x1,x2) = max{x1 + 12, x2 + 16}
          D#!_A(x1) = max{2, x1 - 19}
          *!_A(x1,x2) = max{1, x1 - 145, x2 - 319}
          ln!_A(x1) = max{77, x1 - 18}
          div!_A(x1,x2) = max{79, x1 - 312, x2 - 17}
  
    partial status:
  
      pi(D#) = []
      pi(*) = []
      pi(ln) = []
      pi(div) = []

The following rules are removed:

  p1, p5



-- SCC decomposition.

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

p1: D#(ln(x)) -> D#(x)
p2: D#(div(x,y)) -> D#(y)
p3: 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}


-- Reduction pair.

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

p1: D#(ln(x)) -> D#(x)
p2: D#(div(x,y)) -> D#(x)
p3: D#(div(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:
  
      max/plus interpretation on natural numbers:
      
        D#_A(x1) = 1
        ln_A(x1) = max{31, x1 - 39}
        div_A(x1,x2) = max{31, x1 - 39, x2 - 39}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          D#_A(x1) = max{0, x1 - 6}
          ln_A(x1) = max{141, x1}
          div_A(x1,x2) = max{147, x1 + 1, x2 + 5}
          D#!_A(x1) = max{1, x1 - 145}
          ln!_A(x1) = max{5, x1 - 143}
          div!_A(x1,x2) = max{3, x1 - 78, x2 - 144}
  
    partial status:
  
      pi(D#) = []
      pi(ln) = []
      pi(div) = []

The following rules are removed:

  p2, p3



-- SCC decomposition.

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

p1: D#(ln(x)) -> 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}


-- Reduction pair.

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

p1: D#(ln(x)) -> 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:
  
      max/plus interpretation on natural numbers:
      
        D#_A(x1) = 1
        ln_A(x1) = max{122, x1 - 24}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          D#_A(x1) = max{0, x1 - 54}
          ln_A(x1) = max{56, x1 + 53}
          D#!_A(x1) = max{1, x1 - 54}
          ln!_A(x1) = max{110, x1 - 47}
  
    partial status:
  
      pi(D#) = []
      pi(ln) = []

The following rules are removed:

  p1

Then no dependency pair remains.