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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        div > ln > - > pow > * > + > minus > D#
      
      argument filter:
    
        pi(D#) = [1]
        pi(+) = [1, 2]
        pi(pow) = [1, 2]
        pi(ln) = 1
        pi(div) = [1, 2]
        pi(minus) = 1
        pi(-) = [1, 2]
        pi(*) = [1, 2]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        div > - > + > ln > * > D# > pow > minus
      
      argument filter:
    
        pi(D#) = []
        pi(+) = []
        pi(pow) = []
        pi(ln) = []
        pi(div) = []
        pi(minus) = []
        pi(-) = []
        pi(*) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p5, p6, p8, p9, p10, p11, p12

We remove them from the problem.

-- SCC decomposition.

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

p1: D#(ln(x)) -> D#(x)
p2: D#(minus(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}


-- Reduction pair.

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

p1: D#(ln(x)) -> D#(x)
p2: D#(minus(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 monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        D# > minus > ln
      
      argument filter:
    
        pi(D#) = 1
        pi(ln) = 1
        pi(minus) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        D# > minus > ln
      
      argument filter:
    
        pi(D#) = 1
        pi(ln) = [1]
        pi(minus) = [1]
    

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9

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