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, r2, r3, r4, r5, r6, r7, r8, r9

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:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            D#_A(0) = 1
            D#_A(1) = 1
            +_A(0,0) = 0
            +_A(0,1) = 0
            +_A(1,0) = 0
            +_A(1,1) = 0
            pow_A(0,0) = 0
            pow_A(0,1) = 0
            pow_A(1,0) = 0
            pow_A(1,1) = 0
            ln_A(0) = 0
            ln_A(1) = 0
            div_A(0,0) = 0
            div_A(0,1) = 0
            div_A(1,0) = 0
            div_A(1,1) = 0
            minus_A(0) = 0
            minus_A(1) = 1
            -_A(0,0) = 0
            -_A(0,1) = 0
            -_A(1,0) = 0
            -_A(1,1) = 0
            *_A(0,0) = 0
            *_A(0,1) = 0
            *_A(1,0) = 0
            *_A(1,1) = 0
            D_A(0) = 0
            D_A(1) = 1
            t_A = 1
            |1|_A = 1
            constant_A = 1
            |0|_A = 1
            |2|_A = 1
            D#!_A(0) = 0
            D#!_A(1) = 0
            +!_A(0,0) = 0
            +!_A(0,1) = 0
            +!_A(1,0) = 0
            +!_A(1,1) = 0
            pow!_A(0,0) = 1
            pow!_A(0,1) = 1
            pow!_A(1,0) = 1
            pow!_A(1,1) = 1
            ln!_A(0) = 1
            ln!_A(1) = 1
            div!_A(0,0) = 0
            div!_A(0,1) = 0
            div!_A(1,0) = 0
            div!_A(1,1) = 0
            minus!_A(0) = 1
            minus!_A(1) = 0
            -!_A(0,0) = 0
            -!_A(0,1) = 0
            -!_A(1,0) = 0
            -!_A(1,1) = 0
            *!_A(0,0) = 0
            *!_A(0,1) = 0
            *!_A(1,0) = 0
            *!_A(1,1) = 0
            D!_A(0) = 1
            D!_A(1) = 0
            t!_A = 0
            |1|!_A = 0
            constant!_A = 0
            |0|!_A = 0
            |2|!_A = 0
  
    partial status:
  
      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]
      pi(D) = [1]
      pi(t) = []
      pi(|1|) = []
      pi(constant) = []
      pi(|0|) = []
      pi(|2|) = []

The following rules are removed:

  p2, p4, p5, p7, p8, p9, p11



-- SCC decomposition.

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

p1: D#(+(x,y)) -> D#(x)
p2: D#(pow(x,y)) -> D#(x)
p3: D#(div(x,y)) -> D#(x)
p4: D#(*(x,y)) -> D#(y)
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 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#(+(x,y)) -> D#(y)
p3: D#(*(x,y)) -> D#(y)
p4: D#(div(x,y)) -> D#(x)
p5: 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

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

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:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            D#_A(0) = 0
            D#_A(1) = 0
            +_A(0,0) = 0
            +_A(0,1) = 1
            +_A(1,0) = 1
            +_A(1,1) = 0
            *_A(0,0) = 1
            *_A(0,1) = 0
            *_A(1,0) = 1
            *_A(1,1) = 1
            div_A(0,0) = 0
            div_A(0,1) = 0
            div_A(1,0) = 0
            div_A(1,1) = 0
            pow_A(0,0) = 0
            pow_A(0,1) = 1
            pow_A(1,0) = 0
            pow_A(1,1) = 0
            D_A(0) = 0
            D_A(1) = 0
            t_A = 0
            |1|_A = 0
            constant_A = 1
            |0|_A = 0
            -_A(0,0) = 0
            -_A(0,1) = 0
            -_A(1,0) = 1
            -_A(1,1) = 1
            minus_A(0) = 0
            minus_A(1) = 0
            |2|_A = 1
            ln_A(0) = 1
            ln_A(1) = 0
            D#!_A(0) = 0
            D#!_A(1) = 0
            +!_A(0,0) = 0
            +!_A(0,1) = 0
            +!_A(1,0) = 0
            +!_A(1,1) = 0
            *!_A(0,0) = 0
            *!_A(0,1) = 0
            *!_A(1,0) = 0
            *!_A(1,1) = 0
            div!_A(0,0) = 0
            div!_A(0,1) = 0
            div!_A(1,0) = 0
            div!_A(1,1) = 0
            pow!_A(0,0) = 1
            pow!_A(0,1) = 1
            pow!_A(1,0) = 1
            pow!_A(1,1) = 1
            D!_A(0) = 1
            D!_A(1) = 1
            t!_A = 0
            |1|!_A = 0
            constant!_A = 1
            |0|!_A = 0
            -!_A(0,0) = 0
            -!_A(0,1) = 0
            -!_A(1,0) = 0
            -!_A(1,1) = 0
            minus!_A(0) = 0
            minus!_A(1) = 1
            |2|!_A = 0
            ln!_A(0) = 1
            ln!_A(1) = 0
  
    partial status:
  
      pi(D#) = [1]
      pi(+) = [1, 2]
      pi(*) = [1, 2]
      pi(div) = [1, 2]
      pi(pow) = [1, 2]
      pi(D) = [1]
      pi(t) = []
      pi(|1|) = []
      pi(constant) = []
      pi(|0|) = []
      pi(-) = [2]
      pi(minus) = [1]
      pi(|2|) = []
      pi(ln) = [1]

The following rules are removed:

  p1, p2, p3, p4



-- SCC decomposition.

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

p1: 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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: 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

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

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:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            D#_A(0) = 0
            D#_A(1) = 0
            pow_A(0,0) = 1
            pow_A(0,1) = 1
            pow_A(1,0) = 0
            pow_A(1,1) = 0
            D_A(0) = 0
            D_A(1) = 0
            t_A = 0
            |1|_A = 0
            constant_A = 1
            |0|_A = 0
            +_A(0,0) = 0
            +_A(0,1) = 0
            +_A(1,0) = 0
            +_A(1,1) = 1
            *_A(0,0) = 0
            *_A(0,1) = 1
            *_A(1,0) = 0
            *_A(1,1) = 0
            -_A(0,0) = 0
            -_A(0,1) = 0
            -_A(1,0) = 1
            -_A(1,1) = 0
            minus_A(0) = 0
            minus_A(1) = 1
            div_A(0,0) = 0
            div_A(0,1) = 0
            div_A(1,0) = 1
            div_A(1,1) = 1
            |2|_A = 1
            ln_A(0) = 1
            ln_A(1) = 1
            D#!_A(0) = 1
            D#!_A(1) = 1
            pow!_A(0,0) = 1
            pow!_A(0,1) = 1
            pow!_A(1,0) = 1
            pow!_A(1,1) = 1
            D!_A(0) = 1
            D!_A(1) = 1
            t!_A = 1
            |1|!_A = 1
            constant!_A = 1
            |0|!_A = 1
            +!_A(0,0) = 0
            +!_A(0,1) = 0
            +!_A(1,0) = 0
            +!_A(1,1) = 0
            *!_A(0,0) = 0
            *!_A(0,1) = 1
            *!_A(1,0) = 0
            *!_A(1,1) = 0
            -!_A(0,0) = 1
            -!_A(0,1) = 0
            -!_A(1,0) = 1
            -!_A(1,1) = 1
            minus!_A(0) = 1
            minus!_A(1) = 1
            div!_A(0,0) = 1
            div!_A(0,1) = 1
            div!_A(1,0) = 1
            div!_A(1,1) = 1
            |2|!_A = 0
            ln!_A(0) = 1
            ln!_A(1) = 1
  
    partial status:
  
      pi(D#) = [1]
      pi(pow) = [1, 2]
      pi(D) = []
      pi(t) = []
      pi(|1|) = []
      pi(constant) = []
      pi(|0|) = []
      pi(+) = []
      pi(*) = []
      pi(-) = []
      pi(minus) = []
      pi(div) = []
      pi(|2|) = []
      pi(ln) = []

The following rules are removed:

  p1

Then no dependency pair remains.