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 monotone reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      D#_A(x1) = x1
      +_A(x1,x2) = x1 + x2
      pow_A(x1,x2) = x1 + x2 + 1
      ln_A(x1) = x1
      div_A(x1,x2) = x1 + x2
      minus_A(x1) = x1
      -_A(x1,x2) = x1 + x2
      *_A(x1,x2) = x1 + x2

The next rules are strictly ordered:

  p2, p3
  r1, r2, r3, r4, r5, r6, r7, r8, r9

We remove them from the problem.

-- 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#(minus(x)) -> 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)
p10: D#(+(x,y)) -> D#(y)

and R consists of:

  (no rules)

The estimated dependency graph contains the following SCCs:

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


-- 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#(-(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)

and R consists of:

  (no rules)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      D#_A(x1) = x1
      +_A(x1,x2) = x1 + x2 + 1
      *_A(x1,x2) = x1 + x2
      -_A(x1,x2) = x1 + x2
      minus_A(x1) = x1
      div_A(x1,x2) = x1 + x2
      ln_A(x1) = x1

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.

-- 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#(minus(x)) -> D#(x)
p6: D#(div(x,y)) -> D#(x)
p7: D#(div(x,y)) -> D#(y)
p8: D#(ln(x)) -> D#(x)

and R consists of:

  (no rules)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8}


-- 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#(minus(x)) -> D#(x)
p6: D#(-(x,y)) -> D#(y)
p7: D#(-(x,y)) -> D#(x)
p8: D#(*(x,y)) -> D#(y)

and R consists of:

  (no rules)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      D#_A(x1) = x1
      *_A(x1,x2) = x1 + x2
      ln_A(x1) = x1 + 1
      div_A(x1,x2) = x1 + x2
      minus_A(x1) = x1
      -_A(x1,x2) = x1 + x2

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: D#(*(x,y)) -> D#(x)
p2: D#(div(x,y)) -> D#(y)
p3: D#(div(x,y)) -> D#(x)
p4: D#(minus(x)) -> D#(x)
p5: D#(-(x,y)) -> D#(y)
p6: D#(-(x,y)) -> D#(x)
p7: D#(*(x,y)) -> D#(y)

and R consists of:

  (no rules)

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#(minus(x)) -> D#(x)
p6: D#(div(x,y)) -> D#(x)
p7: D#(div(x,y)) -> D#(y)

and R consists of:

  (no rules)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      D#_A(x1) = x1
      *_A(x1,x2) = x1 + x2 + 1
      -_A(x1,x2) = x1 + x2
      minus_A(x1) = x1
      div_A(x1,x2) = x1 + x2

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.

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

and R consists of:

  (no rules)

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#(div(x,y)) -> D#(y)
p3: D#(div(x,y)) -> D#(x)
p4: D#(minus(x)) -> D#(x)
p5: D#(-(x,y)) -> D#(y)

and R consists of:

  (no rules)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      D#_A(x1) = x1
      -_A(x1,x2) = x1 + x2
      div_A(x1,x2) = x1 + x2
      minus_A(x1) = x1 + 1

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

  (no rules)

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

and R consists of:

  (no rules)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      D#_A(x1) = x1
      -_A(x1,x2) = x1 + x2
      div_A(x1,x2) = x1 + x2 + 1

The next rules are strictly ordered:

  p3, p4

We remove them from the problem.

-- 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)

and R consists of:

  (no rules)

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#(-(x,y)) -> D#(x)
p2: D#(-(x,y)) -> D#(y)

and R consists of:

  (no rules)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      D#_A(x1) = x1
      -_A(x1,x2) = x1 + x2 + 1

The next rules are strictly ordered:

  p1, p2

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