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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        D#_A(x1) = max{2, x1 - 4}
        +_A(x1,x2) = max{7, x1 + 3, x2 + 3}
        pow_A(x1,x2) = max{x1 + 1, x2}
        ln_A(x1) = x1
        div_A(x1,x2) = max{x1 + 1, x2}
        minus_A(x1) = x1
        -_A(x1,x2) = max{x1 + 5, x2}
        *_A(x1,x2) = max{2, x1, x2}
  
    precedence:
    
      D# = + = pow = ln = div = minus = - = *
    
    partial status:
  
      pi(D#) = []
      pi(+) = [2]
      pi(pow) = [2]
      pi(ln) = [1]
      pi(div) = [2]
      pi(minus) = [1]
      pi(-) = [2]
      pi(*) = [2]

The next rules are strictly ordered:

  p12

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#(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)

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}


-- Reduction pair.

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

p1: D#(+(x,y)) -> D#(x)
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#(minus(x)) -> D#(x)
p7: D#(div(x,y)) -> D#(x)
p8: D#(div(x,y)) -> D#(y)
p9: D#(ln(x)) -> D#(x)
p10: D#(pow(x,y)) -> D#(x)
p11: 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 set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        D#_A(x1) = max{4, x1 + 3}
        +_A(x1,x2) = max{x1 + 1, x2}
        *_A(x1,x2) = max{x1, x2 + 2}
        -_A(x1,x2) = max{x1, x2}
        minus_A(x1) = x1
        div_A(x1,x2) = max{x1, x2}
        ln_A(x1) = x1
        pow_A(x1,x2) = max{x1 + 1, x2}
  
    precedence:
    
      D# > + = * = - = minus = div = ln = pow
    
    partial status:
  
      pi(D#) = []
      pi(+) = [2]
      pi(*) = [2]
      pi(-) = [2]
      pi(minus) = [1]
      pi(div) = [2]
      pi(ln) = [1]
      pi(pow) = [2]

The next rules are strictly ordered:

  p3

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#(x)
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)
p9: D#(pow(x,y)) -> D#(x)
p10: 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}


-- 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#(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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        D#_A(x1) = x1 + 4
        +_A(x1,x2) = max{x1, x2}
        pow_A(x1,x2) = max{x1, x2}
        ln_A(x1) = x1
        div_A(x1,x2) = max{x1, x2}
        minus_A(x1) = x1
        -_A(x1,x2) = max{x1 + 5, x2 + 3}
        *_A(x1,x2) = max{x1, x2 + 1}
  
    precedence:
    
      D# = + = pow = ln = div = minus = - = *
    
    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]

The next rules are strictly ordered:

  p1

We remove them from the problem.

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

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        D#_A(x1) = max{4, x1 + 3}
        pow_A(x1,x2) = max{x1, x2 + 2}
        *_A(x1,x2) = max{x1, x2}
        -_A(x1,x2) = max{x1 + 1, x2}
        minus_A(x1) = x1
        div_A(x1,x2) = max{1, x1, x2}
        ln_A(x1) = x1
  
    precedence:
    
      D# = pow = * = - = minus = div = ln
    
    partial status:
  
      pi(D#) = []
      pi(pow) = [2]
      pi(*) = [2]
      pi(-) = [2]
      pi(minus) = [1]
      pi(div) = [2]
      pi(ln) = [1]

The next rules are strictly ordered:

  p1

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#(x)
p3: D#(-(x,y)) -> D#(y)
p4: D#(minus(x)) -> D#(x)
p5: D#(div(x,y)) -> D#(x)
p6: D#(div(x,y)) -> D#(y)
p7: D#(ln(x)) -> D#(x)
p8: 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, 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#(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#(minus(x)) -> D#(x)
p7: D#(-(x,y)) -> D#(y)
p8: 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 set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        D#_A(x1) = x1 + 2
        *_A(x1,x2) = max{x1, x2}
        pow_A(x1,x2) = max{x1, x2}
        ln_A(x1) = x1
        div_A(x1,x2) = max{x1, x2}
        minus_A(x1) = x1 + 1
        -_A(x1,x2) = max{x1, x2}
  
    precedence:
    
      D# > * = pow = ln = div = minus = -
    
    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]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: D#(pow(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)

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#(pow(x,y)) -> D#(x)
p2: D#(-(x,y)) -> D#(x)
p3: D#(-(x,y)) -> D#(y)
p4: D#(minus(x)) -> D#(x)
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 monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        D#_A(x1) = x1 + 3
        pow_A(x1,x2) = max{x1, x2}
        -_A(x1,x2) = max{x1 + 2, x2 + 2}
        minus_A(x1) = x1
        div_A(x1,x2) = max{x1, x2}
        ln_A(x1) = x1
  
    precedence:
    
      D# = pow = - = minus = div = ln
    
    partial status:
  
      pi(D#) = [1]
      pi(pow) = [1, 2]
      pi(-) = [1, 2]
      pi(minus) = [1]
      pi(div) = [1, 2]
      pi(ln) = [1]

The next rules are strictly ordered:

  p1

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)
p6: 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, p6}


-- 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)

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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        D#_A(x1) = max{2, x1 - 3}
        -_A(x1,x2) = max{x1 + 2, x2 + 2}
        ln_A(x1) = x1 + 1
        div_A(x1,x2) = max{x1 + 6, x2 + 2}
        minus_A(x1) = x1
  
    precedence:
    
      D# = - = ln = div = minus
    
    partial status:
  
      pi(D#) = []
      pi(-) = [2]
      pi(ln) = [1]
      pi(div) = [2]
      pi(minus) = [1]

The next rules are strictly ordered:

  p3

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#(x)
p4: D#(minus(x)) -> 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 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#(minus(x)) -> D#(x)
p4: D#(div(x,y)) -> D#(x)
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 set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        D#_A(x1) = x1 + 2
        -_A(x1,x2) = max{x1, x2}
        minus_A(x1) = x1 + 2
        div_A(x1,x2) = max{x1, x2}
        ln_A(x1) = x1
  
    precedence:
    
      D# = - = minus = div = ln
    
    partial status:
  
      pi(D#) = [1]
      pi(-) = [1, 2]
      pi(minus) = [1]
      pi(div) = [1, 2]
      pi(ln) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

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


-- Reduction pair.

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

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

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        D#_A(x1) = x1 + 2
        -_A(x1,x2) = x2
        ln_A(x1) = x1
        div_A(x1,x2) = max{x1, x2}
        minus_A(x1) = x1 + 1
  
    precedence:
    
      D# = - = ln = div = minus
    
    partial status:
  
      pi(D#) = [1]
      pi(-) = [2]
      pi(ln) = [1]
      pi(div) = [1, 2]
      pi(minus) = [1]

The next rules are strictly ordered:

  p1

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


-- 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)
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 set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        D#_A(x1) = x1 + 2
        ln_A(x1) = x1
        minus_A(x1) = x1
        div_A(x1,x2) = max{x1 + 1, x2 + 1}
  
    precedence:
    
      D# = ln = minus = div
    
    partial status:
  
      pi(D#) = [1]
      pi(ln) = [1]
      pi(minus) = [1]
      pi(div) = [1, 2]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

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


-- Reduction pair.

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

p1: D#(minus(x)) -> D#(x)
p2: 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 set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        D#_A(x1) = x1 + 1
        minus_A(x1) = x1
        div_A(x1,x2) = max{x1, x2}
  
    precedence:
    
      D# = minus = div
    
    partial status:
  
      pi(D#) = [1]
      pi(minus) = [1]
      pi(div) = [1, 2]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: 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}


-- Reduction pair.

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

p1: 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 set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        D#_A(x1) = x1 + 3
        div_A(x1,x2) = max{x1 + 1, x2}
  
    precedence:
    
      D# = div
    
    partial status:
  
      pi(D#) = []
      pi(div) = [2]

The next rules are strictly ordered:

  p1

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