YES

We show the termination of the TRS R:

  D(t()) -> s(h())
  D(constant()) -> h()
  D(b(x,y)) -> b(D(x),D(y))
  D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
  D(m(x,y)) -> m(D(x),D(y))
  D(opp(x)) -> opp(D(x))
  D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
  D(ln(x)) -> div(D(x),x)
  D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
  b(h(),x) -> x
  b(x,h()) -> x
  b(s(x),s(y)) -> s(s(b(x,y)))
  b(b(x,y),z) -> b(x,b(y,z))

-- SCC decomposition.

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

p1: D#(b(x,y)) -> b#(D(x),D(y))
p2: D#(b(x,y)) -> D#(x)
p3: D#(b(x,y)) -> D#(y)
p4: D#(c(x,y)) -> b#(c(y,D(x)),c(x,D(y)))
p5: D#(c(x,y)) -> D#(x)
p6: D#(c(x,y)) -> D#(y)
p7: D#(m(x,y)) -> D#(x)
p8: D#(m(x,y)) -> D#(y)
p9: D#(opp(x)) -> D#(x)
p10: D#(div(x,y)) -> D#(x)
p11: D#(div(x,y)) -> D#(y)
p12: D#(ln(x)) -> D#(x)
p13: D#(pow(x,y)) -> b#(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
p14: D#(pow(x,y)) -> D#(x)
p15: D#(pow(x,y)) -> D#(y)
p16: b#(s(x),s(y)) -> b#(x,y)
p17: b#(b(x,y),z) -> b#(x,b(y,z))
p18: b#(b(x,y),z) -> b#(y,z)

and R consists of:

r1: D(t()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

The estimated dependency graph contains the following SCCs:

  {p2, p3, p5, p6, p7, p8, p9, p10, p11, p12, p14, p15}
  {p16, p17, p18}


-- Reduction pair.

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#(opp(x)) -> D#(x)
p7: D#(m(x,y)) -> D#(y)
p8: D#(m(x,y)) -> D#(x)
p9: D#(c(x,y)) -> D#(y)
p10: D#(c(x,y)) -> D#(x)
p11: D#(b(x,y)) -> D#(y)
p12: D#(b(x,y)) -> D#(x)

and R consists of:

r1: D(t()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

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 - 2}
        pow_A(x1,x2) = max{3, x1, x2 + 1}
        ln_A(x1) = max{3, x1}
        div_A(x1,x2) = max{3, x1, x2}
        opp_A(x1) = max{7, x1 + 1}
        m_A(x1,x2) = max{x1, x2}
        c_A(x1,x2) = max{x1 + 3, x2 + 1}
        b_A(x1,x2) = max{3, x1, x2}
  
    precedence:
    
      D# = pow = ln = div = opp = m = c = b
    
    partial status:
  
      pi(D#) = []
      pi(pow) = [2]
      pi(ln) = [1]
      pi(div) = [2]
      pi(opp) = [1]
      pi(m) = [2]
      pi(c) = [2]
      pi(b) = [2]

The next rules are strictly ordered:

  p6

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#(m(x,y)) -> D#(y)
p7: D#(m(x,y)) -> D#(x)
p8: D#(c(x,y)) -> D#(y)
p9: D#(c(x,y)) -> D#(x)
p10: D#(b(x,y)) -> D#(y)
p11: D#(b(x,y)) -> D#(x)

and R consists of:

r1: D(t()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

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#(pow(x,y)) -> D#(y)
p2: D#(b(x,y)) -> D#(x)
p3: D#(b(x,y)) -> D#(y)
p4: D#(c(x,y)) -> D#(x)
p5: D#(c(x,y)) -> D#(y)
p6: D#(m(x,y)) -> D#(x)
p7: D#(m(x,y)) -> D#(y)
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)

and R consists of:

r1: D(t()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

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}
        b_A(x1,x2) = max{x1 + 1, x2}
        c_A(x1,x2) = max{x1, x2}
        m_A(x1,x2) = max{x1 + 1, x2}
        div_A(x1,x2) = max{x1, x2}
        ln_A(x1) = x1
  
    precedence:
    
      D# = pow = b = c = m = div = ln
    
    partial status:
  
      pi(D#) = []
      pi(pow) = [2]
      pi(b) = [2]
      pi(c) = [2]
      pi(m) = [2]
      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#(b(x,y)) -> D#(x)
p2: D#(b(x,y)) -> D#(y)
p3: D#(c(x,y)) -> D#(x)
p4: D#(c(x,y)) -> D#(y)
p5: D#(m(x,y)) -> D#(x)
p6: D#(m(x,y)) -> D#(y)
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)

and R consists of:

r1: D(t()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

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#(b(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#(m(x,y)) -> D#(y)
p7: D#(m(x,y)) -> D#(x)
p8: D#(c(x,y)) -> D#(y)
p9: D#(c(x,y)) -> D#(x)
p10: D#(b(x,y)) -> D#(y)

and R consists of:

r1: D(t()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

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}
        b_A(x1,x2) = max{7, x1, x2 + 3}
        pow_A(x1,x2) = max{x1, x2 + 2}
        ln_A(x1) = max{1, x1}
        div_A(x1,x2) = max{1, x1, x2}
        m_A(x1,x2) = max{x1 + 1, x2}
        c_A(x1,x2) = max{1, x1, x2}
  
    precedence:
    
      pow > D# = b = ln = div = m = c
    
    partial status:
  
      pi(D#) = []
      pi(b) = [2]
      pi(pow) = [2]
      pi(ln) = [1]
      pi(div) = [2]
      pi(m) = [2]
      pi(c) = [2]

The next rules are strictly ordered:

  p10

We remove them from the problem.

-- SCC decomposition.

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

p1: D#(b(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#(m(x,y)) -> D#(y)
p7: D#(m(x,y)) -> D#(x)
p8: D#(c(x,y)) -> D#(y)
p9: D#(c(x,y)) -> D#(x)

and R consists of:

r1: D(t()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

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#(b(x,y)) -> D#(x)
p2: D#(c(x,y)) -> D#(x)
p3: D#(c(x,y)) -> D#(y)
p4: D#(m(x,y)) -> D#(x)
p5: D#(m(x,y)) -> D#(y)
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()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

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}
        b_A(x1,x2) = max{x1, x2}
        c_A(x1,x2) = max{x1, x2 + 2}
        m_A(x1,x2) = max{x1, x2}
        div_A(x1,x2) = max{x1, x2}
        ln_A(x1) = x1 + 1
        pow_A(x1,x2) = max{1, x1, x2}
  
    precedence:
    
      D# = b = c = m = div = ln = pow
    
    partial status:
  
      pi(D#) = []
      pi(b) = [2]
      pi(c) = [2]
      pi(m) = [2]
      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#(b(x,y)) -> D#(x)
p2: D#(c(x,y)) -> D#(x)
p3: D#(m(x,y)) -> D#(x)
p4: D#(m(x,y)) -> D#(y)
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()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

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#(b(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#(m(x,y)) -> D#(y)
p7: D#(m(x,y)) -> D#(x)
p8: D#(c(x,y)) -> D#(x)

and R consists of:

r1: D(t()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

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 + 4
        b_A(x1,x2) = max{x1, x2}
        pow_A(x1,x2) = max{x1 + 1, x2 - 1}
        ln_A(x1) = x1 + 1
        div_A(x1,x2) = max{x1, x2}
        m_A(x1,x2) = max{x1 + 3, x2 + 5}
        c_A(x1,x2) = max{x1, x2 + 1}
  
    precedence:
    
      div = m > D# = b = ln = c > pow
    
    partial status:
  
      pi(D#) = [1]
      pi(b) = [1, 2]
      pi(pow) = []
      pi(ln) = [1]
      pi(div) = [1, 2]
      pi(m) = [1, 2]
      pi(c) = [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#(m(x,y)) -> D#(y)
p6: D#(m(x,y)) -> D#(x)
p7: D#(c(x,y)) -> D#(x)

and R consists of:

r1: D(t()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

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#(c(x,y)) -> D#(x)
p3: D#(m(x,y)) -> D#(x)
p4: D#(m(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()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

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
        pow_A(x1,x2) = max{x1, x2}
        c_A(x1,x2) = max{x1, x2}
        m_A(x1,x2) = max{x1, x2}
        div_A(x1,x2) = max{x1 + 1, x2 + 1}
        ln_A(x1) = x1
  
    precedence:
    
      D# = pow = c = m = div = ln
    
    partial status:
  
      pi(D#) = [1]
      pi(pow) = [1, 2]
      pi(c) = [1, 2]
      pi(m) = [1, 2]
      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#(c(x,y)) -> D#(x)
p2: D#(m(x,y)) -> D#(x)
p3: D#(m(x,y)) -> D#(y)
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()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

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#(c(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#(m(x,y)) -> D#(y)
p6: D#(m(x,y)) -> D#(x)

and R consists of:

r1: D(t()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

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
        c_A(x1,x2) = max{x1 + 4, x2 + 2}
        ln_A(x1) = x1 + 4
        div_A(x1,x2) = max{x1 + 2, x2 + 4}
        m_A(x1,x2) = max{x1 + 4, x2 + 4}
  
    precedence:
    
      div > D# = c = ln = m
    
    partial status:
  
      pi(D#) = [1]
      pi(c) = [1, 2]
      pi(ln) = [1]
      pi(div) = [1, 2]
      pi(m) = [1, 2]

The next rules are strictly ordered:

  p6

We remove them from the problem.

-- SCC decomposition.

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

p1: D#(c(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#(m(x,y)) -> D#(y)

and R consists of:

r1: D(t()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

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#(c(x,y)) -> D#(x)
p2: D#(m(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()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

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}
        c_A(x1,x2) = max{x1 + 1, x2 + 1}
        m_A(x1,x2) = x2
        div_A(x1,x2) = max{x1, x2 + 2}
        ln_A(x1) = x1
  
    precedence:
    
      D# = c = m = div = ln
    
    partial status:
  
      pi(D#) = []
      pi(c) = [2]
      pi(m) = [2]
      pi(div) = [2]
      pi(ln) = [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#(c(x,y)) -> D#(x)
p2: D#(m(x,y)) -> D#(y)
p3: D#(div(x,y)) -> D#(x)
p4: D#(ln(x)) -> D#(x)

and R consists of:

r1: D(t()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

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

and R consists of:

r1: D(t()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

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{3, x1 + 2}
        c_A(x1,x2) = max{x1, x2}
        ln_A(x1) = x1
        div_A(x1,x2) = max{x1 + 3, x2 + 3}
        m_A(x1,x2) = max{x1 + 1, x2}
  
    precedence:
    
      D# = c = ln = div = m
    
    partial status:
  
      pi(D#) = []
      pi(c) = [2]
      pi(ln) = [1]
      pi(div) = [2]
      pi(m) = [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#(c(x,y)) -> D#(x)
p2: D#(ln(x)) -> D#(x)
p3: D#(m(x,y)) -> D#(y)

and R consists of:

r1: D(t()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

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#(c(x,y)) -> D#(x)
p2: D#(m(x,y)) -> D#(y)
p3: D#(ln(x)) -> D#(x)

and R consists of:

r1: D(t()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

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
        c_A(x1,x2) = max{x1, x2}
        m_A(x1,x2) = x2
        ln_A(x1) = x1
  
    precedence:
    
      c = m = ln > D#
    
    partial status:
  
      pi(D#) = [1]
      pi(c) = [1, 2]
      pi(m) = [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#(m(x,y)) -> D#(y)
p2: D#(ln(x)) -> D#(x)

and R consists of:

r1: D(t()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

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

and R consists of:

r1: D(t()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

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{5, x1 + 2}
        m_A(x1,x2) = x2 + 3
        ln_A(x1) = max{3, x1}
  
    precedence:
    
      D# = m = ln
    
    partial status:
  
      pi(D#) = [1]
      pi(m) = [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#(ln(x)) -> D#(x)

and R consists of:

r1: D(t()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

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()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

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 + 2
  
    precedence:
    
      D# = ln
    
    partial status:
  
      pi(D#) = [1]
      pi(ln) = [1]

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: b#(s(x),s(y)) -> b#(x,y)
p2: b#(b(x,y),z) -> b#(y,z)
p3: b#(b(x,y),z) -> b#(x,b(y,z))

and R consists of:

r1: D(t()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

The set of usable rules consists of

  r10, r11, r12, r13

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        b#_A(x1,x2) = max{x1 + 3, x2 + 2}
        s_A(x1) = x1
        b_A(x1,x2) = max{x1 + 1, x2}
        h_A = 0
  
    precedence:
    
      s = b > b# = h
    
    partial status:
  
      pi(b#) = [1, 2]
      pi(s) = []
      pi(b) = []
      pi(h) = []

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: b#(s(x),s(y)) -> b#(x,y)
p2: b#(b(x,y),z) -> b#(y,z)

and R consists of:

r1: D(t()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: b#(s(x),s(y)) -> b#(x,y)
p2: b#(b(x,y),z) -> b#(y,z)

and R consists of:

r1: D(t()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        b#_A(x1,x2) = max{0, x1 - 2}
        s_A(x1) = max{2, x1}
        b_A(x1,x2) = max{3, x1 - 1, x2 + 1}
  
    precedence:
    
      b# = s = b
    
    partial status:
  
      pi(b#) = []
      pi(s) = [1]
      pi(b) = [2]

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: b#(s(x),s(y)) -> b#(x,y)

and R consists of:

r1: D(t()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: b#(s(x),s(y)) -> b#(x,y)

and R consists of:

r1: D(t()) -> s(h())
r2: D(constant()) -> h()
r3: D(b(x,y)) -> b(D(x),D(y))
r4: D(c(x,y)) -> b(c(y,D(x)),c(x,D(y)))
r5: D(m(x,y)) -> m(D(x),D(y))
r6: D(opp(x)) -> opp(D(x))
r7: D(div(x,y)) -> m(div(D(x),y),div(c(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> b(c(c(y,pow(x,m(y,|1|()))),D(x)),c(c(pow(x,y),ln(x)),D(y)))
r10: b(h(),x) -> x
r11: b(x,h()) -> x
r12: b(s(x),s(y)) -> s(s(b(x,y)))
r13: b(b(x,y),z) -> b(x,b(y,z))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        b#_A(x1,x2) = max{0, x1 - 2, x2 - 2}
        s_A(x1) = max{3, x1 + 1}
  
    precedence:
    
      b# = s
    
    partial status:
  
      pi(b#) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

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