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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          D#_A(x1) = ((1,0),(0,0)) x1
          pow_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (1,1)
          ln_A(x1) = ((1,1),(1,1)) x1 + (1,1)
          div_A(x1,x2) = x1 + x2 + (1,1)
          opp_A(x1) = x1 + (1,1)
          m_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (1,1)
          c_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (1,1)
          b_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (0,1)
  
    precedence:
    
      m > opp > c > div > pow > D# = ln = b
    
    partial status:
  
      pi(D#) = []
      pi(pow) = []
      pi(ln) = [1]
      pi(div) = []
      pi(opp) = []
      pi(m) = []
      pi(c) = []
      pi(b) = [1]

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#(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#(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#(y)
p5: D#(m(x,y)) -> D#(x)
p6: D#(m(x,y)) -> D#(y)
p7: D#(opp(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)

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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          D#_A(x1) = ((0,1),(0,0)) x1 + (2,2)
          pow_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,0),(0,1)) x2 + (1,1)
          b_A(x1,x2) = x1 + ((1,1),(1,1)) x2 + (3,2)
          c_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,0),(0,1)) x2 + (1,1)
          m_A(x1,x2) = ((0,0),(0,1)) x1 + x2 + (1,1)
          opp_A(x1) = ((0,0),(0,1)) x1 + (1,1)
          div_A(x1,x2) = ((0,0),(0,1)) x1 + x2 + (1,1)
          ln_A(x1) = ((0,0),(0,1)) x1 + (1,1)
  
    precedence:
    
      m > opp > b > c > D# = pow = div > ln
    
    partial status:
  
      pi(D#) = []
      pi(pow) = []
      pi(b) = [1]
      pi(c) = []
      pi(m) = []
      pi(opp) = []
      pi(div) = [2]
      pi(ln) = []

The next rules are strictly ordered:

  p11

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

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#(pow(x,y)) -> D#(y)
p2: D#(ln(x)) -> D#(x)
p3: D#(div(x,y)) -> D#(y)
p4: D#(div(x,y)) -> D#(x)
p5: D#(opp(x)) -> 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#(b(x,y)) -> D#(y)
p10: 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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          D#_A(x1) = ((0,1),(0,1)) x1 + (2,2)
          pow_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (3,2)
          ln_A(x1) = ((1,1),(0,1)) x1 + (3,2)
          div_A(x1,x2) = ((0,0),(1,1)) x1 + ((1,1),(1,1)) x2 + (3,2)
          opp_A(x1) = x1 + (1,1)
          m_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,0),(0,1)) x2 + (1,1)
          c_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,0),(0,1)) x2 + (1,1)
          b_A(x1,x2) = ((0,0),(1,1)) x1 + ((0,0),(0,1)) x2 + (1,1)
  
    precedence:
    
      div > D# = pow = opp > ln > m = c = b
    
    partial status:
  
      pi(D#) = []
      pi(pow) = [2]
      pi(ln) = [1]
      pi(div) = [2]
      pi(opp) = []
      pi(m) = []
      pi(c) = []
      pi(b) = []

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#(y)
p3: D#(div(x,y)) -> D#(x)
p4: D#(opp(x)) -> D#(x)
p5: D#(m(x,y)) -> D#(y)
p6: D#(m(x,y)) -> D#(x)
p7: D#(c(x,y)) -> D#(y)
p8: D#(b(x,y)) -> D#(y)
p9: 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}


-- Reduction pair.

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

p1: D#(ln(x)) -> D#(x)
p2: D#(b(x,y)) -> D#(x)
p3: D#(b(x,y)) -> D#(y)
p4: D#(c(x,y)) -> D#(y)
p5: D#(m(x,y)) -> D#(x)
p6: D#(m(x,y)) -> D#(y)
p7: D#(opp(x)) -> D#(x)
p8: D#(div(x,y)) -> D#(x)
p9: D#(div(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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          D#_A(x1) = ((0,1),(0,0)) x1 + (2,2)
          ln_A(x1) = ((0,0),(0,1)) x1 + (1,1)
          b_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (3,2)
          c_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,0),(0,1)) x2 + (1,1)
          m_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,0),(0,1)) x2 + (1,1)
          opp_A(x1) = ((0,0),(0,1)) x1 + (1,1)
          div_A(x1,x2) = ((1,0),(1,1)) x1 + ((0,0),(0,1)) x2 + (3,1)
  
    precedence:
    
      c > m > b > opp > D# = ln = div
    
    partial status:
  
      pi(D#) = []
      pi(ln) = []
      pi(b) = []
      pi(c) = [1]
      pi(m) = []
      pi(opp) = []
      pi(div) = [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#(ln(x)) -> D#(x)
p2: D#(b(x,y)) -> D#(x)
p3: D#(b(x,y)) -> D#(y)
p4: D#(m(x,y)) -> D#(x)
p5: D#(m(x,y)) -> D#(y)
p6: D#(opp(x)) -> D#(x)
p7: D#(div(x,y)) -> D#(x)
p8: D#(div(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, p6, p7, p8}


-- Reduction pair.

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

p1: D#(ln(x)) -> D#(x)
p2: D#(div(x,y)) -> D#(y)
p3: D#(div(x,y)) -> D#(x)
p4: D#(opp(x)) -> D#(x)
p5: D#(m(x,y)) -> D#(y)
p6: D#(m(x,y)) -> D#(x)
p7: D#(b(x,y)) -> D#(y)
p8: 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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          D#_A(x1) = ((1,1),(0,0)) x1 + (2,2)
          ln_A(x1) = ((1,1),(0,0)) x1 + (1,1)
          div_A(x1,x2) = ((1,1),(0,0)) x1 + x2 + (1,1)
          opp_A(x1) = ((1,1),(0,0)) x1 + (1,1)
          m_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,1),(0,0)) x2 + (1,1)
          b_A(x1,x2) = ((0,0),(1,1)) x1 + ((0,0),(1,1)) x2 + (3,2)
  
    precedence:
    
      div > m > opp > D# = b > ln
    
    partial status:
  
      pi(D#) = []
      pi(ln) = []
      pi(div) = []
      pi(opp) = []
      pi(m) = []
      pi(b) = []

The next rules are strictly ordered:

  p5

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#(y)
p3: D#(div(x,y)) -> D#(x)
p4: D#(opp(x)) -> D#(x)
p5: D#(m(x,y)) -> D#(x)
p6: D#(b(x,y)) -> D#(y)
p7: 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}


-- Reduction pair.

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

p1: D#(ln(x)) -> D#(x)
p2: D#(b(x,y)) -> D#(x)
p3: D#(b(x,y)) -> D#(y)
p4: D#(m(x,y)) -> D#(x)
p5: D#(opp(x)) -> D#(x)
p6: D#(div(x,y)) -> D#(x)
p7: D#(div(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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          D#_A(x1) = ((0,1),(0,0)) x1 + (2,2)
          ln_A(x1) = x1 + (1,1)
          b_A(x1,x2) = ((1,1),(0,1)) x1 + ((0,0),(0,1)) x2 + (3,2)
          m_A(x1,x2) = x1 + ((1,1),(1,1)) x2 + (1,1)
          opp_A(x1) = x1 + (1,1)
          div_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,0),(1,1)) x2 + (1,1)
  
    precedence:
    
      D# > b > m > ln > opp = div
    
    partial status:
  
      pi(D#) = []
      pi(ln) = []
      pi(b) = []
      pi(m) = []
      pi(opp) = []
      pi(div) = []

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#(ln(x)) -> D#(x)
p2: D#(b(x,y)) -> D#(x)
p3: D#(b(x,y)) -> D#(y)
p4: D#(opp(x)) -> D#(x)
p5: D#(div(x,y)) -> D#(x)
p6: D#(div(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, p6}


-- Reduction pair.

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

p1: D#(ln(x)) -> D#(x)
p2: D#(div(x,y)) -> D#(y)
p3: D#(div(x,y)) -> D#(x)
p4: D#(opp(x)) -> D#(x)
p5: D#(b(x,y)) -> D#(y)
p6: 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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          D#_A(x1) = ((0,1),(0,1)) x1 + (2,2)
          ln_A(x1) = x1 + (1,1)
          div_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,0),(1,1)) x2 + (1,1)
          opp_A(x1) = ((0,0),(0,1)) x1 + (1,1)
          b_A(x1,x2) = ((0,0),(0,1)) x1 + ((1,0),(1,1)) x2 + (1,1)
  
    precedence:
    
      D# = div > opp > ln > b
    
    partial status:
  
      pi(D#) = []
      pi(ln) = []
      pi(div) = []
      pi(opp) = []
      pi(b) = [2]

The next rules are strictly ordered:

  p5

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


-- Reduction pair.

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

p1: D#(ln(x)) -> D#(x)
p2: D#(b(x,y)) -> D#(x)
p3: D#(opp(x)) -> D#(x)
p4: D#(div(x,y)) -> D#(x)
p5: D#(div(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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          D#_A(x1) = ((0,1),(0,1)) x1
          ln_A(x1) = x1 + (1,1)
          b_A(x1,x2) = ((0,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
          opp_A(x1) = ((0,0),(0,1)) x1 + (1,1)
          div_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,0),(0,1)) x2 + (1,1)
  
    precedence:
    
      ln > D# = div > b = opp
    
    partial status:
  
      pi(D#) = []
      pi(ln) = []
      pi(b) = []
      pi(opp) = []
      pi(div) = []

The next rules are strictly ordered:

  p5

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#(b(x,y)) -> D#(x)
p3: D#(opp(x)) -> D#(x)
p4: D#(div(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}


-- Reduction pair.

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#(opp(x)) -> D#(x)
p4: 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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          D#_A(x1) = ((0,1),(0,1)) x1 + (2,2)
          ln_A(x1) = ((1,1),(0,1)) x1 + (3,2)
          div_A(x1,x2) = x1 + ((1,1),(1,1)) x2 + (1,1)
          opp_A(x1) = ((0,0),(0,1)) x1 + (1,1)
          b_A(x1,x2) = ((0,0),(0,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
  
    precedence:
    
      ln = div = opp > D# = b
    
    partial status:
  
      pi(D#) = []
      pi(ln) = [1]
      pi(div) = []
      pi(opp) = []
      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: D#(ln(x)) -> D#(x)
p2: D#(opp(x)) -> D#(x)
p3: 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}


-- Reduction pair.

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

p1: D#(ln(x)) -> D#(x)
p2: D#(b(x,y)) -> D#(x)
p3: D#(opp(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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          D#_A(x1) = ((0,1),(0,0)) x1 + (2,2)
          ln_A(x1) = x1 + (1,1)
          b_A(x1,x2) = ((0,0),(0,1)) x1 + ((1,1),(1,1)) x2 + (0,1)
          opp_A(x1) = ((0,0),(0,1)) x1 + (1,1)
  
    precedence:
    
      D# = ln = b = opp
    
    partial status:
  
      pi(D#) = []
      pi(ln) = [1]
      pi(b) = [2]
      pi(opp) = []

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#(ln(x)) -> D#(x)
p2: D#(opp(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#(ln(x)) -> D#(x)
p2: D#(opp(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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          D#_A(x1) = ((0,1),(0,0)) x1 + (2,2)
          ln_A(x1) = ((0,0),(0,1)) x1 + (1,1)
          opp_A(x1) = ((0,1),(0,1)) x1 + (1,1)
  
    precedence:
    
      D# = ln = opp
    
    partial status:
  
      pi(D#) = []
      pi(ln) = []
      pi(opp) = []

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#(opp(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#(opp(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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          D#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
          opp_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      D# = opp
    
    partial status:
  
      pi(D#) = []
      pi(opp) = []

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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          b#_A(x1,x2) = ((0,1),(0,0)) x1 + (1,1)
          s_A(x1) = ((0,1),(0,1)) x1 + (1,1)
          b_A(x1,x2) = ((1,1),(0,1)) x1 + x2 + (4,2)
          h_A() = (1,1)
  
    precedence:
    
      h > b# = s = b
    
    partial status:
  
      pi(b#) = []
      pi(s) = []
      pi(b) = [1, 2]
      pi(h) = []

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)
p2: 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 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#(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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          b#_A(x1,x2) = ((1,1),(1,1)) x1 + (0,2)
          s_A(x1) = x1 + (1,1)
          b_A(x1,x2) = ((1,1),(1,1)) x1 + x2 + (2,2)
          h_A() = (1,1)
  
    precedence:
    
      s = b > b# = h
    
    partial status:
  
      pi(b#) = [1]
      pi(s) = []
      pi(b) = []
      pi(h) = []

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: 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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: 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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          b#_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,1),(0,1)) x2 + (1,1)
          b_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (1,1)
          h_A() = (1,1)
          s_A(x1) = (0,0)
  
    precedence:
    
      b > b# = h > s
    
    partial status:
  
      pi(b#) = [1]
      pi(b) = []
      pi(h) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

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