YES

We show the termination of the TRS R:

  dx(X) -> one()
  dx(a()) -> zero()
  dx(plus(ALPHA,BETA)) -> plus(dx(ALPHA),dx(BETA))
  dx(times(ALPHA,BETA)) -> plus(times(BETA,dx(ALPHA)),times(ALPHA,dx(BETA)))
  dx(minus(ALPHA,BETA)) -> minus(dx(ALPHA),dx(BETA))
  dx(neg(ALPHA)) -> neg(dx(ALPHA))
  dx(div(ALPHA,BETA)) -> minus(div(dx(ALPHA),BETA),times(ALPHA,div(dx(BETA),exp(BETA,two()))))
  dx(ln(ALPHA)) -> div(dx(ALPHA),ALPHA)
  dx(exp(ALPHA,BETA)) -> plus(times(BETA,times(exp(ALPHA,minus(BETA,one())),dx(ALPHA))),times(exp(ALPHA,BETA),times(ln(ALPHA),dx(BETA))))

-- SCC decomposition.

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

p1: dx#(plus(ALPHA,BETA)) -> dx#(ALPHA)
p2: dx#(plus(ALPHA,BETA)) -> dx#(BETA)
p3: dx#(times(ALPHA,BETA)) -> dx#(ALPHA)
p4: dx#(times(ALPHA,BETA)) -> dx#(BETA)
p5: dx#(minus(ALPHA,BETA)) -> dx#(ALPHA)
p6: dx#(minus(ALPHA,BETA)) -> dx#(BETA)
p7: dx#(neg(ALPHA)) -> dx#(ALPHA)
p8: dx#(div(ALPHA,BETA)) -> dx#(ALPHA)
p9: dx#(div(ALPHA,BETA)) -> dx#(BETA)
p10: dx#(ln(ALPHA)) -> dx#(ALPHA)
p11: dx#(exp(ALPHA,BETA)) -> dx#(ALPHA)
p12: dx#(exp(ALPHA,BETA)) -> dx#(BETA)

and R consists of:

r1: dx(X) -> one()
r2: dx(a()) -> zero()
r3: dx(plus(ALPHA,BETA)) -> plus(dx(ALPHA),dx(BETA))
r4: dx(times(ALPHA,BETA)) -> plus(times(BETA,dx(ALPHA)),times(ALPHA,dx(BETA)))
r5: dx(minus(ALPHA,BETA)) -> minus(dx(ALPHA),dx(BETA))
r6: dx(neg(ALPHA)) -> neg(dx(ALPHA))
r7: dx(div(ALPHA,BETA)) -> minus(div(dx(ALPHA),BETA),times(ALPHA,div(dx(BETA),exp(BETA,two()))))
r8: dx(ln(ALPHA)) -> div(dx(ALPHA),ALPHA)
r9: dx(exp(ALPHA,BETA)) -> plus(times(BETA,times(exp(ALPHA,minus(BETA,one())),dx(ALPHA))),times(exp(ALPHA,BETA),times(ln(ALPHA),dx(BETA))))

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: dx#(plus(ALPHA,BETA)) -> dx#(ALPHA)
p2: dx#(exp(ALPHA,BETA)) -> dx#(BETA)
p3: dx#(exp(ALPHA,BETA)) -> dx#(ALPHA)
p4: dx#(ln(ALPHA)) -> dx#(ALPHA)
p5: dx#(div(ALPHA,BETA)) -> dx#(BETA)
p6: dx#(div(ALPHA,BETA)) -> dx#(ALPHA)
p7: dx#(neg(ALPHA)) -> dx#(ALPHA)
p8: dx#(minus(ALPHA,BETA)) -> dx#(BETA)
p9: dx#(minus(ALPHA,BETA)) -> dx#(ALPHA)
p10: dx#(times(ALPHA,BETA)) -> dx#(BETA)
p11: dx#(times(ALPHA,BETA)) -> dx#(ALPHA)
p12: dx#(plus(ALPHA,BETA)) -> dx#(BETA)

and R consists of:

r1: dx(X) -> one()
r2: dx(a()) -> zero()
r3: dx(plus(ALPHA,BETA)) -> plus(dx(ALPHA),dx(BETA))
r4: dx(times(ALPHA,BETA)) -> plus(times(BETA,dx(ALPHA)),times(ALPHA,dx(BETA)))
r5: dx(minus(ALPHA,BETA)) -> minus(dx(ALPHA),dx(BETA))
r6: dx(neg(ALPHA)) -> neg(dx(ALPHA))
r7: dx(div(ALPHA,BETA)) -> minus(div(dx(ALPHA),BETA),times(ALPHA,div(dx(BETA),exp(BETA,two()))))
r8: dx(ln(ALPHA)) -> div(dx(ALPHA),ALPHA)
r9: dx(exp(ALPHA,BETA)) -> plus(times(BETA,times(exp(ALPHA,minus(BETA,one())),dx(ALPHA))),times(exp(ALPHA,BETA),times(ln(ALPHA),dx(BETA))))

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:
          dx#_A(x1) = ((1,1),(1,1)) x1 + (2,2)
          plus_A(x1,x2) = x1 + ((1,1),(0,0)) x2 + (1,1)
          exp_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,1),(0,0)) x2 + (1,1)
          ln_A(x1) = ((1,1),(0,0)) x1 + (1,1)
          div_A(x1,x2) = ((1,1),(0,0)) x1 + x2 + (1,1)
          neg_A(x1) = x1 + (0,1)
          minus_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,1),(0,0)) x2 + (1,1)
          times_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,1),(0,1)) x2 + (1,1)
  
    precedence:
    
      neg > minus > ln > dx# = plus = exp = div = times
    
    partial status:
  
      pi(dx#) = [1]
      pi(plus) = [1]
      pi(exp) = []
      pi(ln) = []
      pi(div) = [2]
      pi(neg) = [1]
      pi(minus) = []
      pi(times) = [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: dx#(exp(ALPHA,BETA)) -> dx#(BETA)
p2: dx#(exp(ALPHA,BETA)) -> dx#(ALPHA)
p3: dx#(ln(ALPHA)) -> dx#(ALPHA)
p4: dx#(div(ALPHA,BETA)) -> dx#(BETA)
p5: dx#(div(ALPHA,BETA)) -> dx#(ALPHA)
p6: dx#(neg(ALPHA)) -> dx#(ALPHA)
p7: dx#(minus(ALPHA,BETA)) -> dx#(BETA)
p8: dx#(minus(ALPHA,BETA)) -> dx#(ALPHA)
p9: dx#(times(ALPHA,BETA)) -> dx#(BETA)
p10: dx#(times(ALPHA,BETA)) -> dx#(ALPHA)
p11: dx#(plus(ALPHA,BETA)) -> dx#(BETA)

and R consists of:

r1: dx(X) -> one()
r2: dx(a()) -> zero()
r3: dx(plus(ALPHA,BETA)) -> plus(dx(ALPHA),dx(BETA))
r4: dx(times(ALPHA,BETA)) -> plus(times(BETA,dx(ALPHA)),times(ALPHA,dx(BETA)))
r5: dx(minus(ALPHA,BETA)) -> minus(dx(ALPHA),dx(BETA))
r6: dx(neg(ALPHA)) -> neg(dx(ALPHA))
r7: dx(div(ALPHA,BETA)) -> minus(div(dx(ALPHA),BETA),times(ALPHA,div(dx(BETA),exp(BETA,two()))))
r8: dx(ln(ALPHA)) -> div(dx(ALPHA),ALPHA)
r9: dx(exp(ALPHA,BETA)) -> plus(times(BETA,times(exp(ALPHA,minus(BETA,one())),dx(ALPHA))),times(exp(ALPHA,BETA),times(ln(ALPHA),dx(BETA))))

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: dx#(exp(ALPHA,BETA)) -> dx#(BETA)
p2: dx#(plus(ALPHA,BETA)) -> dx#(BETA)
p3: dx#(times(ALPHA,BETA)) -> dx#(ALPHA)
p4: dx#(times(ALPHA,BETA)) -> dx#(BETA)
p5: dx#(minus(ALPHA,BETA)) -> dx#(ALPHA)
p6: dx#(minus(ALPHA,BETA)) -> dx#(BETA)
p7: dx#(neg(ALPHA)) -> dx#(ALPHA)
p8: dx#(div(ALPHA,BETA)) -> dx#(ALPHA)
p9: dx#(div(ALPHA,BETA)) -> dx#(BETA)
p10: dx#(ln(ALPHA)) -> dx#(ALPHA)
p11: dx#(exp(ALPHA,BETA)) -> dx#(ALPHA)

and R consists of:

r1: dx(X) -> one()
r2: dx(a()) -> zero()
r3: dx(plus(ALPHA,BETA)) -> plus(dx(ALPHA),dx(BETA))
r4: dx(times(ALPHA,BETA)) -> plus(times(BETA,dx(ALPHA)),times(ALPHA,dx(BETA)))
r5: dx(minus(ALPHA,BETA)) -> minus(dx(ALPHA),dx(BETA))
r6: dx(neg(ALPHA)) -> neg(dx(ALPHA))
r7: dx(div(ALPHA,BETA)) -> minus(div(dx(ALPHA),BETA),times(ALPHA,div(dx(BETA),exp(BETA,two()))))
r8: dx(ln(ALPHA)) -> div(dx(ALPHA),ALPHA)
r9: dx(exp(ALPHA,BETA)) -> plus(times(BETA,times(exp(ALPHA,minus(BETA,one())),dx(ALPHA))),times(exp(ALPHA,BETA),times(ln(ALPHA),dx(BETA))))

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:
          dx#_A(x1) = ((0,1),(0,0)) x1 + (2,2)
          exp_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,0),(0,1)) x2 + (1,1)
          plus_A(x1,x2) = ((1,1),(1,1)) x1 + x2 + (1,1)
          times_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,0),(0,1)) x2 + (3,2)
          minus_A(x1,x2) = ((0,0),(0,1)) x1 + x2 + (1,1)
          neg_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:
    
      minus > neg > dx# = exp = times > plus > div = ln
    
    partial status:
  
      pi(dx#) = []
      pi(exp) = []
      pi(plus) = []
      pi(times) = []
      pi(minus) = []
      pi(neg) = []
      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: dx#(exp(ALPHA,BETA)) -> dx#(BETA)
p2: dx#(plus(ALPHA,BETA)) -> dx#(BETA)
p3: dx#(times(ALPHA,BETA)) -> dx#(ALPHA)
p4: dx#(times(ALPHA,BETA)) -> dx#(BETA)
p5: dx#(minus(ALPHA,BETA)) -> dx#(ALPHA)
p6: dx#(minus(ALPHA,BETA)) -> dx#(BETA)
p7: dx#(neg(ALPHA)) -> dx#(ALPHA)
p8: dx#(div(ALPHA,BETA)) -> dx#(ALPHA)
p9: dx#(div(ALPHA,BETA)) -> dx#(BETA)
p10: dx#(ln(ALPHA)) -> dx#(ALPHA)

and R consists of:

r1: dx(X) -> one()
r2: dx(a()) -> zero()
r3: dx(plus(ALPHA,BETA)) -> plus(dx(ALPHA),dx(BETA))
r4: dx(times(ALPHA,BETA)) -> plus(times(BETA,dx(ALPHA)),times(ALPHA,dx(BETA)))
r5: dx(minus(ALPHA,BETA)) -> minus(dx(ALPHA),dx(BETA))
r6: dx(neg(ALPHA)) -> neg(dx(ALPHA))
r7: dx(div(ALPHA,BETA)) -> minus(div(dx(ALPHA),BETA),times(ALPHA,div(dx(BETA),exp(BETA,two()))))
r8: dx(ln(ALPHA)) -> div(dx(ALPHA),ALPHA)
r9: dx(exp(ALPHA,BETA)) -> plus(times(BETA,times(exp(ALPHA,minus(BETA,one())),dx(ALPHA))),times(exp(ALPHA,BETA),times(ln(ALPHA),dx(BETA))))

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: dx#(exp(ALPHA,BETA)) -> dx#(BETA)
p2: dx#(ln(ALPHA)) -> dx#(ALPHA)
p3: dx#(div(ALPHA,BETA)) -> dx#(BETA)
p4: dx#(div(ALPHA,BETA)) -> dx#(ALPHA)
p5: dx#(neg(ALPHA)) -> dx#(ALPHA)
p6: dx#(minus(ALPHA,BETA)) -> dx#(BETA)
p7: dx#(minus(ALPHA,BETA)) -> dx#(ALPHA)
p8: dx#(times(ALPHA,BETA)) -> dx#(BETA)
p9: dx#(times(ALPHA,BETA)) -> dx#(ALPHA)
p10: dx#(plus(ALPHA,BETA)) -> dx#(BETA)

and R consists of:

r1: dx(X) -> one()
r2: dx(a()) -> zero()
r3: dx(plus(ALPHA,BETA)) -> plus(dx(ALPHA),dx(BETA))
r4: dx(times(ALPHA,BETA)) -> plus(times(BETA,dx(ALPHA)),times(ALPHA,dx(BETA)))
r5: dx(minus(ALPHA,BETA)) -> minus(dx(ALPHA),dx(BETA))
r6: dx(neg(ALPHA)) -> neg(dx(ALPHA))
r7: dx(div(ALPHA,BETA)) -> minus(div(dx(ALPHA),BETA),times(ALPHA,div(dx(BETA),exp(BETA,two()))))
r8: dx(ln(ALPHA)) -> div(dx(ALPHA),ALPHA)
r9: dx(exp(ALPHA,BETA)) -> plus(times(BETA,times(exp(ALPHA,minus(BETA,one())),dx(ALPHA))),times(exp(ALPHA,BETA),times(ln(ALPHA),dx(BETA))))

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:
          dx#_A(x1) = ((1,1),(1,1)) x1 + (2,2)
          exp_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (3,2)
          ln_A(x1) = x1 + (3,1)
          div_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,1),(1,1)) x2 + (3,2)
          neg_A(x1) = ((0,1),(1,0)) x1 + (1,1)
          minus_A(x1,x2) = x1 + ((1,1),(0,0)) x2 + (1,1)
          times_A(x1,x2) = ((0,1),(1,0)) x1 + ((1,1),(0,0)) x2 + (1,1)
          plus_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,0),(1,1)) x2 + (3,1)
  
    precedence:
    
      dx# = div > ln > exp > neg = minus = times = plus
    
    partial status:
  
      pi(dx#) = []
      pi(exp) = [2]
      pi(ln) = [1]
      pi(div) = []
      pi(neg) = []
      pi(minus) = []
      pi(times) = []
      pi(plus) = []

The next rules are strictly ordered:

  p8

We remove them from the problem.

-- SCC decomposition.

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

p1: dx#(exp(ALPHA,BETA)) -> dx#(BETA)
p2: dx#(ln(ALPHA)) -> dx#(ALPHA)
p3: dx#(div(ALPHA,BETA)) -> dx#(BETA)
p4: dx#(div(ALPHA,BETA)) -> dx#(ALPHA)
p5: dx#(neg(ALPHA)) -> dx#(ALPHA)
p6: dx#(minus(ALPHA,BETA)) -> dx#(BETA)
p7: dx#(minus(ALPHA,BETA)) -> dx#(ALPHA)
p8: dx#(times(ALPHA,BETA)) -> dx#(ALPHA)
p9: dx#(plus(ALPHA,BETA)) -> dx#(BETA)

and R consists of:

r1: dx(X) -> one()
r2: dx(a()) -> zero()
r3: dx(plus(ALPHA,BETA)) -> plus(dx(ALPHA),dx(BETA))
r4: dx(times(ALPHA,BETA)) -> plus(times(BETA,dx(ALPHA)),times(ALPHA,dx(BETA)))
r5: dx(minus(ALPHA,BETA)) -> minus(dx(ALPHA),dx(BETA))
r6: dx(neg(ALPHA)) -> neg(dx(ALPHA))
r7: dx(div(ALPHA,BETA)) -> minus(div(dx(ALPHA),BETA),times(ALPHA,div(dx(BETA),exp(BETA,two()))))
r8: dx(ln(ALPHA)) -> div(dx(ALPHA),ALPHA)
r9: dx(exp(ALPHA,BETA)) -> plus(times(BETA,times(exp(ALPHA,minus(BETA,one())),dx(ALPHA))),times(exp(ALPHA,BETA),times(ln(ALPHA),dx(BETA))))

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: dx#(exp(ALPHA,BETA)) -> dx#(BETA)
p2: dx#(plus(ALPHA,BETA)) -> dx#(BETA)
p3: dx#(times(ALPHA,BETA)) -> dx#(ALPHA)
p4: dx#(minus(ALPHA,BETA)) -> dx#(ALPHA)
p5: dx#(minus(ALPHA,BETA)) -> dx#(BETA)
p6: dx#(neg(ALPHA)) -> dx#(ALPHA)
p7: dx#(div(ALPHA,BETA)) -> dx#(ALPHA)
p8: dx#(div(ALPHA,BETA)) -> dx#(BETA)
p9: dx#(ln(ALPHA)) -> dx#(ALPHA)

and R consists of:

r1: dx(X) -> one()
r2: dx(a()) -> zero()
r3: dx(plus(ALPHA,BETA)) -> plus(dx(ALPHA),dx(BETA))
r4: dx(times(ALPHA,BETA)) -> plus(times(BETA,dx(ALPHA)),times(ALPHA,dx(BETA)))
r5: dx(minus(ALPHA,BETA)) -> minus(dx(ALPHA),dx(BETA))
r6: dx(neg(ALPHA)) -> neg(dx(ALPHA))
r7: dx(div(ALPHA,BETA)) -> minus(div(dx(ALPHA),BETA),times(ALPHA,div(dx(BETA),exp(BETA,two()))))
r8: dx(ln(ALPHA)) -> div(dx(ALPHA),ALPHA)
r9: dx(exp(ALPHA,BETA)) -> plus(times(BETA,times(exp(ALPHA,minus(BETA,one())),dx(ALPHA))),times(exp(ALPHA,BETA),times(ln(ALPHA),dx(BETA))))

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:
          dx#_A(x1) = ((0,1),(0,0)) x1 + (2,2)
          exp_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (3,2)
          plus_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(0,1)) x2 + (3,2)
          times_A(x1,x2) = ((0,0),(0,1)) x1 + ((1,1),(1,1)) x2 + (3,1)
          minus_A(x1,x2) = ((0,0),(0,1)) x1 + x2 + (1,1)
          neg_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)
          ln_A(x1) = ((0,0),(0,1)) x1 + (1,1)
  
    precedence:
    
      times > dx# = exp > plus > minus = neg = div = ln
    
    partial status:
  
      pi(dx#) = []
      pi(exp) = [2]
      pi(plus) = [2]
      pi(times) = [2]
      pi(minus) = [2]
      pi(neg) = []
      pi(div) = []
      pi(ln) = []

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: dx#(exp(ALPHA,BETA)) -> dx#(BETA)
p2: dx#(plus(ALPHA,BETA)) -> dx#(BETA)
p3: dx#(minus(ALPHA,BETA)) -> dx#(ALPHA)
p4: dx#(minus(ALPHA,BETA)) -> dx#(BETA)
p5: dx#(neg(ALPHA)) -> dx#(ALPHA)
p6: dx#(div(ALPHA,BETA)) -> dx#(ALPHA)
p7: dx#(div(ALPHA,BETA)) -> dx#(BETA)
p8: dx#(ln(ALPHA)) -> dx#(ALPHA)

and R consists of:

r1: dx(X) -> one()
r2: dx(a()) -> zero()
r3: dx(plus(ALPHA,BETA)) -> plus(dx(ALPHA),dx(BETA))
r4: dx(times(ALPHA,BETA)) -> plus(times(BETA,dx(ALPHA)),times(ALPHA,dx(BETA)))
r5: dx(minus(ALPHA,BETA)) -> minus(dx(ALPHA),dx(BETA))
r6: dx(neg(ALPHA)) -> neg(dx(ALPHA))
r7: dx(div(ALPHA,BETA)) -> minus(div(dx(ALPHA),BETA),times(ALPHA,div(dx(BETA),exp(BETA,two()))))
r8: dx(ln(ALPHA)) -> div(dx(ALPHA),ALPHA)
r9: dx(exp(ALPHA,BETA)) -> plus(times(BETA,times(exp(ALPHA,minus(BETA,one())),dx(ALPHA))),times(exp(ALPHA,BETA),times(ln(ALPHA),dx(BETA))))

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: dx#(exp(ALPHA,BETA)) -> dx#(BETA)
p2: dx#(ln(ALPHA)) -> dx#(ALPHA)
p3: dx#(div(ALPHA,BETA)) -> dx#(BETA)
p4: dx#(div(ALPHA,BETA)) -> dx#(ALPHA)
p5: dx#(neg(ALPHA)) -> dx#(ALPHA)
p6: dx#(minus(ALPHA,BETA)) -> dx#(BETA)
p7: dx#(minus(ALPHA,BETA)) -> dx#(ALPHA)
p8: dx#(plus(ALPHA,BETA)) -> dx#(BETA)

and R consists of:

r1: dx(X) -> one()
r2: dx(a()) -> zero()
r3: dx(plus(ALPHA,BETA)) -> plus(dx(ALPHA),dx(BETA))
r4: dx(times(ALPHA,BETA)) -> plus(times(BETA,dx(ALPHA)),times(ALPHA,dx(BETA)))
r5: dx(minus(ALPHA,BETA)) -> minus(dx(ALPHA),dx(BETA))
r6: dx(neg(ALPHA)) -> neg(dx(ALPHA))
r7: dx(div(ALPHA,BETA)) -> minus(div(dx(ALPHA),BETA),times(ALPHA,div(dx(BETA),exp(BETA,two()))))
r8: dx(ln(ALPHA)) -> div(dx(ALPHA),ALPHA)
r9: dx(exp(ALPHA,BETA)) -> plus(times(BETA,times(exp(ALPHA,minus(BETA,one())),dx(ALPHA))),times(exp(ALPHA,BETA),times(ln(ALPHA),dx(BETA))))

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:
          dx#_A(x1) = ((0,1),(0,1)) x1 + (2,2)
          exp_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,0),(0,1)) x2 + (1,1)
          ln_A(x1) = ((1,1),(1,1)) x1 + (3,2)
          div_A(x1,x2) = ((0,0),(0,1)) x1 + ((1,1),(1,1)) x2 + (3,2)
          neg_A(x1) = x1 + (0,1)
          minus_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,0),(0,1)) x2 + (1,1)
          plus_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,0),(0,1)) x2 + (0,1)
  
    precedence:
    
      dx# = ln > div > exp = minus > neg = plus
    
    partial status:
  
      pi(dx#) = []
      pi(exp) = []
      pi(ln) = [1]
      pi(div) = [2]
      pi(neg) = [1]
      pi(minus) = []
      pi(plus) = [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: dx#(exp(ALPHA,BETA)) -> dx#(BETA)
p2: dx#(ln(ALPHA)) -> dx#(ALPHA)
p3: dx#(div(ALPHA,BETA)) -> dx#(BETA)
p4: dx#(neg(ALPHA)) -> dx#(ALPHA)
p5: dx#(minus(ALPHA,BETA)) -> dx#(BETA)
p6: dx#(minus(ALPHA,BETA)) -> dx#(ALPHA)
p7: dx#(plus(ALPHA,BETA)) -> dx#(BETA)

and R consists of:

r1: dx(X) -> one()
r2: dx(a()) -> zero()
r3: dx(plus(ALPHA,BETA)) -> plus(dx(ALPHA),dx(BETA))
r4: dx(times(ALPHA,BETA)) -> plus(times(BETA,dx(ALPHA)),times(ALPHA,dx(BETA)))
r5: dx(minus(ALPHA,BETA)) -> minus(dx(ALPHA),dx(BETA))
r6: dx(neg(ALPHA)) -> neg(dx(ALPHA))
r7: dx(div(ALPHA,BETA)) -> minus(div(dx(ALPHA),BETA),times(ALPHA,div(dx(BETA),exp(BETA,two()))))
r8: dx(ln(ALPHA)) -> div(dx(ALPHA),ALPHA)
r9: dx(exp(ALPHA,BETA)) -> plus(times(BETA,times(exp(ALPHA,minus(BETA,one())),dx(ALPHA))),times(exp(ALPHA,BETA),times(ln(ALPHA),dx(BETA))))

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: dx#(exp(ALPHA,BETA)) -> dx#(BETA)
p2: dx#(plus(ALPHA,BETA)) -> dx#(BETA)
p3: dx#(minus(ALPHA,BETA)) -> dx#(ALPHA)
p4: dx#(minus(ALPHA,BETA)) -> dx#(BETA)
p5: dx#(neg(ALPHA)) -> dx#(ALPHA)
p6: dx#(div(ALPHA,BETA)) -> dx#(BETA)
p7: dx#(ln(ALPHA)) -> dx#(ALPHA)

and R consists of:

r1: dx(X) -> one()
r2: dx(a()) -> zero()
r3: dx(plus(ALPHA,BETA)) -> plus(dx(ALPHA),dx(BETA))
r4: dx(times(ALPHA,BETA)) -> plus(times(BETA,dx(ALPHA)),times(ALPHA,dx(BETA)))
r5: dx(minus(ALPHA,BETA)) -> minus(dx(ALPHA),dx(BETA))
r6: dx(neg(ALPHA)) -> neg(dx(ALPHA))
r7: dx(div(ALPHA,BETA)) -> minus(div(dx(ALPHA),BETA),times(ALPHA,div(dx(BETA),exp(BETA,two()))))
r8: dx(ln(ALPHA)) -> div(dx(ALPHA),ALPHA)
r9: dx(exp(ALPHA,BETA)) -> plus(times(BETA,times(exp(ALPHA,minus(BETA,one())),dx(ALPHA))),times(exp(ALPHA,BETA),times(ln(ALPHA),dx(BETA))))

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:
          dx#_A(x1) = ((0,1),(0,1)) x1 + (2,2)
          exp_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,1),(1,1)) x2 + (3,2)
          plus_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (3,2)
          minus_A(x1,x2) = ((0,0),(1,1)) x1 + x2 + (1,1)
          neg_A(x1) = ((0,0),(0,1)) x1 + (1,1)
          div_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,0),(0,1)) x2 + (1,1)
          ln_A(x1) = ((0,0),(0,1)) x1 + (1,1)
  
    precedence:
    
      dx# = exp = plus > minus = neg > div = ln
    
    partial status:
  
      pi(dx#) = []
      pi(exp) = [1]
      pi(plus) = [2]
      pi(minus) = [2]
      pi(neg) = []
      pi(div) = []
      pi(ln) = []

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: dx#(exp(ALPHA,BETA)) -> dx#(BETA)
p2: dx#(plus(ALPHA,BETA)) -> dx#(BETA)
p3: dx#(minus(ALPHA,BETA)) -> dx#(ALPHA)
p4: dx#(neg(ALPHA)) -> dx#(ALPHA)
p5: dx#(div(ALPHA,BETA)) -> dx#(BETA)
p6: dx#(ln(ALPHA)) -> dx#(ALPHA)

and R consists of:

r1: dx(X) -> one()
r2: dx(a()) -> zero()
r3: dx(plus(ALPHA,BETA)) -> plus(dx(ALPHA),dx(BETA))
r4: dx(times(ALPHA,BETA)) -> plus(times(BETA,dx(ALPHA)),times(ALPHA,dx(BETA)))
r5: dx(minus(ALPHA,BETA)) -> minus(dx(ALPHA),dx(BETA))
r6: dx(neg(ALPHA)) -> neg(dx(ALPHA))
r7: dx(div(ALPHA,BETA)) -> minus(div(dx(ALPHA),BETA),times(ALPHA,div(dx(BETA),exp(BETA,two()))))
r8: dx(ln(ALPHA)) -> div(dx(ALPHA),ALPHA)
r9: dx(exp(ALPHA,BETA)) -> plus(times(BETA,times(exp(ALPHA,minus(BETA,one())),dx(ALPHA))),times(exp(ALPHA,BETA),times(ln(ALPHA),dx(BETA))))

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: dx#(exp(ALPHA,BETA)) -> dx#(BETA)
p2: dx#(ln(ALPHA)) -> dx#(ALPHA)
p3: dx#(div(ALPHA,BETA)) -> dx#(BETA)
p4: dx#(neg(ALPHA)) -> dx#(ALPHA)
p5: dx#(minus(ALPHA,BETA)) -> dx#(ALPHA)
p6: dx#(plus(ALPHA,BETA)) -> dx#(BETA)

and R consists of:

r1: dx(X) -> one()
r2: dx(a()) -> zero()
r3: dx(plus(ALPHA,BETA)) -> plus(dx(ALPHA),dx(BETA))
r4: dx(times(ALPHA,BETA)) -> plus(times(BETA,dx(ALPHA)),times(ALPHA,dx(BETA)))
r5: dx(minus(ALPHA,BETA)) -> minus(dx(ALPHA),dx(BETA))
r6: dx(neg(ALPHA)) -> neg(dx(ALPHA))
r7: dx(div(ALPHA,BETA)) -> minus(div(dx(ALPHA),BETA),times(ALPHA,div(dx(BETA),exp(BETA,two()))))
r8: dx(ln(ALPHA)) -> div(dx(ALPHA),ALPHA)
r9: dx(exp(ALPHA,BETA)) -> plus(times(BETA,times(exp(ALPHA,minus(BETA,one())),dx(ALPHA))),times(exp(ALPHA,BETA),times(ln(ALPHA),dx(BETA))))

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:
          dx#_A(x1) = ((1,1),(0,1)) x1 + (2,2)
          exp_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (3,2)
          ln_A(x1) = ((1,1),(1,1)) x1 + (3,2)
          div_A(x1,x2) = ((1,1),(1,1)) x1 + x2 + (1,1)
          neg_A(x1) = x1 + (1,1)
          minus_A(x1,x2) = x1 + ((1,1),(1,1)) x2 + (1,1)
          plus_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(0,1)) x2 + (3,2)
  
    precedence:
    
      neg > minus > dx# > div = plus > exp = ln
    
    partial status:
  
      pi(dx#) = [1]
      pi(exp) = [2]
      pi(ln) = []
      pi(div) = [2]
      pi(neg) = [1]
      pi(minus) = []
      pi(plus) = [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: dx#(exp(ALPHA,BETA)) -> dx#(BETA)
p2: dx#(ln(ALPHA)) -> dx#(ALPHA)
p3: dx#(div(ALPHA,BETA)) -> dx#(BETA)
p4: dx#(neg(ALPHA)) -> dx#(ALPHA)
p5: dx#(minus(ALPHA,BETA)) -> dx#(ALPHA)

and R consists of:

r1: dx(X) -> one()
r2: dx(a()) -> zero()
r3: dx(plus(ALPHA,BETA)) -> plus(dx(ALPHA),dx(BETA))
r4: dx(times(ALPHA,BETA)) -> plus(times(BETA,dx(ALPHA)),times(ALPHA,dx(BETA)))
r5: dx(minus(ALPHA,BETA)) -> minus(dx(ALPHA),dx(BETA))
r6: dx(neg(ALPHA)) -> neg(dx(ALPHA))
r7: dx(div(ALPHA,BETA)) -> minus(div(dx(ALPHA),BETA),times(ALPHA,div(dx(BETA),exp(BETA,two()))))
r8: dx(ln(ALPHA)) -> div(dx(ALPHA),ALPHA)
r9: dx(exp(ALPHA,BETA)) -> plus(times(BETA,times(exp(ALPHA,minus(BETA,one())),dx(ALPHA))),times(exp(ALPHA,BETA),times(ln(ALPHA),dx(BETA))))

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: dx#(exp(ALPHA,BETA)) -> dx#(BETA)
p2: dx#(minus(ALPHA,BETA)) -> dx#(ALPHA)
p3: dx#(neg(ALPHA)) -> dx#(ALPHA)
p4: dx#(div(ALPHA,BETA)) -> dx#(BETA)
p5: dx#(ln(ALPHA)) -> dx#(ALPHA)

and R consists of:

r1: dx(X) -> one()
r2: dx(a()) -> zero()
r3: dx(plus(ALPHA,BETA)) -> plus(dx(ALPHA),dx(BETA))
r4: dx(times(ALPHA,BETA)) -> plus(times(BETA,dx(ALPHA)),times(ALPHA,dx(BETA)))
r5: dx(minus(ALPHA,BETA)) -> minus(dx(ALPHA),dx(BETA))
r6: dx(neg(ALPHA)) -> neg(dx(ALPHA))
r7: dx(div(ALPHA,BETA)) -> minus(div(dx(ALPHA),BETA),times(ALPHA,div(dx(BETA),exp(BETA,two()))))
r8: dx(ln(ALPHA)) -> div(dx(ALPHA),ALPHA)
r9: dx(exp(ALPHA,BETA)) -> plus(times(BETA,times(exp(ALPHA,minus(BETA,one())),dx(ALPHA))),times(exp(ALPHA,BETA),times(ln(ALPHA),dx(BETA))))

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:
          dx#_A(x1) = ((1,0),(0,0)) x1 + (2,2)
          exp_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (3,2)
          minus_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,1),(1,1)) x2 + (1,1)
          neg_A(x1) = ((1,1),(0,0)) x1 + (1,0)
          div_A(x1,x2) = ((1,1),(1,1)) x1 + x2 + (1,1)
          ln_A(x1) = x1 + (1,1)
  
    precedence:
    
      minus > exp = neg > div = ln > dx#
    
    partial status:
  
      pi(dx#) = []
      pi(exp) = [2]
      pi(minus) = []
      pi(neg) = []
      pi(div) = [1, 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: dx#(exp(ALPHA,BETA)) -> dx#(BETA)
p2: dx#(minus(ALPHA,BETA)) -> dx#(ALPHA)
p3: dx#(neg(ALPHA)) -> dx#(ALPHA)
p4: dx#(ln(ALPHA)) -> dx#(ALPHA)

and R consists of:

r1: dx(X) -> one()
r2: dx(a()) -> zero()
r3: dx(plus(ALPHA,BETA)) -> plus(dx(ALPHA),dx(BETA))
r4: dx(times(ALPHA,BETA)) -> plus(times(BETA,dx(ALPHA)),times(ALPHA,dx(BETA)))
r5: dx(minus(ALPHA,BETA)) -> minus(dx(ALPHA),dx(BETA))
r6: dx(neg(ALPHA)) -> neg(dx(ALPHA))
r7: dx(div(ALPHA,BETA)) -> minus(div(dx(ALPHA),BETA),times(ALPHA,div(dx(BETA),exp(BETA,two()))))
r8: dx(ln(ALPHA)) -> div(dx(ALPHA),ALPHA)
r9: dx(exp(ALPHA,BETA)) -> plus(times(BETA,times(exp(ALPHA,minus(BETA,one())),dx(ALPHA))),times(exp(ALPHA,BETA),times(ln(ALPHA),dx(BETA))))

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: dx#(exp(ALPHA,BETA)) -> dx#(BETA)
p2: dx#(ln(ALPHA)) -> dx#(ALPHA)
p3: dx#(neg(ALPHA)) -> dx#(ALPHA)
p4: dx#(minus(ALPHA,BETA)) -> dx#(ALPHA)

and R consists of:

r1: dx(X) -> one()
r2: dx(a()) -> zero()
r3: dx(plus(ALPHA,BETA)) -> plus(dx(ALPHA),dx(BETA))
r4: dx(times(ALPHA,BETA)) -> plus(times(BETA,dx(ALPHA)),times(ALPHA,dx(BETA)))
r5: dx(minus(ALPHA,BETA)) -> minus(dx(ALPHA),dx(BETA))
r6: dx(neg(ALPHA)) -> neg(dx(ALPHA))
r7: dx(div(ALPHA,BETA)) -> minus(div(dx(ALPHA),BETA),times(ALPHA,div(dx(BETA),exp(BETA,two()))))
r8: dx(ln(ALPHA)) -> div(dx(ALPHA),ALPHA)
r9: dx(exp(ALPHA,BETA)) -> plus(times(BETA,times(exp(ALPHA,minus(BETA,one())),dx(ALPHA))),times(exp(ALPHA,BETA),times(ln(ALPHA),dx(BETA))))

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:
          dx#_A(x1) = ((0,1),(0,0)) x1 + (2,2)
          exp_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(0,1)) x2 + (3,2)
          ln_A(x1) = x1 + (1,1)
          neg_A(x1) = ((0,0),(0,1)) x1 + (1,1)
          minus_A(x1,x2) = ((0,0),(0,1)) x1 + ((1,1),(1,1)) x2 + (3,1)
  
    precedence:
    
      exp = ln = neg > dx# > minus
    
    partial status:
  
      pi(dx#) = []
      pi(exp) = [2]
      pi(ln) = [1]
      pi(neg) = []
      pi(minus) = []

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: dx#(exp(ALPHA,BETA)) -> dx#(BETA)
p2: dx#(neg(ALPHA)) -> dx#(ALPHA)
p3: dx#(minus(ALPHA,BETA)) -> dx#(ALPHA)

and R consists of:

r1: dx(X) -> one()
r2: dx(a()) -> zero()
r3: dx(plus(ALPHA,BETA)) -> plus(dx(ALPHA),dx(BETA))
r4: dx(times(ALPHA,BETA)) -> plus(times(BETA,dx(ALPHA)),times(ALPHA,dx(BETA)))
r5: dx(minus(ALPHA,BETA)) -> minus(dx(ALPHA),dx(BETA))
r6: dx(neg(ALPHA)) -> neg(dx(ALPHA))
r7: dx(div(ALPHA,BETA)) -> minus(div(dx(ALPHA),BETA),times(ALPHA,div(dx(BETA),exp(BETA,two()))))
r8: dx(ln(ALPHA)) -> div(dx(ALPHA),ALPHA)
r9: dx(exp(ALPHA,BETA)) -> plus(times(BETA,times(exp(ALPHA,minus(BETA,one())),dx(ALPHA))),times(exp(ALPHA,BETA),times(ln(ALPHA),dx(BETA))))

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: dx#(exp(ALPHA,BETA)) -> dx#(BETA)
p2: dx#(minus(ALPHA,BETA)) -> dx#(ALPHA)
p3: dx#(neg(ALPHA)) -> dx#(ALPHA)

and R consists of:

r1: dx(X) -> one()
r2: dx(a()) -> zero()
r3: dx(plus(ALPHA,BETA)) -> plus(dx(ALPHA),dx(BETA))
r4: dx(times(ALPHA,BETA)) -> plus(times(BETA,dx(ALPHA)),times(ALPHA,dx(BETA)))
r5: dx(minus(ALPHA,BETA)) -> minus(dx(ALPHA),dx(BETA))
r6: dx(neg(ALPHA)) -> neg(dx(ALPHA))
r7: dx(div(ALPHA,BETA)) -> minus(div(dx(ALPHA),BETA),times(ALPHA,div(dx(BETA),exp(BETA,two()))))
r8: dx(ln(ALPHA)) -> div(dx(ALPHA),ALPHA)
r9: dx(exp(ALPHA,BETA)) -> plus(times(BETA,times(exp(ALPHA,minus(BETA,one())),dx(ALPHA))),times(exp(ALPHA,BETA),times(ln(ALPHA),dx(BETA))))

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:
          dx#_A(x1) = ((0,1),(0,1)) x1 + (2,2)
          exp_A(x1,x2) = ((1,1),(1,1)) x1 + x2 + (0,1)
          minus_A(x1,x2) = ((0,0),(0,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
          neg_A(x1) = ((0,0),(0,1)) x1 + (1,1)
  
    precedence:
    
      minus > dx# = exp = neg
    
    partial status:
  
      pi(dx#) = []
      pi(exp) = [1, 2]
      pi(minus) = [2]
      pi(neg) = []

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: dx#(exp(ALPHA,BETA)) -> dx#(BETA)
p2: dx#(neg(ALPHA)) -> dx#(ALPHA)

and R consists of:

r1: dx(X) -> one()
r2: dx(a()) -> zero()
r3: dx(plus(ALPHA,BETA)) -> plus(dx(ALPHA),dx(BETA))
r4: dx(times(ALPHA,BETA)) -> plus(times(BETA,dx(ALPHA)),times(ALPHA,dx(BETA)))
r5: dx(minus(ALPHA,BETA)) -> minus(dx(ALPHA),dx(BETA))
r6: dx(neg(ALPHA)) -> neg(dx(ALPHA))
r7: dx(div(ALPHA,BETA)) -> minus(div(dx(ALPHA),BETA),times(ALPHA,div(dx(BETA),exp(BETA,two()))))
r8: dx(ln(ALPHA)) -> div(dx(ALPHA),ALPHA)
r9: dx(exp(ALPHA,BETA)) -> plus(times(BETA,times(exp(ALPHA,minus(BETA,one())),dx(ALPHA))),times(exp(ALPHA,BETA),times(ln(ALPHA),dx(BETA))))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: dx#(exp(ALPHA,BETA)) -> dx#(BETA)
p2: dx#(neg(ALPHA)) -> dx#(ALPHA)

and R consists of:

r1: dx(X) -> one()
r2: dx(a()) -> zero()
r3: dx(plus(ALPHA,BETA)) -> plus(dx(ALPHA),dx(BETA))
r4: dx(times(ALPHA,BETA)) -> plus(times(BETA,dx(ALPHA)),times(ALPHA,dx(BETA)))
r5: dx(minus(ALPHA,BETA)) -> minus(dx(ALPHA),dx(BETA))
r6: dx(neg(ALPHA)) -> neg(dx(ALPHA))
r7: dx(div(ALPHA,BETA)) -> minus(div(dx(ALPHA),BETA),times(ALPHA,div(dx(BETA),exp(BETA,two()))))
r8: dx(ln(ALPHA)) -> div(dx(ALPHA),ALPHA)
r9: dx(exp(ALPHA,BETA)) -> plus(times(BETA,times(exp(ALPHA,minus(BETA,one())),dx(ALPHA))),times(exp(ALPHA,BETA),times(ln(ALPHA),dx(BETA))))

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:
          dx#_A(x1) = ((1,0),(0,0)) x1 + (2,2)
          exp_A(x1,x2) = ((1,1),(1,1)) x1 + x2 + (1,1)
          neg_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      dx# = exp = neg
    
    partial status:
  
      pi(dx#) = []
      pi(exp) = [2]
      pi(neg) = []

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: dx#(neg(ALPHA)) -> dx#(ALPHA)

and R consists of:

r1: dx(X) -> one()
r2: dx(a()) -> zero()
r3: dx(plus(ALPHA,BETA)) -> plus(dx(ALPHA),dx(BETA))
r4: dx(times(ALPHA,BETA)) -> plus(times(BETA,dx(ALPHA)),times(ALPHA,dx(BETA)))
r5: dx(minus(ALPHA,BETA)) -> minus(dx(ALPHA),dx(BETA))
r6: dx(neg(ALPHA)) -> neg(dx(ALPHA))
r7: dx(div(ALPHA,BETA)) -> minus(div(dx(ALPHA),BETA),times(ALPHA,div(dx(BETA),exp(BETA,two()))))
r8: dx(ln(ALPHA)) -> div(dx(ALPHA),ALPHA)
r9: dx(exp(ALPHA,BETA)) -> plus(times(BETA,times(exp(ALPHA,minus(BETA,one())),dx(ALPHA))),times(exp(ALPHA,BETA),times(ln(ALPHA),dx(BETA))))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: dx#(neg(ALPHA)) -> dx#(ALPHA)

and R consists of:

r1: dx(X) -> one()
r2: dx(a()) -> zero()
r3: dx(plus(ALPHA,BETA)) -> plus(dx(ALPHA),dx(BETA))
r4: dx(times(ALPHA,BETA)) -> plus(times(BETA,dx(ALPHA)),times(ALPHA,dx(BETA)))
r5: dx(minus(ALPHA,BETA)) -> minus(dx(ALPHA),dx(BETA))
r6: dx(neg(ALPHA)) -> neg(dx(ALPHA))
r7: dx(div(ALPHA,BETA)) -> minus(div(dx(ALPHA),BETA),times(ALPHA,div(dx(BETA),exp(BETA,two()))))
r8: dx(ln(ALPHA)) -> div(dx(ALPHA),ALPHA)
r9: dx(exp(ALPHA,BETA)) -> plus(times(BETA,times(exp(ALPHA,minus(BETA,one())),dx(ALPHA))),times(exp(ALPHA,BETA),times(ln(ALPHA),dx(BETA))))

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:
          dx#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
          neg_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      dx# = neg
    
    partial status:
  
      pi(dx#) = []
      pi(neg) = []

The next rules are strictly ordered:

  p1

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