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 monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      times > dx# > minus > neg > div > ln > exp > plus
    
    argument filter:
  
      pi(dx#) = 1
      pi(plus) = [1, 2]
      pi(exp) = [1, 2]
      pi(ln) = 1
      pi(div) = [1, 2]
      pi(neg) = 1
      pi(minus) = [1, 2]
      pi(times) = [1, 2]

The next rules are strictly ordered:

  p1, p2, p3, p5, p6, p8, p9, p10, p11, p12
  r1, r2, r3, r4, r5, r6, r7, r8, r9

We remove them from the problem.

-- SCC decomposition.

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

p1: dx#(ln(ALPHA)) -> dx#(ALPHA)
p2: dx#(neg(ALPHA)) -> dx#(ALPHA)

and R consists of:

  (no rules)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: dx#(ln(ALPHA)) -> dx#(ALPHA)
p2: dx#(neg(ALPHA)) -> dx#(ALPHA)

and R consists of:

  (no rules)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      dx# > neg > ln
    
    argument filter:
  
      pi(dx#) = 1
      pi(ln) = [1]
      pi(neg) = [1]

The next rules are strictly ordered:

  p1, p2

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