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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      dx#_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1
      plus_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x2 + (1,1,1,1)
      exp_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x2 + (1,1,1,1)
      ln_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + (1,1,1,1)
      div_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x2 + (1,1,1,1)
      neg_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + (1,1,1,1)
      minus_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x2 + (1,1,1,1)
      times_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x2 + (1,1,1,1)

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12
  r1, r2, r3, r4, r5, r6, r7, r8, r9

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