YES

TRS:

  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))

linear polynomial interpretations on N:

  D_A(x1) = 0
  D#_A(x1) = 3
  t_A = 1
  t#_A = 0
  s_A(x1) = 0
  s#_A(x1) = 1
  h_A = 0
  h#_A = 1
  constant_A = 1
  constant#_A = 2
  b_A(x1,x2) = x1 + x2
  b#_A(x1,x2) = 2
  c_A(x1,x2) = 0
  c#_A(x1,x2) = 0
  m_A(x1,x2) = x2
  m#_A(x1,x2) = 2
  opp_A(x1) = 0
  opp#_A(x1) = 2
  div_A(x1,x2) = 0
  div#_A(x1,x2) = 2
  pow_A(x1,x2) = 1
  pow#_A(x1,x2) = 1
  2_A = 1
  2#_A = 0
  ln_A(x1) = x1 + 1
  ln#_A(x1) = 0
  1_A = 1
  1#_A = 2

precedence:

  b > D = t > s = h = opp = div = 1 > constant = pow = 2 > c = m = ln