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 monotone reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      D#_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1
      pow_A(x1,x2) = ((1,0,0),(1,1,0),(1,1,1)) x1 + ((1,0,0),(1,1,0),(1,1,1)) x2 + (1,1,1)
      ln_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (1,1,1)
      div_A(x1,x2) = ((1,0,0),(1,1,0),(1,1,1)) x1 + ((1,0,0),(1,1,0),(1,1,1)) x2 + (1,1,1)
      opp_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (1,1,1)
      m_A(x1,x2) = ((1,0,0),(1,1,0),(1,1,1)) x1 + ((1,0,0),(1,1,0),(1,1,1)) x2 + (1,1,1)
      c_A(x1,x2) = ((1,0,0),(1,1,0),(1,1,1)) x1 + ((1,0,0),(1,1,0),(1,1,1)) x2 + (1,1,1)
      b_A(x1,x2) = ((1,0,0),(1,1,0),(1,1,1)) x1 + ((1,0,0),(1,1,0),(1,1,1)) x2 + (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, r10, r11, r12, r13

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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      b#_A(x1,x2) = ((1,0,0),(1,1,0),(1,0,1)) x1 + ((1,0,0),(0,1,0),(1,1,1)) x2
      s_A(x1) = ((1,0,0),(0,1,0),(1,0,0)) x1 + (1,1,1)
      b_A(x1,x2) = ((1,0,0),(1,1,0),(1,1,1)) x1 + ((1,0,0),(0,1,0),(1,1,1)) x2 + (6,1,1)
      h_A() = (1,1,1)

The next rules are strictly ordered:

  p1, p2, p3

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