YES
We show the termination of the TRS R:

  D(t()) -> |1|()
  D(constant()) -> |0|()
  D(+(x,y)) -> +(D(x),D(y))
  D(*(x,y)) -> +(*(y,D(x)),*(x,D(y)))
  D(-(x,y)) -> -(D(x),D(y))


-- SCC decomposition.

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

p1: D#(+(x,y)) -> D#(x)
p2: D#(+(x,y)) -> D#(y)
p3: D#(*(x,y)) -> D#(x)
p4: D#(*(x,y)) -> D#(y)
p5: D#(-(x,y)) -> D#(x)
p6: D#(-(x,y)) -> D#(y)

and R consists of:

r1: D(t()) -> |1|()
r2: D(constant()) -> |0|()
r3: D(+(x,y)) -> +(D(x),D(y))
r4: D(*(x,y)) -> +(*(y,D(x)),*(x,D(y)))
r5: D(-(x,y)) -> -(D(x),D(y))

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: D#(+(x,y)) -> D#(x)
p2: D#(-(x,y)) -> D#(y)
p3: D#(-(x,y)) -> D#(x)
p4: D#(*(x,y)) -> D#(y)
p5: D#(*(x,y)) -> D#(x)
p6: D#(+(x,y)) -> D#(y)

and R consists of:

r1: D(t()) -> |1|()
r2: D(constant()) -> |0|()
r3: D(+(x,y)) -> +(D(x),D(y))
r4: D(*(x,y)) -> +(*(y,D(x)),*(x,D(y)))
r5: D(-(x,y)) -> -(D(x),D(y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        D#_A(x1) = max{7, x1 + 4}
        +_A(x1,x2) = max{23, x1 + 6, x2 + 4}
        -_A(x1,x2) = max{38, x1 + 19, x2 + 4}
        *_A(x1,x2) = max{x1 + 36, x2 + 9}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          D#_A(x1) = 0
          +_A(x1,x2) = max{x1 + 19, x2 + 20}
          -_A(x1,x2) = max{17, x1 - 2, x2 - 1}
          *_A(x1,x2) = max{x1 - 2, x2 + 7}
          D#!_A(x1) = 1
          +!_A(x1,x2) = max{3, x1 - 8, x2 - 2}
          -!_A(x1,x2) = max{3, x1 - 2, x2 - 17}
          *!_A(x1,x2) = max{4, x1 - 9}
  
    partial status:
  
      pi(D#) = [1]
      pi(+) = []
      pi(-) = [1, 2]
      pi(*) = [1, 2]

The following rules are removed:

  p1, p2, p3, p4, p5, p6

Then no dependency pair remains.