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 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
      +_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)
      -_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)
      *_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
  r1, r2, r3, r4, r5

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