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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            D#_A(x1) = 0
            +_A(x1,x2) = max{2, x1 - 3, x2 - 9}
            -_A(x1,x2) = max{2, x1 - 4, x2 - 2}
            *_A(x1,x2) = max{x1 - 6, x2 + 1}
            D#!_A(x1) = max{1, x1 - 27}
            +!_A(x1,x2) = max{2, x1 - 20, x2 - 3}
            -!_A(x1,x2) = max{3, x1 - 19, x2 - 21}
            *!_A(x1,x2) = max{1, x1 - 19, x2 - 27}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            D#_A(x1) = 0
            +_A(x1,x2) = max{5, x1 - 13, x2 - 57}
            -_A(x1,x2) = max{5, x1 - 26, x2 - 26}
            *_A(x1,x2) = max{12, x1 - 13, x2 + 1}
            D#!_A(x1) = max{1, x1 - 39}
            +!_A(x1,x2) = max{5, x1 - 39, x2 - 36}
            -!_A(x1,x2) = max{2, x1 - 5, x2 - 8}
            *!_A(x1,x2) = max{1, x1 - 23, x2 - 26}
        
  
    partial status:
  
      pi(D#) = [1]
      pi(+) = [1, 2]
      pi(-) = [1, 2]
      pi(*) = [1, 2]

The following rules are removed:

  p1, p2, p3, p5, p6



-- SCC decomposition.

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

p1: 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}


-- Reduction pair.

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

p1: 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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            D#_A(x1) = x1
            *_A(x1,x2) = max{x1 - 7, x2 + 4}
            D#!_A(x1) = max{1, x1 - 1}
            *!_A(x1,x2) = max{2, x1 - 4, x2 - 36}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            D#_A(x1) = x1
            *_A(x1,x2) = max{7, x1 - 9, x2}
            D#!_A(x1) = max{2, x1 - 44}
            *!_A(x1,x2) = max{1, x1 - 42, x2 - 43}
        
  
    partial status:
  
      pi(D#) = []
      pi(*) = [2]

The following rules are removed:

  p1

Then no dependency pair remains.