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

  r1, r2, r3, r4, r5

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:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            D#_A(0) = 0
            D#_A(1) = 0
            +_A(0,0) = 0
            +_A(0,1) = 0
            +_A(1,0) = 1
            +_A(1,1) = 1
            -_A(0,0) = 0
            -_A(0,1) = 0
            -_A(1,0) = 1
            -_A(1,1) = 1
            *_A(0,0) = 1
            *_A(0,1) = 0
            *_A(1,0) = 0
            *_A(1,1) = 1
            D_A(0) = 1
            D_A(1) = 0
            t_A = 1
            |1|_A = 0
            constant_A = 1
            |0|_A = 0
            D#!_A(0) = 0
            D#!_A(1) = 0
            +!_A(0,0) = 0
            +!_A(0,1) = 0
            +!_A(1,0) = 0
            +!_A(1,1) = 0
            -!_A(0,0) = 1
            -!_A(0,1) = 1
            -!_A(1,0) = 1
            -!_A(1,1) = 1
            *!_A(0,0) = 1
            *!_A(0,1) = 1
            *!_A(1,0) = 1
            *!_A(1,1) = 1
            D!_A(0) = 1
            D!_A(1) = 1
            t!_A = 1
            |1|!_A = 0
            constant!_A = 1
            |0|!_A = 1
  
    partial status:
  
      pi(D#) = [1]
      pi(+) = [1, 2]
      pi(-) = [1, 2]
      pi(*) = [1, 2]
      pi(D) = [1]
      pi(t) = []
      pi(|1|) = []
      pi(constant) = []
      pi(|0|) = []

The following rules are removed:

  p1, p4



-- SCC decomposition.

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

p1: D#(-(x,y)) -> D#(y)
p2: D#(-(x,y)) -> D#(x)
p3: D#(*(x,y)) -> D#(x)
p4: 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}


-- Reduction pair.

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

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

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

  r1, r2, r3, r4, r5

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:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            D#_A(0) = 0
            D#_A(1) = 0
            -_A(0,0) = 0
            -_A(0,1) = 0
            -_A(1,0) = 1
            -_A(1,1) = 1
            +_A(0,0) = 0
            +_A(0,1) = 0
            +_A(1,0) = 1
            +_A(1,1) = 1
            *_A(0,0) = 1
            *_A(0,1) = 0
            *_A(1,0) = 0
            *_A(1,1) = 1
            D_A(0) = 1
            D_A(1) = 0
            t_A = 1
            |1|_A = 0
            constant_A = 1
            |0|_A = 0
            D#!_A(0) = 1
            D#!_A(1) = 1
            -!_A(0,0) = 0
            -!_A(0,1) = 0
            -!_A(1,0) = 0
            -!_A(1,1) = 0
            +!_A(0,0) = 0
            +!_A(0,1) = 0
            +!_A(1,0) = 0
            +!_A(1,1) = 0
            *!_A(0,0) = 0
            *!_A(0,1) = 0
            *!_A(1,0) = 1
            *!_A(1,1) = 1
            D!_A(0) = 0
            D!_A(1) = 0
            t!_A = 1
            |1|!_A = 0
            constant!_A = 1
            |0|!_A = 0
  
    partial status:
  
      pi(D#) = [1]
      pi(-) = [1, 2]
      pi(+) = [2]
      pi(*) = [1]
      pi(D) = [1]
      pi(t) = []
      pi(|1|) = []
      pi(constant) = []
      pi(|0|) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

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

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}


-- Reduction pair.

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

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

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

  r1, r2, r3, r4, r5

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:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            D#_A(0) = 0
            D#_A(1) = 0
            +_A(0,0) = 1
            +_A(0,1) = 1
            +_A(1,0) = 0
            +_A(1,1) = 0
            -_A(0,0) = 0
            -_A(0,1) = 0
            -_A(1,0) = 1
            -_A(1,1) = 1
            *_A(0,0) = 1
            *_A(0,1) = 1
            *_A(1,0) = 1
            *_A(1,1) = 1
            D_A(0) = 1
            D_A(1) = 0
            t_A = 1
            |1|_A = 0
            constant_A = 0
            |0|_A = 1
            D#!_A(0) = 0
            D#!_A(1) = 0
            +!_A(0,0) = 0
            +!_A(0,1) = 0
            +!_A(1,0) = 0
            +!_A(1,1) = 0
            -!_A(0,0) = 1
            -!_A(0,1) = 1
            -!_A(1,0) = 1
            -!_A(1,1) = 1
            *!_A(0,0) = 0
            *!_A(0,1) = 0
            *!_A(1,0) = 0
            *!_A(1,1) = 0
            D!_A(0) = 1
            D!_A(1) = 1
            t!_A = 1
            |1|!_A = 0
            constant!_A = 1
            |0|!_A = 1
  
    partial status:
  
      pi(D#) = [1]
      pi(+) = [2]
      pi(-) = [1, 2]
      pi(*) = [1]
      pi(D) = [1]
      pi(t) = []
      pi(|1|) = []
      pi(constant) = []
      pi(|0|) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: D#(-(x,y)) -> D#(x)
p2: D#(*(x,y)) -> D#(x)

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}


-- Reduction pair.

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

p1: D#(-(x,y)) -> D#(x)
p2: D#(*(x,y)) -> D#(x)

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

  r1, r2, r3, r4, r5

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:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            D#_A(0) = 0
            D#_A(1) = 0
            -_A(0,0) = 0
            -_A(0,1) = 0
            -_A(1,0) = 1
            -_A(1,1) = 1
            *_A(0,0) = 1
            *_A(0,1) = 0
            *_A(1,0) = 0
            *_A(1,1) = 1
            D_A(0) = 1
            D_A(1) = 0
            t_A = 1
            |1|_A = 0
            constant_A = 1
            |0|_A = 0
            +_A(0,0) = 0
            +_A(0,1) = 0
            +_A(1,0) = 1
            +_A(1,1) = 1
            D#!_A(0) = 0
            D#!_A(1) = 0
            -!_A(0,0) = 1
            -!_A(0,1) = 1
            -!_A(1,0) = 1
            -!_A(1,1) = 1
            *!_A(0,0) = 0
            *!_A(0,1) = 0
            *!_A(1,0) = 1
            *!_A(1,1) = 1
            D!_A(0) = 1
            D!_A(1) = 1
            t!_A = 0
            |1|!_A = 1
            constant!_A = 1
            |0|!_A = 1
            +!_A(0,0) = 0
            +!_A(0,1) = 1
            +!_A(1,0) = 0
            +!_A(1,1) = 0
  
    partial status:
  
      pi(D#) = [1]
      pi(-) = [1]
      pi(*) = [1]
      pi(D) = [1]
      pi(t) = []
      pi(|1|) = []
      pi(constant) = []
      pi(|0|) = []
      pi(+) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: D#(*(x,y)) -> D#(x)

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#(x)

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

  r1, r2, r3, r4, r5

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:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            D#_A(0) = 0
            D#_A(1) = 0
            *_A(0,0) = 1
            *_A(0,1) = 0
            *_A(1,0) = 0
            *_A(1,1) = 1
            D_A(0) = 1
            D_A(1) = 0
            t_A = 1
            |1|_A = 0
            constant_A = 1
            |0|_A = 0
            +_A(0,0) = 0
            +_A(0,1) = 0
            +_A(1,0) = 1
            +_A(1,1) = 1
            -_A(0,0) = 0
            -_A(0,1) = 0
            -_A(1,0) = 1
            -_A(1,1) = 1
            D#!_A(0) = 1
            D#!_A(1) = 1
            *!_A(0,0) = 1
            *!_A(0,1) = 1
            *!_A(1,0) = 1
            *!_A(1,1) = 1
            D!_A(0) = 0
            D!_A(1) = 0
            t!_A = 0
            |1|!_A = 0
            constant!_A = 1
            |0|!_A = 0
            +!_A(0,0) = 1
            +!_A(0,1) = 0
            +!_A(1,0) = 0
            +!_A(1,1) = 1
            -!_A(0,0) = 0
            -!_A(0,1) = 0
            -!_A(1,0) = 0
            -!_A(1,1) = 0
  
    partial status:
  
      pi(D#) = [1]
      pi(*) = [1]
      pi(D) = [1]
      pi(t) = []
      pi(|1|) = []
      pi(constant) = []
      pi(|0|) = []
      pi(+) = []
      pi(-) = [2]

The following rules are removed:

  p1

Then no dependency pair remains.