YES
We show the termination of the TRS R:

  |0|(|#|()) -> |#|()
  +(x,|#|()) -> x
  +(|#|(),x) -> x
  +(|0|(x),|0|(y)) -> |0|(+(x,y))
  +(|0|(x),|1|(y)) -> |1|(+(x,y))
  +(|1|(x),|0|(y)) -> |1|(+(x,y))
  +(|1|(x),|1|(y)) -> |0|(+(+(x,y),|1|(|#|())))
  *(|#|(),x) -> |#|()
  *(|0|(x),y) -> |0|(*(x,y))
  *(|1|(x),y) -> +(|0|(*(x,y)),y)
  sum(nil()) -> |0|(|#|())
  sum(cons(x,l)) -> +(x,sum(l))
  prod(nil()) -> |1|(|#|())
  prod(cons(x,l)) -> *(x,prod(l))


-- SCC decomposition.

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

p1: +#(|0|(x),|0|(y)) -> |0|#(+(x,y))
p2: +#(|0|(x),|0|(y)) -> +#(x,y)
p3: +#(|0|(x),|1|(y)) -> +#(x,y)
p4: +#(|1|(x),|0|(y)) -> +#(x,y)
p5: +#(|1|(x),|1|(y)) -> |0|#(+(+(x,y),|1|(|#|())))
p6: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|()))
p7: +#(|1|(x),|1|(y)) -> +#(x,y)
p8: *#(|0|(x),y) -> |0|#(*(x,y))
p9: *#(|0|(x),y) -> *#(x,y)
p10: *#(|1|(x),y) -> +#(|0|(*(x,y)),y)
p11: *#(|1|(x),y) -> |0|#(*(x,y))
p12: *#(|1|(x),y) -> *#(x,y)
p13: sum#(nil()) -> |0|#(|#|())
p14: sum#(cons(x,l)) -> +#(x,sum(l))
p15: sum#(cons(x,l)) -> sum#(l)
p16: prod#(cons(x,l)) -> *#(x,prod(l))
p17: prod#(cons(x,l)) -> prod#(l)

and R consists of:

r1: |0|(|#|()) -> |#|()
r2: +(x,|#|()) -> x
r3: +(|#|(),x) -> x
r4: +(|0|(x),|0|(y)) -> |0|(+(x,y))
r5: +(|0|(x),|1|(y)) -> |1|(+(x,y))
r6: +(|1|(x),|0|(y)) -> |1|(+(x,y))
r7: +(|1|(x),|1|(y)) -> |0|(+(+(x,y),|1|(|#|())))
r8: *(|#|(),x) -> |#|()
r9: *(|0|(x),y) -> |0|(*(x,y))
r10: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r11: sum(nil()) -> |0|(|#|())
r12: sum(cons(x,l)) -> +(x,sum(l))
r13: prod(nil()) -> |1|(|#|())
r14: prod(cons(x,l)) -> *(x,prod(l))

The estimated dependency graph contains the following SCCs:

  {p15}
  {p17}
  {p9, p12}
  {p2, p3, p4, p6, p7}


-- Reduction pair.

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

p1: sum#(cons(x,l)) -> sum#(l)

and R consists of:

r1: |0|(|#|()) -> |#|()
r2: +(x,|#|()) -> x
r3: +(|#|(),x) -> x
r4: +(|0|(x),|0|(y)) -> |0|(+(x,y))
r5: +(|0|(x),|1|(y)) -> |1|(+(x,y))
r6: +(|1|(x),|0|(y)) -> |1|(+(x,y))
r7: +(|1|(x),|1|(y)) -> |0|(+(+(x,y),|1|(|#|())))
r8: *(|#|(),x) -> |#|()
r9: *(|0|(x),y) -> |0|(*(x,y))
r10: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r11: sum(nil()) -> |0|(|#|())
r12: sum(cons(x,l)) -> +(x,sum(l))
r13: prod(nil()) -> |1|(|#|())
r14: prod(cons(x,l)) -> *(x,prod(l))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          sum#_A(x1) = x1
          cons_A(x1,x2) = x2 + 7
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          sum#_A(x1) = 0
          cons_A(x1,x2) = max{0, x1 - 29, x2 - 29}
          sum#!_A(x1) = 5
          cons!_A(x1,x2) = max{6, x1 - 110, x2 + 4}
  
    partial status:
  
      pi(sum#) = []
      pi(cons) = [2]

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

p1: prod#(cons(x,l)) -> prod#(l)

and R consists of:

r1: |0|(|#|()) -> |#|()
r2: +(x,|#|()) -> x
r3: +(|#|(),x) -> x
r4: +(|0|(x),|0|(y)) -> |0|(+(x,y))
r5: +(|0|(x),|1|(y)) -> |1|(+(x,y))
r6: +(|1|(x),|0|(y)) -> |1|(+(x,y))
r7: +(|1|(x),|1|(y)) -> |0|(+(+(x,y),|1|(|#|())))
r8: *(|#|(),x) -> |#|()
r9: *(|0|(x),y) -> |0|(*(x,y))
r10: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r11: sum(nil()) -> |0|(|#|())
r12: sum(cons(x,l)) -> +(x,sum(l))
r13: prod(nil()) -> |1|(|#|())
r14: prod(cons(x,l)) -> *(x,prod(l))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          prod#_A(x1) = x1
          cons_A(x1,x2) = x2 + 6
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          prod#_A(x1) = x1
          cons_A(x1,x2) = max{x1 + 50, x2}
          prod#!_A(x1) = max{2, x1 - 51}
          cons!_A(x1,x2) = 1
  
    partial status:
  
      pi(prod#) = []
      pi(cons) = [2]

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

p1: *#(|1|(x),y) -> *#(x,y)
p2: *#(|0|(x),y) -> *#(x,y)

and R consists of:

r1: |0|(|#|()) -> |#|()
r2: +(x,|#|()) -> x
r3: +(|#|(),x) -> x
r4: +(|0|(x),|0|(y)) -> |0|(+(x,y))
r5: +(|0|(x),|1|(y)) -> |1|(+(x,y))
r6: +(|1|(x),|0|(y)) -> |1|(+(x,y))
r7: +(|1|(x),|1|(y)) -> |0|(+(+(x,y),|1|(|#|())))
r8: *(|#|(),x) -> |#|()
r9: *(|0|(x),y) -> |0|(*(x,y))
r10: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r11: sum(nil()) -> |0|(|#|())
r12: sum(cons(x,l)) -> +(x,sum(l))
r13: prod(nil()) -> |1|(|#|())
r14: prod(cons(x,l)) -> *(x,prod(l))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          *#_A(x1,x2) = x1 + x2 + 5
          |1|_A(x1) = x1 + 4
          |0|_A(x1) = x1 + 2
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          *#_A(x1,x2) = 0
          |1|_A(x1) = max{3, x1 - 13}
          |0|_A(x1) = max{7, x1 - 3}
          *#!_A(x1,x2) = max{7, x1 - 8, x2 - 1}
          |1|!_A(x1) = max{2, x1 - 9}
          |0|!_A(x1) = max{6, x1 - 23}
  
    partial status:
  
      pi(*#) = [1, 2]
      pi(|1|) = [1]
      pi(|0|) = []

The following rules are removed:

  p1, p2

Then no dependency pair remains.

-- Reduction pair.

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

p1: +#(|0|(x),|0|(y)) -> +#(x,y)
p2: +#(|1|(x),|1|(y)) -> +#(x,y)
p3: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|()))
p4: +#(|0|(x),|1|(y)) -> +#(x,y)
p5: +#(|1|(x),|0|(y)) -> +#(x,y)

and R consists of:

r1: |0|(|#|()) -> |#|()
r2: +(x,|#|()) -> x
r3: +(|#|(),x) -> x
r4: +(|0|(x),|0|(y)) -> |0|(+(x,y))
r5: +(|0|(x),|1|(y)) -> |1|(+(x,y))
r6: +(|1|(x),|0|(y)) -> |1|(+(x,y))
r7: +(|1|(x),|1|(y)) -> |0|(+(+(x,y),|1|(|#|())))
r8: *(|#|(),x) -> |#|()
r9: *(|0|(x),y) -> |0|(*(x,y))
r10: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r11: sum(nil()) -> |0|(|#|())
r12: sum(cons(x,l)) -> +(x,sum(l))
r13: prod(nil()) -> |1|(|#|())
r14: prod(cons(x,l)) -> *(x,prod(l))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          +#_A(x1,x2) = x1 + x2 + 7
          |0|_A(x1) = x1 + 1
          |1|_A(x1) = x1 + 6
          +_A(x1,x2) = x1 + x2 + 2
          |#|_A() = 2
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          +#_A(x1,x2) = max{0, x1 - 15, x2 - 10}
          |0|_A(x1) = max{7, x1}
          |1|_A(x1) = max{3, x1}
          +_A(x1,x2) = max{14, x1, x2 + 2}
          |#|_A = 9
          +#!_A(x1,x2) = max{0, x1 - 18, x2 - 8}
          |0|!_A(x1) = max{13, x1 + 11}
          |1|!_A(x1) = max{13, x1 - 15}
          +!_A(x1,x2) = max{13, x2 - 15}
          |#|!_A = 11
  
    partial status:
  
      pi(+#) = [2]
      pi(|0|) = []
      pi(|1|) = [1]
      pi(+) = []
      pi(|#|) = []

The following rules are removed:

  p1, p2, p3, p4, p5

Then no dependency pair remains.