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,y),z) -> +(x,+(y,z))
  *(|#|(),x) -> |#|()
  *(|0|(x),y) -> |0|(*(x,y))
  *(|1|(x),y) -> +(|0|(*(x,y)),y)
  *(*(x,y),z) -> *(x,*(y,z))
  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: +#(+(x,y),z) -> +#(x,+(y,z))
p9: +#(+(x,y),z) -> +#(y,z)
p10: *#(|0|(x),y) -> |0|#(*(x,y))
p11: *#(|0|(x),y) -> *#(x,y)
p12: *#(|1|(x),y) -> +#(|0|(*(x,y)),y)
p13: *#(|1|(x),y) -> |0|#(*(x,y))
p14: *#(|1|(x),y) -> *#(x,y)
p15: *#(*(x,y),z) -> *#(x,*(y,z))
p16: *#(*(x,y),z) -> *#(y,z)
p17: sum#(nil()) -> |0|#(|#|())
p18: sum#(cons(x,l)) -> +#(x,sum(l))
p19: sum#(cons(x,l)) -> sum#(l)
p20: prod#(cons(x,l)) -> *#(x,prod(l))
p21: 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,y),z) -> +(x,+(y,z))
r9: *(|#|(),x) -> |#|()
r10: *(|0|(x),y) -> |0|(*(x,y))
r11: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r12: *(*(x,y),z) -> *(x,*(y,z))
r13: sum(nil()) -> |0|(|#|())
r14: sum(cons(x,l)) -> +(x,sum(l))
r15: prod(nil()) -> |1|(|#|())
r16: prod(cons(x,l)) -> *(x,prod(l))

The estimated dependency graph contains the following SCCs:

  {p19}
  {p21}
  {p11, p14, p15, p16}
  {p2, p3, p4, p6, p7, p8, p9}


-- 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,y),z) -> +(x,+(y,z))
r9: *(|#|(),x) -> |#|()
r10: *(|0|(x),y) -> |0|(*(x,y))
r11: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r12: *(*(x,y),z) -> *(x,*(y,z))
r13: sum(nil()) -> |0|(|#|())
r14: sum(cons(x,l)) -> +(x,sum(l))
r15: prod(nil()) -> |1|(|#|())
r16: 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 + 9
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          sum#_A(x1) = x1
          cons_A(x1,x2) = max{x1 + 92, x2 + 91}
          sum#!_A(x1) = max{2, x1 - 93}
          cons!_A(x1,x2) = max{1, x1 - 45}
  
    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,y),z) -> +(x,+(y,z))
r9: *(|#|(),x) -> |#|()
r10: *(|0|(x),y) -> |0|(*(x,y))
r11: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r12: *(*(x,y),z) -> *(x,*(y,z))
r13: sum(nil()) -> |0|(|#|())
r14: sum(cons(x,l)) -> +(x,sum(l))
r15: prod(nil()) -> |1|(|#|())
r16: 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 + 7
  
    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{122, x1 + 107, x2 + 1}
          prod#!_A(x1) = max{5, x1 - 123}
          cons!_A(x1,x2) = max{x1 + 4, x2 + 6}
  
    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: *#(*(x,y),z) -> *#(y,z)
p2: *#(*(x,y),z) -> *#(x,*(y,z))
p3: *#(|1|(x),y) -> *#(x,y)
p4: *#(|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,y),z) -> +(x,+(y,z))
r9: *(|#|(),x) -> |#|()
r10: *(|0|(x),y) -> |0|(*(x,y))
r11: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r12: *(*(x,y),z) -> *(x,*(y,z))
r13: sum(nil()) -> |0|(|#|())
r14: sum(cons(x,l)) -> +(x,sum(l))
r15: prod(nil()) -> |1|(|#|())
r16: prod(cons(x,l)) -> *(x,prod(l))

The set of usable rules consists of

  r2, r3

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 + 1
          *_A(x1,x2) = x1 + x2 + 7
          |1|_A(x1) = x1 + 6
          |0|_A(x1) = x1 + 2
          +_A(x1,x2) = x1 + x2 + 3
          |#|_A() = 0
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          *#_A(x1,x2) = 0
          *_A(x1,x2) = max{4, x2 - 56}
          |1|_A(x1) = max{9, x1 - 56}
          |0|_A(x1) = max{34, x1 - 56}
          +_A(x1,x2) = max{33, x1 + 22, x2 + 16}
          |#|_A = 16
          *#!_A(x1,x2) = max{1, x1 - 56}
          *!_A(x1,x2) = max{14, x1 - 56, x2 + 11}
          |1|!_A(x1) = max{2, x1 - 17}
          |0|!_A(x1) = max{10, x1 - 52}
          +!_A(x1,x2) = max{15, x1, x2 - 35}
          |#|!_A = 1
  
    partial status:
  
      pi(*#) = []
      pi(*) = [2]
      pi(|1|) = []
      pi(|0|) = []
      pi(+) = [1, 2]
      pi(|#|) = []

The following rules are removed:

  p1, p2, p3, p4

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: +#(+(x,y),z) -> +#(y,z)
p3: +#(+(x,y),z) -> +#(x,+(y,z))
p4: +#(|1|(x),|1|(y)) -> +#(x,y)
p5: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|()))
p6: +#(|0|(x),|1|(y)) -> +#(x,y)
p7: +#(|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,y),z) -> +(x,+(y,z))
r9: *(|#|(),x) -> |#|()
r10: *(|0|(x),y) -> |0|(*(x,y))
r11: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r12: *(*(x,y),z) -> *(x,*(y,z))
r13: sum(nil()) -> |0|(|#|())
r14: sum(cons(x,l)) -> +(x,sum(l))
r15: prod(nil()) -> |1|(|#|())
r16: prod(cons(x,l)) -> *(x,prod(l))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8

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 + 3
          |0|_A(x1) = x1 + 2
          +_A(x1,x2) = x1 + x2 + 1
          |1|_A(x1) = x1 + 7
          |#|_A() = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          +#_A(x1,x2) = 0
          |0|_A(x1) = 5
          +_A(x1,x2) = max{2, x1, x2}
          |1|_A(x1) = 26
          |#|_A = 5
          +#!_A(x1,x2) = 7
          |0|!_A(x1) = max{82, x1 - 6}
          +!_A(x1,x2) = max{60, x1 + 55}
          |1|!_A(x1) = max{6, x1 + 3}
          |#|!_A = 2
  
    partial status:
  
      pi(+#) = []
      pi(|0|) = []
      pi(+) = []
      pi(|1|) = []
      pi(|#|) = []

The following rules are removed:

  p1, p2, p4, p5, p6, p7



-- SCC decomposition.

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

p1: +#(+(x,y),z) -> +#(x,+(y,z))

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,y),z) -> +(x,+(y,z))
r9: *(|#|(),x) -> |#|()
r10: *(|0|(x),y) -> |0|(*(x,y))
r11: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r12: *(*(x,y),z) -> *(x,*(y,z))
r13: sum(nil()) -> |0|(|#|())
r14: sum(cons(x,l)) -> +(x,sum(l))
r15: prod(nil()) -> |1|(|#|())
r16: prod(cons(x,l)) -> *(x,prod(l))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: +#(+(x,y),z) -> +#(x,+(y,z))

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,y),z) -> +(x,+(y,z))
r9: *(|#|(),x) -> |#|()
r10: *(|0|(x),y) -> |0|(*(x,y))
r11: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r12: *(*(x,y),z) -> *(x,*(y,z))
r13: sum(nil()) -> |0|(|#|())
r14: sum(cons(x,l)) -> +(x,sum(l))
r15: prod(nil()) -> |1|(|#|())
r16: prod(cons(x,l)) -> *(x,prod(l))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9

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 + 1
          +_A(x1,x2) = x1 + x2 + 2
          |0|_A(x1) = x1 + 1
          |#|_A() = 1
          |1|_A(x1) = x1 + 5
          *_A(x1,x2) = 6
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          +#_A(x1,x2) = max{0, x1 - 64, x2 - 95}
          +_A(x1,x2) = max{63, x1 + 31, x2}
          |0|_A(x1) = 33
          |#|_A = 5
          |1|_A(x1) = 77
          *_A(x1,x2) = max{63, x1 - 78, x2 - 78}
          +#!_A(x1,x2) = max{79, x1 - 64, x2 - 95}
          +!_A(x1,x2) = max{79, x1 - 34, x2 - 67}
          |0|!_A(x1) = max{60, x1 - 83}
          |#|!_A = 1
          |1|!_A(x1) = max{1, x1 - 2}
          *!_A(x1,x2) = max{6, x2 - 68}
  
    partial status:
  
      pi(+#) = [1]
      pi(+) = [1]
      pi(|0|) = [1]
      pi(|#|) = []
      pi(|1|) = [1]
      pi(*) = []

The following rules are removed:

  p1

Then no dependency pair remains.