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 monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        sum#_A(x1) = ((1,1),(1,1)) x1
        cons_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        sum#_A(x1) = ((1,0),(1,1)) x1
        cons_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,0),(1,1)) x2 + (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        sum#_A(x1) = ((1,1),(1,1)) x1
        cons_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
    

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14

We remove them from the problem.  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 monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        prod#_A(x1) = ((1,1),(1,0)) x1
        cons_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        prod#_A(x1) = x1
        cons_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,1),(0,0)) x2 + (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        prod#_A(x1) = ((1,1),(1,1)) x1
        cons_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
    

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14

We remove them from the problem.  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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        *#_A(x1,x2) = ((1,1),(1,1)) x1
        |1|_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        |0|_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        *#_A(x1,x2) = ((1,1),(1,0)) x1
        |1|_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        |0|_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        *#_A(x1,x2) = ((1,1),(1,1)) x1
        |1|_A(x1) = ((0,1),(1,1)) x1 + (1,1)
        |0|_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.  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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        +#_A(x1,x2) = x1 + x2
        |0|_A(x1) = x1 + (1,1)
        |1|_A(x1) = x1 + (4,1)
        +_A(x1,x2) = x1 + x2 + (1,1)
        |#|_A() = (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        +#_A(x1,x2) = ((0,1),(0,0)) x1 + ((0,1),(1,0)) x2
        |0|_A(x1) = ((1,0),(1,0)) x1 + (1,1)
        |1|_A(x1) = x1 + (1,1)
        +_A(x1,x2) = ((1,1),(0,0)) x1 + x2 + (1,0)
        |#|_A() = (1,3)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        +#_A(x1,x2) = (0,0)
        |0|_A(x1) = (4,1)
        |1|_A(x1) = ((1,1),(1,0)) x1 + (1,4)
        +_A(x1,x2) = ((0,0),(1,1)) x1 + (3,1)
        |#|_A() = (1,1)
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5

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