YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: +#(s(x),s(y)) -> +#(x,y)
p2: +#(+(x,y),z) -> +#(x,+(y,z))
p3: +#(+(x,y),z) -> +#(y,z)
p4: *#(s(x),s(y)) -> +#(*(x,y),+(x,y))
p5: *#(s(x),s(y)) -> *#(x,y)
p6: *#(s(x),s(y)) -> +#(x,y)
p7: *#(*(x,y),z) -> *#(x,*(y,z))
p8: *#(*(x,y),z) -> *#(y,z)
p9: sum#(cons(x,l)) -> +#(x,sum(l))
p10: sum#(cons(x,l)) -> sum#(l)
p11: prod#(cons(x,l)) -> *#(x,prod(l))
p12: prod#(cons(x,l)) -> prod#(l)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p10}
  {p12}
  {p5, p7, p8}
  {p1, p2, p3}


-- 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: +(x,|0|()) -> x
r2: +(|0|(),x) -> x
r3: +(s(x),s(y)) -> s(s(+(x,y)))
r4: +(+(x,y),z) -> +(x,+(y,z))
r5: *(x,|0|()) -> |0|()
r6: *(|0|(),x) -> |0|()
r7: *(s(x),s(y)) -> s(+(*(x,y),+(x,y)))
r8: *(*(x,y),z) -> *(x,*(y,z))
r9: sum(nil()) -> |0|()
r10: sum(cons(x,l)) -> +(x,sum(l))
r11: prod(nil()) -> s(|0|())
r12: 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. lexicographic path order with precedence:
    
      precedence:
      
        cons > sum#
      
      argument filter:
    
        pi(sum#) = 1
        pi(cons) = [1, 2]
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        sum#_A(x1) = x1
        cons_A(x1,x2) = x1 + x2 + 1
    

The next rules are strictly ordered:

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

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: +(x,|0|()) -> x
r2: +(|0|(),x) -> x
r3: +(s(x),s(y)) -> s(s(+(x,y)))
r4: +(+(x,y),z) -> +(x,+(y,z))
r5: *(x,|0|()) -> |0|()
r6: *(|0|(),x) -> |0|()
r7: *(s(x),s(y)) -> s(+(*(x,y),+(x,y)))
r8: *(*(x,y),z) -> *(x,*(y,z))
r9: sum(nil()) -> |0|()
r10: sum(cons(x,l)) -> +(x,sum(l))
r11: prod(nil()) -> s(|0|())
r12: 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. lexicographic path order with precedence:
    
      precedence:
      
        cons > prod#
      
      argument filter:
    
        pi(prod#) = 1
        pi(cons) = [1, 2]
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        prod#_A(x1) = x1
        cons_A(x1,x2) = x1 + x2 + 1
    

The next rules are strictly ordered:

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

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: *#(*(x,y),z) -> *#(y,z)
p2: *#(*(x,y),z) -> *#(x,*(y,z))
p3: *#(s(x),s(y)) -> *#(x,y)

and R consists of:

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

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        |0| > * > + > *# > s
      
      argument filter:
    
        pi(*#) = 1
        pi(*) = [1, 2]
        pi(s) = [1]
        pi(+) = [1, 2]
        pi(|0|) = []
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        *#_A(x1,x2) = 0
        *_A(x1,x2) = 1
        s_A(x1) = 2
        +_A(x1,x2) = x2 + 1
        |0|_A() = 2
    

The next rules are strictly ordered:

  p1, p2, p3

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: +#(s(x),s(y)) -> +#(x,y)
p2: +#(+(x,y),z) -> +#(y,z)
p3: +#(+(x,y),z) -> +#(x,+(y,z))

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3, r4

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        |0| > s > + > +#
      
      argument filter:
    
        pi(+#) = [1]
        pi(s) = 1
        pi(+) = [1, 2]
        pi(|0|) = []
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        +#_A(x1,x2) = 0
        s_A(x1) = 1
        +_A(x1,x2) = 2
        |0|_A() = 1
    

The next rules are strictly ordered:

  p2, p3

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

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

and R consists of:

r1: +(x,|0|()) -> x
r2: +(|0|(),x) -> x
r3: +(s(x),s(y)) -> s(s(+(x,y)))
r4: +(+(x,y),z) -> +(x,+(y,z))
r5: *(x,|0|()) -> |0|()
r6: *(|0|(),x) -> |0|()
r7: *(s(x),s(y)) -> s(+(*(x,y),+(x,y)))
r8: *(*(x,y),z) -> *(x,*(y,z))
r9: sum(nil()) -> |0|()
r10: sum(cons(x,l)) -> +(x,sum(l))
r11: prod(nil()) -> s(|0|())
r12: 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. lexicographic path order with precedence:
    
      precedence:
      
        s > +#
      
      argument filter:
    
        pi(+#) = 2
        pi(s) = [1]
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        +#_A(x1,x2) = 0
        s_A(x1) = x1 + 1
    

The next rules are strictly ordered:

  p1

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