YES
We show the termination of the TRS R:

  +(x,|0|()) -> x
  +(|0|(),x) -> x
  +(s(x),s(y)) -> s(s(+(x,y)))
  *(x,|0|()) -> |0|()
  *(|0|(),x) -> |0|()
  *(s(x),s(y)) -> s(+(*(x,y),+(x,y)))
  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: *#(s(x),s(y)) -> +#(*(x,y),+(x,y))
p3: *#(s(x),s(y)) -> *#(x,y)
p4: *#(s(x),s(y)) -> +#(x,y)
p5: sum#(cons(x,l)) -> +#(x,sum(l))
p6: sum#(cons(x,l)) -> sum#(l)
p7: prod#(cons(x,l)) -> *#(x,prod(l))
p8: 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,|0|()) -> |0|()
r5: *(|0|(),x) -> |0|()
r6: *(s(x),s(y)) -> s(+(*(x,y),+(x,y)))
r7: sum(nil()) -> |0|()
r8: sum(cons(x,l)) -> +(x,sum(l))
r9: prod(nil()) -> s(|0|())
r10: prod(cons(x,l)) -> *(x,prod(l))

The estimated dependency graph contains the following SCCs:

  {p6}
  {p8}
  {p3}
  {p1}


-- 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,|0|()) -> |0|()
r5: *(|0|(),x) -> |0|()
r6: *(s(x),s(y)) -> s(+(*(x,y),+(x,y)))
r7: sum(nil()) -> |0|()
r8: sum(cons(x,l)) -> +(x,sum(l))
r9: prod(nil()) -> s(|0|())
r10: 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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              sum#_A(x1) = x1
              cons_A(x1,x2) = x1 + x2 + 1
              sum#!_A(x1) = x1 + 1
              cons!_A(x1,x2) = x2 + 2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1) = x1
            cons_A(x1,x2) = max{25, x1 - 3, x2 + 1}
            sum#!_A(x1) = max{1, x1 - 28}
            cons!_A(x1,x2) = max{x1 + 4, x2 - 27}
        
  
    partial status:
  
      pi(sum#) = [1]
      pi(cons) = [1, 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: +(x,|0|()) -> x
r2: +(|0|(),x) -> x
r3: +(s(x),s(y)) -> s(s(+(x,y)))
r4: *(x,|0|()) -> |0|()
r5: *(|0|(),x) -> |0|()
r6: *(s(x),s(y)) -> s(+(*(x,y),+(x,y)))
r7: sum(nil()) -> |0|()
r8: sum(cons(x,l)) -> +(x,sum(l))
r9: prod(nil()) -> s(|0|())
r10: 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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              prod#_A(x1) = x1
              cons_A(x1,x2) = x1 + x2 + 1
              prod#!_A(x1) = x1 + 2
              cons!_A(x1,x2) = x2 + 2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            prod#_A(x1) = x1
            cons_A(x1,x2) = max{x1, x2}
            prod#!_A(x1) = max{4, x1 - 1}
            cons!_A(x1,x2) = max{4, x1 + 1, x2 + 3}
        
  
    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: *#(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,|0|()) -> |0|()
r5: *(|0|(),x) -> |0|()
r6: *(s(x),s(y)) -> s(+(*(x,y),+(x,y)))
r7: sum(nil()) -> |0|()
r8: sum(cons(x,l)) -> +(x,sum(l))
r9: prod(nil()) -> s(|0|())
r10: 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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              *#_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 1
              *#!_A(x1,x2) = 3
              s!_A(x1) = 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            *#_A(x1,x2) = max{x1, x2}
            s_A(x1) = max{17, x1 + 14}
            *#!_A(x1,x2) = 7
            s!_A(x1) = 7
        
  
    partial status:
  
      pi(*#) = [2]
      pi(s) = [1]

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: +#(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,|0|()) -> |0|()
r5: *(|0|(),x) -> |0|()
r6: *(s(x),s(y)) -> s(+(*(x,y),+(x,y)))
r7: sum(nil()) -> |0|()
r8: sum(cons(x,l)) -> +(x,sum(l))
r9: prod(nil()) -> s(|0|())
r10: 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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              +#_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 1
              +#!_A(x1,x2) = 2
              s!_A(x1) = 2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            +#_A(x1,x2) = 0
            s_A(x1) = 12
            +#!_A(x1,x2) = 4
            s!_A(x1) = 4
        
  
    partial status:
  
      pi(+#) = [2]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.