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

  r5

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        sum#_A(x1) = 0
        cons_A(x1,x2) = max{15, x1 - 2, x2 - 2}
        *_A(x1,x2) = max{9, x1, x2 - 11}
        |0|_A = 0
  
    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{12, x1 + 8, x2 + 9}
          *_A(x1,x2) = max{8, x1 - 5, x2 - 5}
          |0|_A = 4
          sum#!_A(x1) = max{1, x1 - 10}
          cons!_A(x1,x2) = max{5, x1 - 9, x2 - 9}
          *!_A(x1,x2) = max{3, x1 - 10, x2 - 5}
          |0|!_A = 0
  
    partial status:
  
      pi(sum#) = []
      pi(cons) = []
      pi(*) = []
      pi(|0|) = []

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

  r5

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        prod#_A(x1) = 0
        cons_A(x1,x2) = max{10, x1 - 4, x2 - 4}
        *_A(x1,x2) = max{4, x1 - 6, x2 - 3}
        |0|_A = 0
  
    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{10, x1 + 7, x2 + 1}
          *_A(x1,x2) = max{9, x1 - 5, x2 - 5}
          |0|_A = 1
          prod#!_A(x1) = max{1, x1 - 8}
          cons!_A(x1,x2) = max{6, x1 - 7, x2 - 7}
          *!_A(x1,x2) = max{4, x1 - 18, x2 - 2}
          |0|!_A = 2
  
    partial status:
  
      pi(prod#) = []
      pi(cons) = []
      pi(*) = []
      pi(|0|) = []

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:
  
      max/plus interpretation on natural numbers:
      
        *#_A(x1,x2) = x2
        s_A(x1) = x1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          *#_A(x1,x2) = 0
          s_A(x1) = max{26, x1 - 103}
          *#!_A(x1,x2) = 0
          s!_A(x1) = max{52, x1 - 52}
  
    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:
  
      max/plus interpretation on natural numbers:
      
        +#_A(x1,x2) = x2 + 7
        s_A(x1) = x1 + 7
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          +#_A(x1,x2) = 0
          s_A(x1) = max{10, x1 - 53}
          +#!_A(x1,x2) = 0
          s!_A(x1) = max{5, x1 - 25}
  
    partial status:
  
      pi(+#) = [2]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.