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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          sum#_A(x1) = x1
          cons_A(x1,x2) = x2 + 5
  
    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 - 1, x2}
          sum#!_A(x1) = max{1, x1 - 41}
          cons!_A(x1,x2) = max{3, x1 - 40, x2 - 40}
  
    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: +(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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          prod#_A(x1) = x1
          cons_A(x1,x2) = x2 + 4
          *_A(x1,x2) = 3
          |0|_A() = 1
  
    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{x1 - 1, x2 + 1}
          *_A(x1,x2) = max{52, x1 - 51, x2 - 51}
          |0|_A = 4
          prod#!_A(x1) = max{1, x1 - 74}
          cons!_A(x1,x2) = max{3, x1 - 4, x2 - 73}
          *!_A(x1,x2) = max{2, x1 - 5, x2 - 7}
          |0|!_A = 1
  
    partial status:
  
      pi(prod#) = []
      pi(cons) = [2]
      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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          *#_A(x1,x2) = 0
          s_A(x1) = 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          *#_A(x1,x2) = max{x1, x2}
          s_A(x1) = x1 + 13
          *#!_A(x1,x2) = max{8, x1 - 1, x2 - 47}
          s!_A(x1) = max{7, x1 - 48}
  
    partial status:
  
      pi(*#) = []
      pi(s) = []

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

  r5

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) = 0
          s_A(x1) = 3
          *_A(x1,x2) = 2
          |0|_A() = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          +#_A(x1,x2) = 0
          s_A(x1) = max{14, x1 + 4}
          *_A(x1,x2) = max{10, x1 - 16, x2 - 37}
          |0|_A = 5
          +#!_A(x1,x2) = max{1, x2 - 2}
          s!_A(x1) = max{0, x1 - 43}
          *!_A(x1,x2) = max{1, x1 - 79}
          |0|!_A = 2
  
    partial status:
  
      pi(+#) = []
      pi(s) = []
      pi(*) = []
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.