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, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        sum#_A(x1) = max{2, x1 - 16}
        cons_A(x1,x2) = max{19, x1 - 36, x2 + 15}
        *_A(x1,x2) = max{18, x1 - 54, x2 - 20}
        |0|_A = 13
  
    base order 2:
  
      precedence:
      
        cons > sum# > * > |0|
      
  
    partial status:
  
      pi(sum#) = []
      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: 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, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        prod#_A(x1) = max{2, x1 - 16}
        cons_A(x1,x2) = max{19, x1 - 36, x2 + 15}
        *_A(x1,x2) = max{18, x1 - 54, x2 - 20}
        |0|_A = 13
  
    base order 2:
  
      precedence:
      
        cons > prod# > * > |0|
      
  
    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

  r1, r5

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        *#_A(x1,x2) = max{x1 - 3, x2 + 5}
        s_A(x1) = x1 + 6
        +_A(x1,x2) = max{x1 + 23, x2 - 17}
        |0|_A = 1
        *_A(x1,x2) = max{7, x1 - 41, x2 - 39}
  
    base order 2:
  
      precedence:
      
        + > *# = * > s = |0|
      
  
    partial status:
  
      pi(*#) = [2]
      pi(s) = [1]
      pi(+) = [1]
      pi(|0|) = []
      pi(*) = []

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

  r1, r5

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        +#_A(x1,x2) = max{x1 - 1, x2 + 6}
        s_A(x1) = max{9, x1 + 5}
        +_A(x1,x2) = max{x1 + 24, x2 - 19}
        |0|_A = 2
        *_A(x1,x2) = max{10, x1 - 39, x2 - 38}
  
    base order 2:
  
      precedence:
      
        + > * > +# = s > |0|
      
  
    partial status:
  
      pi(+#) = [2]
      pi(s) = [1]
      pi(+) = [1]
      pi(|0|) = []
      pi(*) = []

The following rules are removed:

  p1

Then no dependency pair remains.