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

  r2, r3

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        sum#_A(x1) = x1 + 12
        cons_A(x1,x2) = x2 + 13
        +_A(x1,x2) = max{17, x1 - 18, x2 + 4}
        |0|_A = 10
        s_A(x1) = 13
  
    base order 2:
  
      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
            +_A(x1,x2) = x2 + 2
            |0|_A() = 4
            s_A(x1) = 4
            sum#!_A(x1) = 1
            cons!_A(x1,x2) = 2
            +!_A(x1,x2) = x1 + x2 + 11
            |0|!_A() = 0
            s!_A(x1) = 0
  
    partial status:
  
      pi(sum#) = [1]
      pi(cons) = [2]
      pi(+) = []
      pi(|0|) = []
      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: 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

  r2, r3

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        prod#_A(x1) = max{11, x1 + 9}
        cons_A(x1,x2) = x2 + 11
        +_A(x1,x2) = max{21, x1 - 22, x2 + 5}
        |0|_A = 12
        s_A(x1) = 11
  
    base order 2:
  
      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
            +_A(x1,x2) = x2 + 2
            |0|_A() = 8
            s_A(x1) = 7
            prod#!_A(x1) = 4
            cons!_A(x1,x2) = 5
            +!_A(x1,x2) = x1 + x2
            |0|!_A() = 3
            s!_A(x1) = 1
  
    partial status:
  
      pi(prod#) = [1]
      pi(cons) = [2]
      pi(+) = []
      pi(|0|) = []
      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: *#(*(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, r6

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        *#_A(x1,x2) = max{11, x1 + 9}
        *_A(x1,x2) = max{16, x1 + 10, x2 + 12}
        s_A(x1) = max{12, x1 + 10}
        +_A(x1,x2) = max{x1 + 9, x2 + 28}
        |0|_A = 0
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            *#_A(x1,x2) = 0
            *_A(x1,x2) = x1 + 3
            s_A(x1) = x1 + 7
            +_A(x1,x2) = x1 + x2 + 2
            |0|_A() = 4
            *#!_A(x1,x2) = x1 + 9
            *!_A(x1,x2) = 3
            s!_A(x1) = x1 + 10
            +!_A(x1,x2) = x1 + x2 + 4
            |0|!_A() = 0
  
    partial status:
  
      pi(*#) = [1]
      pi(*) = []
      pi(s) = []
      pi(+) = [1]
      pi(|0|) = []

The following rules are removed:

  p1, p2, p3

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

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        +#_A(x1,x2) = 0
        s_A(x1) = x1 + 12
        +_A(x1,x2) = 12
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            +#_A(x1,x2) = 0
            s_A(x1) = x1 + 1
            +_A(x1,x2) = x1 + x2 + 2
            +#!_A(x1,x2) = x1 + 1
            s!_A(x1) = 3
            +!_A(x1,x2) = 3
  
    partial status:
  
      pi(+#) = []
      pi(s) = [1]
      pi(+) = []

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.