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

  r5, r6

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        sum#_A(x1) = max{4, x1 + 2}
        cons_A(x1,x2) = max{7, x1 - 9, x2 + 3}
        *_A(x1,x2) = max{4, x1 - 19, x2 - 6}
        |0|_A = 1
  
    base order 2:
  
      precedence:
      
        sum# > cons > * > |0|
      
  
    partial status:
  
      pi(sum#) = [1]
      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,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

  r5, r6

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        prod#_A(x1) = max{4, x1 + 2}
        cons_A(x1,x2) = max{7, x1 - 9, x2 + 3}
        *_A(x1,x2) = max{4, x1 - 19, x2 - 6}
        |0|_A = 1
  
    base order 2:
  
      precedence:
      
        prod# > cons > * > |0|
      
  
    partial status:
  
      pi(prod#) = [1]
      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: *#(*(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

  r5, r6

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        *#_A(x1,x2) = max{1, x1 - 1}
        *_A(x1,x2) = max{15, x1 + 8, x2 + 8}
        s_A(x1) = max{14, x1 + 4}
        |0|_A = 16
  
    base order 2:
  
      precedence:
      
        * > *# > s > |0|
      
  
    partial status:
  
      pi(*#) = []
      pi(*) = [1]
      pi(s) = [1]
      pi(|0|) = []

The following rules are removed:

  p2



-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: *#(*(x,y),z) -> *#(y,z)
p2: *#(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 estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: *#(*(x,y),z) -> *#(y,z)
p2: *#(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

  r5, r6

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        *#_A(x1,x2) = max{11, x2 + 2}
        *_A(x1,x2) = max{4, x1 - 16, x2 - 37}
        s_A(x1) = max{13, x1 + 10}
        |0|_A = 0
  
    base order 2:
  
      precedence:
      
        *# = s > * > |0|
      
  
    partial status:
  
      pi(*#) = []
      pi(*) = []
      pi(s) = [1]
      pi(|0|) = []

The following rules are removed:

  p2



-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: *#(*(x,y),z) -> *#(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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: *#(*(x,y),z) -> *#(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

  r5, r6

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        *#_A(x1,x2) = max{x1 + 47, x2 + 24}
        *_A(x1,x2) = max{22, x1 - 23, x2 + 20}
        |0|_A = 2
  
    base order 2:
  
      precedence:
      
        *# = * > |0|
      
  
    partial status:
  
      pi(*#) = [1]
      pi(*) = [2]
      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)
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

  r1, r2, r5, r6

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        +#_A(x1,x2) = max{9, x1 + 2}
        s_A(x1) = max{10, x1 + 8}
        +_A(x1,x2) = max{11, x1 + 8, x2 + 8}
        |0|_A = 0
        *_A(x1,x2) = max{3, x1 - 22, x2 - 26}
  
    base order 2:
  
      precedence:
      
        + > * > +# = s > |0|
      
  
    partial status:
  
      pi(+#) = [1]
      pi(s) = [1]
      pi(+) = []
      pi(|0|) = []
      pi(*) = []

The following rules are removed:

  p3



-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: +#(s(x),s(y)) -> +#(x,y)
p2: +#(+(x,y),z) -> +#(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 estimated dependency graph contains the following SCCs:

  {p1, p2}


-- 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)

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, r5, r6

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        +#_A(x1,x2) = x1
        s_A(x1) = max{10, x1 + 6}
        +_A(x1,x2) = max{x1 - 13, x2}
        |0|_A = 4
        *_A(x1,x2) = max{5, x1 - 20, x2 - 13}
  
    base order 2:
  
      precedence:
      
        |0| = * > s = + > +#
      
  
    partial status:
  
      pi(+#) = [1]
      pi(s) = []
      pi(+) = [2]
      pi(|0|) = []
      pi(*) = []

The following rules are removed:

  p2



-- SCC decomposition.

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,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:

  {p1}


-- 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,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

  r5, r6

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        +#_A(x1,x2) = max{1, x1 - 1, x2 - 12}
        s_A(x1) = x1 + 11
        *_A(x1,x2) = max{4, x1 - 33, x2 - 6}
        |0|_A = 0
  
    base order 2:
  
      precedence:
      
        +# > s > * > |0|
      
  
    partial status:
  
      pi(+#) = []
      pi(s) = [1]
      pi(*) = []
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.