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))
  app(nil(),l) -> l
  app(cons(x,l1),l2) -> cons(x,app(l1,l2))
  sum(nil()) -> |0|()
  sum(cons(x,l)) -> +(x,sum(l))
  sum(app(l1,l2)) -> +(sum(l1),sum(l2))
  prod(nil()) -> s(|0|())
  prod(cons(x,l)) -> *(x,prod(l))
  prod(app(l1,l2)) -> *(prod(l1),prod(l2))


-- 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: app#(cons(x,l1),l2) -> app#(l1,l2)
p10: sum#(cons(x,l)) -> +#(x,sum(l))
p11: sum#(cons(x,l)) -> sum#(l)
p12: sum#(app(l1,l2)) -> +#(sum(l1),sum(l2))
p13: sum#(app(l1,l2)) -> sum#(l1)
p14: sum#(app(l1,l2)) -> sum#(l2)
p15: prod#(cons(x,l)) -> *#(x,prod(l))
p16: prod#(cons(x,l)) -> prod#(l)
p17: prod#(app(l1,l2)) -> *#(prod(l1),prod(l2))
p18: prod#(app(l1,l2)) -> prod#(l1)
p19: prod#(app(l1,l2)) -> prod#(l2)

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: app(nil(),l) -> l
r10: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r11: sum(nil()) -> |0|()
r12: sum(cons(x,l)) -> +(x,sum(l))
r13: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r14: prod(nil()) -> s(|0|())
r15: prod(cons(x,l)) -> *(x,prod(l))
r16: prod(app(l1,l2)) -> *(prod(l1),prod(l2))

The estimated dependency graph contains the following SCCs:

  {p11, p13, p14}
  {p16, p18, p19}
  {p5, p7, p8}
  {p1, p2, p3}
  {p9}


-- Reduction pair.

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

p1: sum#(app(l1,l2)) -> sum#(l2)
p2: sum#(app(l1,l2)) -> sum#(l1)
p3: 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: app(nil(),l) -> l
r10: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r11: sum(nil()) -> |0|()
r12: sum(cons(x,l)) -> +(x,sum(l))
r13: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r14: prod(nil()) -> s(|0|())
r15: prod(cons(x,l)) -> *(x,prod(l))
r16: prod(app(l1,l2)) -> *(prod(l1),prod(l2))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              sum#_A(x1) = 0
              app_A(x1,x2) = 1
              cons_A(x1,x2) = 0
              sum#!_A(x1) = 4
              app!_A(x1,x2) = 4
              cons!_A(x1,x2) = 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1) = 0
            app_A(x1,x2) = max{x1 + 14, x2 + 3}
            cons_A(x1,x2) = max{19, x1 - 26, x2 + 4}
            sum#!_A(x1) = max{1, x1 - 9}
            app!_A(x1,x2) = max{4, x1 - 8}
            cons!_A(x1,x2) = max{0, x1 - 1, x2 - 40}
        
  
    partial status:
  
      pi(sum#) = []
      pi(app) = [1, 2]
      pi(cons) = [2]

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: prod#(app(l1,l2)) -> prod#(l2)
p2: prod#(app(l1,l2)) -> prod#(l1)
p3: 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: app(nil(),l) -> l
r10: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r11: sum(nil()) -> |0|()
r12: sum(cons(x,l)) -> +(x,sum(l))
r13: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r14: prod(nil()) -> s(|0|())
r15: prod(cons(x,l)) -> *(x,prod(l))
r16: prod(app(l1,l2)) -> *(prod(l1),prod(l2))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. 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
              app_A(x1,x2) = x1 + x2 + 2
              cons_A(x1,x2) = x1 + x2 + 2
              prod#!_A(x1) = x1 + 3
              app!_A(x1,x2) = x2 + 2
              cons!_A(x1,x2) = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            prod#_A(x1) = 0
            app_A(x1,x2) = max{0, x1 - 28, x2 - 26}
            cons_A(x1,x2) = max{2, x1 - 28, x2}
            prod#!_A(x1) = max{2, x1 - 22}
            app!_A(x1,x2) = max{0, x1 - 21, x2 - 20}
            cons!_A(x1,x2) = max{1, x1 - 29, x2 - 48}
        
  
    partial status:
  
      pi(prod#) = []
      pi(app) = [1, 2]
      pi(cons) = [2]

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: *#(*(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: app(nil(),l) -> l
r10: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r11: sum(nil()) -> |0|()
r12: sum(cons(x,l)) -> +(x,sum(l))
r13: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r14: prod(nil()) -> s(|0|())
r15: prod(cons(x,l)) -> *(x,prod(l))
r16: prod(app(l1,l2)) -> *(prod(l1),prod(l2))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. 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 + 2
              s_A(x1) = x1 + 2
              +_A(x1,x2) = x1 + x2 + 1
              |0|_A() = 0
              *#!_A(x1,x2) = 0
              *!_A(x1,x2) = 1
              s!_A(x1) = x1 + 4
              +!_A(x1,x2) = 3
              |0|!_A() = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            *#_A(x1,x2) = 0
            *_A(x1,x2) = 104
            s_A(x1) = max{78, x1 + 25}
            +_A(x1,x2) = max{80, x1 + 24, x2 + 52}
            |0|_A = 105
            *#!_A(x1,x2) = 1
            *!_A(x1,x2) = max{x1 - 123, x2 + 50}
            s!_A(x1) = max{129, x1 + 2}
            +!_A(x1,x2) = x1 + 3
            |0|!_A = 50
        
  
    partial status:
  
      pi(*#) = [1]
      pi(*) = [1, 2]
      pi(s) = [1]
      pi(+) = [1, 2]
      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: app(nil(),l) -> l
r10: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r11: sum(nil()) -> |0|()
r12: sum(cons(x,l)) -> +(x,sum(l))
r13: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r14: prod(nil()) -> s(|0|())
r15: prod(cons(x,l)) -> *(x,prod(l))
r16: prod(app(l1,l2)) -> *(prod(l1),prod(l2))

The set of usable rules consists of

  r1, r2, r3, r4

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              +#_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 3
              +_A(x1,x2) = x1 + x2 + 1
              |0|_A() = 3
              +#!_A(x1,x2) = x1 + 4
              s!_A(x1) = 2
              +!_A(x1,x2) = x1 + 3
              |0|!_A() = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            +#_A(x1,x2) = max{0, x1 - 13, x2 - 14}
            s_A(x1) = max{2, x1}
            +_A(x1,x2) = max{12, x1 + 1, x2}
            |0|_A = 6
            +#!_A(x1,x2) = max{x1 + 20, x2 + 7}
            s!_A(x1) = max{0, x1 - 3}
            +!_A(x1,x2) = max{5, x1 - 6}
            |0|!_A = 1
        
  
    partial status:
  
      pi(+#) = []
      pi(s) = []
      pi(+) = [1, 2]
      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: app#(cons(x,l1),l2) -> app#(l1,l2)

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: app(nil(),l) -> l
r10: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r11: sum(nil()) -> |0|()
r12: sum(cons(x,l)) -> +(x,sum(l))
r13: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r14: prod(nil()) -> s(|0|())
r15: prod(cons(x,l)) -> *(x,prod(l))
r16: prod(app(l1,l2)) -> *(prod(l1),prod(l2))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              app#_A(x1,x2) = x1
              cons_A(x1,x2) = x2 + 1
              app#!_A(x1,x2) = x1 + 1
              cons!_A(x1,x2) = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = 0
            cons_A(x1,x2) = max{0, x1 - 17, x2 - 17}
            app#!_A(x1,x2) = max{x1 - 1, x2 + 2}
            cons!_A(x1,x2) = 0
        
  
    partial status:
  
      pi(app#) = [2]
      pi(cons) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.