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!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1) = 0
            app_A(x1,x2) = max{1, x1 - 44, x2 - 42}
            cons_A(x1,x2) = max{2, x1 - 42, x2 - 42}
            sum#!_A(x1) = max{1, x1 - 89}
            app!_A(x1,x2) = max{2, x1 - 47, x2 - 47}
            cons!_A(x1,x2) = max{1, x1 - 89, x2 - 89}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1) = 0
            app_A(x1,x2) = max{1, x1 - 24, x2 - 21}
            cons_A(x1,x2) = max{3, x1 - 2, x2 - 21}
            sum#!_A(x1) = max{0, x1 - 56}
            app!_A(x1,x2) = max{1, x1 - 32, x2 - 55}
            cons!_A(x1,x2) = max{0, x1 - 33, x2 - 33}
        
  
    partial status:
  
      pi(sum#) = [1]
      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!):
        
          max/plus interpretation on natural numbers:
          
            prod#_A(x1) = 0
            app_A(x1,x2) = max{1, x1 - 28, x2 - 26}
            cons_A(x1,x2) = max{2, x1 - 1, x2 - 26}
            prod#!_A(x1) = max{0, x1 - 11}
            app!_A(x1,x2) = max{1, x1 - 11, x2 - 8}
            cons!_A(x1,x2) = max{0, x1 - 10, x2 - 8}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            prod#_A(x1) = 0
            app_A(x1,x2) = max{1, x1 - 20, x2 - 31}
            cons_A(x1,x2) = max{2, x1 - 20, x2 - 18}
            prod#!_A(x1) = max{1, x1 - 110}
            app!_A(x1,x2) = max{2, x1 - 110, x2 - 109}
            cons!_A(x1,x2) = max{1, x1 - 80, x2 - 77}
        
  
    partial status:
  
      pi(prod#) = [1]
      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, r3, r4, r5, r6, r7, r8

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!):
        
          max/plus interpretation on natural numbers:
          
            *#_A(x1,x2) = max{0, x1 - 10, x2 - 19}
            *_A(x1,x2) = max{x1 + 9, x2}
            s_A(x1) = max{6, x1}
            +_A(x1,x2) = max{10, x1, x2}
            |0|_A = 10
            *#!_A(x1,x2) = x1 + 2
            *!_A(x1,x2) = max{12, x1 - 11}
            s!_A(x1) = 12
            +!_A(x1,x2) = max{9, x1 + 4, x2 - 22}
            |0|!_A = 10
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            *#_A(x1,x2) = max{0, x1 - 4, x2 - 7}
            *_A(x1,x2) = max{x1 + 3, x2}
            s_A(x1) = max{3, x1}
            +_A(x1,x2) = max{3, x1, x2}
            |0|_A = 3
            *#!_A(x1,x2) = max{1, x2 - 1}
            *!_A(x1,x2) = max{7, x1 - 6, x2 - 10}
            s!_A(x1) = 5
            +!_A(x1,x2) = max{11, x1 + 1, x2 - 4}
            |0|!_A = 6
        
  
    partial status:
  
      pi(*#) = []
      pi(*) = [1, 2]
      pi(s) = []
      pi(+) = [1, 2]
      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: 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:

  {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: 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!):
        
          max/plus interpretation on natural numbers:
          
            *#_A(x1,x2) = 0
            *_A(x1,x2) = max{5, x1 - 18, x2 - 17}
            s_A(x1) = max{6, x1 - 7}
            *#!_A(x1,x2) = max{0, x1 - 40}
            *!_A(x1,x2) = max{4, x1 - 28, x2 - 39}
            s!_A(x1) = max{2, x1 - 38}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            *#_A(x1,x2) = 0
            *_A(x1,x2) = max{3, x1 - 11, x2 - 10}
            s_A(x1) = max{8, x1 - 11}
            *#!_A(x1,x2) = max{9, x1 - 9, x2 - 9}
            *!_A(x1,x2) = max{10, x1 - 5, x2 - 7}
            s!_A(x1) = 8
        
  
    partial status:
  
      pi(*#) = [1]
      pi(*) = [1, 2]
      pi(s) = [1]

The following rules are removed:

  p1, p2

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!):
        
          max/plus interpretation on natural numbers:
          
            +#_A(x1,x2) = max{0, x1 - 11, x2 - 21}
            s_A(x1) = max{15, x1}
            +_A(x1,x2) = max{x1 + 10, x2}
            |0|_A = 13
            +#!_A(x1,x2) = 1
            s!_A(x1) = 4
            +!_A(x1,x2) = 40
            |0|!_A = 38
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            +#_A(x1,x2) = max{0, x1 - 34, x2 - 50}
            s_A(x1) = max{14, x1}
            +_A(x1,x2) = max{33, x1 + 16, x2}
            |0|_A = 5
            +#!_A(x1,x2) = max{0, x1 - 15}
            s!_A(x1) = max{1, x1 - 40}
            +!_A(x1,x2) = 0
            |0|!_A = 3
        
  
    partial status:
  
      pi(+#) = []
      pi(s) = []
      pi(+) = [1, 2]
      pi(|0|) = []

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: 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:

  {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: 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!):
        
          max/plus interpretation on natural numbers:
          
            +#_A(x1,x2) = 0
            s_A(x1) = max{2, x1 - 3}
            +_A(x1,x2) = max{1, x1 - 19, x2 - 8}
            +#!_A(x1,x2) = max{1, x1 - 84}
            s!_A(x1) = max{6, x1 - 74}
            +!_A(x1,x2) = max{4, x1 - 45, x2 - 83}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            +#_A(x1,x2) = 0
            s_A(x1) = max{12, x1 - 14}
            +_A(x1,x2) = max{13, x1 - 14, x2 - 14}
            +#!_A(x1,x2) = max{5, x1 - 14, x2 - 13}
            s!_A(x1) = 2
            +!_A(x1,x2) = max{4, x1 - 63, x2 - 14}
        
  
    partial status:
  
      pi(+#) = [1]
      pi(s) = [1]
      pi(+) = [1, 2]

The following rules are removed:

  p1, p2

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!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = 0
            cons_A(x1,x2) = max{2, x1 - 65, x2 - 26}
            app#!_A(x1,x2) = max{0, x1 - 5}
            cons!_A(x1,x2) = max{4, x1 - 108, x2 - 4}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = 0
            cons_A(x1,x2) = max{2, x1 - 46, x2 - 43}
            app#!_A(x1,x2) = max{2, x1 - 93, x2 - 1}
            cons!_A(x1,x2) = max{1, x1 - 92, x2 - 92}
        
  
    partial status:
  
      pi(app#) = [1]
      pi(cons) = [1, 2]

The following rules are removed:

  p1

Then no dependency pair remains.