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

  r14

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        sum#_A(x1) = 2
        app_A(x1,x2) = max{x1 - 15, x2 + 3}
        cons_A(x1,x2) = max{6, x1 + 1, x2 - 15}
        prod_A(x1) = max{15, x1 - 4}
        nil_A = 15
        s_A(x1) = max{14, x1 + 4}
        |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:
            sum#_A(x1) = x1
            app_A(x1,x2) = x1 + x2 + 7
            cons_A(x1,x2) = x1 + x2 + 1
            prod_A(x1) = x1 + 6
            nil_A() = 1
            s_A(x1) = 0
            |0|_A() = 3
            sum#!_A(x1) = x1 + 14
            app!_A(x1,x2) = 13
            cons!_A(x1,x2) = x2 + 3
            prod!_A(x1) = x1 + 4
            nil!_A() = 2
            s!_A(x1) = 1
            |0|!_A() = 0
  
    partial status:
  
      pi(sum#) = []
      pi(app) = [2]
      pi(cons) = [1]
      pi(prod) = []
      pi(nil) = []
      pi(s) = []
      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: 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:
  
      max/plus interpretation on natural numbers:
      
        prod#_A(x1) = max{1, x1 - 1}
        app_A(x1,x2) = max{46, x1 + 3, x2 + 40}
        cons_A(x1,x2) = max{51, x1 - 5, x2 + 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:
            prod#_A(x1) = x1
            app_A(x1,x2) = x1 + x2 + 2
            cons_A(x1,x2) = x1 + x2 + 2
            prod#!_A(x1) = 1
            app!_A(x1,x2) = 2
            cons!_A(x1,x2) = x2 + 1
  
    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, r5, r6

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        *#_A(x1,x2) = x1 + 4
        *_A(x1,x2) = max{21, x1 + 11, x2 + 16}
        s_A(x1) = max{16, x1 + 11}
        +_A(x1,x2) = max{9, x1, x2 + 8}
        |0|_A = 7
  
    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) = x1
            *_A(x1,x2) = x1 + x2 + 4
            s_A(x1) = x1 + 1
            +_A(x1,x2) = x1 + x2 + 2
            |0|_A() = 7
            *#!_A(x1,x2) = x1
            *!_A(x1,x2) = x1 + x2 + 5
            s!_A(x1) = 6
            +!_A(x1,x2) = x2 + 8
            |0|!_A() = 0
  
    partial status:
  
      pi(*#) = [1]
      pi(*) = [2]
      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: 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

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        +#_A(x1,x2) = max{6, x1 - 8, x2 - 11}
        s_A(x1) = max{5, x1}
        +_A(x1,x2) = max{7, x1 + 3, x2}
        |0|_A = 2
        *_A(x1,x2) = max{3, x1 - 13, x2 - 30}
  
    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) = x1 + x2
            s_A(x1) = x1 + 1
            +_A(x1,x2) = x1 + x2 + 5
            |0|_A() = 6
            *_A(x1,x2) = x2 + 5
            +#!_A(x1,x2) = x1 + x2 + 2
            s!_A(x1) = 1
            +!_A(x1,x2) = x1 + x2 + 8
            |0|!_A() = 0
            *!_A(x1,x2) = x1 + x2
  
    partial status:
  
      pi(+#) = []
      pi(s) = []
      pi(+) = [1, 2]
      pi(|0|) = []
      pi(*) = []

The following rules are removed:

  p1, p2



-- SCC decomposition.

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

p1: +#(+(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 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) -> +#(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:
  
      max/plus interpretation on natural numbers:
      
        +#_A(x1,x2) = max{1, x1 - 1, x2 - 2}
        +_A(x1,x2) = max{x1, x2}
        |0|_A = 4
        s_A(x1) = 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) = x1 + x2
            +_A(x1,x2) = x1 + x2 + 2
            |0|_A() = 11
            s_A(x1) = 11
            +#!_A(x1,x2) = x1 + 6
            +!_A(x1,x2) = x1 + 5
            |0|!_A() = 11
            s!_A(x1) = 9
  
    partial status:
  
      pi(+#) = []
      pi(+) = [1, 2]
      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: 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:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = x1 + 20
        cons_A(x1,x2) = max{x1 - 21, x2 + 21}
  
    base order 2:
  
      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) = x1 + x2 + 4
            app#!_A(x1,x2) = 8
            cons!_A(x1,x2) = x2 + 7
  
    partial status:
  
      pi(app#) = [1]
      pi(cons) = [2]

The following rules are removed:

  p1

Then no dependency pair remains.