YES
We show the termination of the TRS R:

  |0|(|#|()) -> |#|()
  +(x,|#|()) -> x
  +(|#|(),x) -> x
  +(|0|(x),|0|(y)) -> |0|(+(x,y))
  +(|0|(x),|1|(y)) -> |1|(+(x,y))
  +(|1|(x),|0|(y)) -> |1|(+(x,y))
  +(|1|(x),|1|(y)) -> |0|(+(+(x,y),|1|(|#|())))
  +(+(x,y),z) -> +(x,+(y,z))
  *(|#|(),x) -> |#|()
  *(|0|(x),y) -> |0|(*(x,y))
  *(|1|(x),y) -> +(|0|(*(x,y)),y)
  *(*(x,y),z) -> *(x,*(y,z))
  *(x,+(y,z)) -> +(*(x,y),*(x,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()) -> |1|(|#|())
  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: +#(|0|(x),|0|(y)) -> |0|#(+(x,y))
p2: +#(|0|(x),|0|(y)) -> +#(x,y)
p3: +#(|0|(x),|1|(y)) -> +#(x,y)
p4: +#(|1|(x),|0|(y)) -> +#(x,y)
p5: +#(|1|(x),|1|(y)) -> |0|#(+(+(x,y),|1|(|#|())))
p6: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|()))
p7: +#(|1|(x),|1|(y)) -> +#(x,y)
p8: +#(+(x,y),z) -> +#(x,+(y,z))
p9: +#(+(x,y),z) -> +#(y,z)
p10: *#(|0|(x),y) -> |0|#(*(x,y))
p11: *#(|0|(x),y) -> *#(x,y)
p12: *#(|1|(x),y) -> +#(|0|(*(x,y)),y)
p13: *#(|1|(x),y) -> |0|#(*(x,y))
p14: *#(|1|(x),y) -> *#(x,y)
p15: *#(*(x,y),z) -> *#(x,*(y,z))
p16: *#(*(x,y),z) -> *#(y,z)
p17: *#(x,+(y,z)) -> +#(*(x,y),*(x,z))
p18: *#(x,+(y,z)) -> *#(x,y)
p19: *#(x,+(y,z)) -> *#(x,z)
p20: app#(cons(x,l1),l2) -> app#(l1,l2)
p21: sum#(nil()) -> |0|#(|#|())
p22: sum#(cons(x,l)) -> +#(x,sum(l))
p23: sum#(cons(x,l)) -> sum#(l)
p24: sum#(app(l1,l2)) -> +#(sum(l1),sum(l2))
p25: sum#(app(l1,l2)) -> sum#(l1)
p26: sum#(app(l1,l2)) -> sum#(l2)
p27: prod#(cons(x,l)) -> *#(x,prod(l))
p28: prod#(cons(x,l)) -> prod#(l)
p29: prod#(app(l1,l2)) -> *#(prod(l1),prod(l2))
p30: prod#(app(l1,l2)) -> prod#(l1)
p31: prod#(app(l1,l2)) -> prod#(l2)

and R consists of:

r1: |0|(|#|()) -> |#|()
r2: +(x,|#|()) -> x
r3: +(|#|(),x) -> x
r4: +(|0|(x),|0|(y)) -> |0|(+(x,y))
r5: +(|0|(x),|1|(y)) -> |1|(+(x,y))
r6: +(|1|(x),|0|(y)) -> |1|(+(x,y))
r7: +(|1|(x),|1|(y)) -> |0|(+(+(x,y),|1|(|#|())))
r8: +(+(x,y),z) -> +(x,+(y,z))
r9: *(|#|(),x) -> |#|()
r10: *(|0|(x),y) -> |0|(*(x,y))
r11: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r12: *(*(x,y),z) -> *(x,*(y,z))
r13: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r14: app(nil(),l) -> l
r15: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r16: sum(nil()) -> |0|(|#|())
r17: sum(cons(x,l)) -> +(x,sum(l))
r18: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r19: prod(nil()) -> |1|(|#|())
r20: prod(cons(x,l)) -> *(x,prod(l))
r21: prod(app(l1,l2)) -> *(prod(l1),prod(l2))

The estimated dependency graph contains the following SCCs:

  {p23, p25, p26}
  {p28, p30, p31}
  {p11, p14, p15, p16, p18, p19}
  {p2, p3, p4, p6, p7, p8, p9}
  {p20}


-- 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: |0|(|#|()) -> |#|()
r2: +(x,|#|()) -> x
r3: +(|#|(),x) -> x
r4: +(|0|(x),|0|(y)) -> |0|(+(x,y))
r5: +(|0|(x),|1|(y)) -> |1|(+(x,y))
r6: +(|1|(x),|0|(y)) -> |1|(+(x,y))
r7: +(|1|(x),|1|(y)) -> |0|(+(+(x,y),|1|(|#|())))
r8: +(+(x,y),z) -> +(x,+(y,z))
r9: *(|#|(),x) -> |#|()
r10: *(|0|(x),y) -> |0|(*(x,y))
r11: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r12: *(*(x,y),z) -> *(x,*(y,z))
r13: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r14: app(nil(),l) -> l
r15: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r16: sum(nil()) -> |0|(|#|())
r17: sum(cons(x,l)) -> +(x,sum(l))
r18: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r19: prod(nil()) -> |1|(|#|())
r20: prod(cons(x,l)) -> *(x,prod(l))
r21: 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 - 12, x2 - 10}
            cons_A(x1,x2) = max{2, x1 - 10, x2 - 10}
            sum#!_A(x1) = 0
            app!_A(x1,x2) = 1
            cons!_A(x1,x2) = max{26, x1 - 11, x2 - 4}
        
        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 + 2
              cons_A(x1,x2) = x1 + x2 + 1
              sum#!_A(x1) = x1
              app!_A(x1,x2) = x1 + 30
              cons!_A(x1,x2) = x2 + 31
        
  
    partial status:
  
      pi(sum#) = []
      pi(app) = [1]
      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: |0|(|#|()) -> |#|()
r2: +(x,|#|()) -> x
r3: +(|#|(),x) -> x
r4: +(|0|(x),|0|(y)) -> |0|(+(x,y))
r5: +(|0|(x),|1|(y)) -> |1|(+(x,y))
r6: +(|1|(x),|0|(y)) -> |1|(+(x,y))
r7: +(|1|(x),|1|(y)) -> |0|(+(+(x,y),|1|(|#|())))
r8: +(+(x,y),z) -> +(x,+(y,z))
r9: *(|#|(),x) -> |#|()
r10: *(|0|(x),y) -> |0|(*(x,y))
r11: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r12: *(*(x,y),z) -> *(x,*(y,z))
r13: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r14: app(nil(),l) -> l
r15: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r16: sum(nil()) -> |0|(|#|())
r17: sum(cons(x,l)) -> +(x,sum(l))
r18: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r19: prod(nil()) -> |1|(|#|())
r20: prod(cons(x,l)) -> *(x,prod(l))
r21: 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{0, x1 - 12, x2 - 2}
            cons_A(x1,x2) = max{x1 + 9, x2 - 2}
            prod#!_A(x1) = 6
            app!_A(x1,x2) = 7
            cons!_A(x1,x2) = max{7, x1 + 5}
        
        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 + 1
              cons_A(x1,x2) = x1 + x2 + 1
              prod#!_A(x1) = x1 + 7
              app!_A(x1,x2) = 5
              cons!_A(x1,x2) = x2 + 6
        
  
    partial status:
  
      pi(prod#) = []
      pi(app) = [1]
      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)) -> *#(x,z)
p2: *#(x,+(y,z)) -> *#(x,y)
p3: *#(*(x,y),z) -> *#(y,z)
p4: *#(*(x,y),z) -> *#(x,*(y,z))
p5: *#(|1|(x),y) -> *#(x,y)
p6: *#(|0|(x),y) -> *#(x,y)

and R consists of:

r1: |0|(|#|()) -> |#|()
r2: +(x,|#|()) -> x
r3: +(|#|(),x) -> x
r4: +(|0|(x),|0|(y)) -> |0|(+(x,y))
r5: +(|0|(x),|1|(y)) -> |1|(+(x,y))
r6: +(|1|(x),|0|(y)) -> |1|(+(x,y))
r7: +(|1|(x),|1|(y)) -> |0|(+(+(x,y),|1|(|#|())))
r8: +(+(x,y),z) -> +(x,+(y,z))
r9: *(|#|(),x) -> |#|()
r10: *(|0|(x),y) -> |0|(*(x,y))
r11: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r12: *(*(x,y),z) -> *(x,*(y,z))
r13: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r14: app(nil(),l) -> l
r15: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r16: sum(nil()) -> |0|(|#|())
r17: sum(cons(x,l)) -> +(x,sum(l))
r18: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r19: prod(nil()) -> |1|(|#|())
r20: prod(cons(x,l)) -> *(x,prod(l))
r21: prod(app(l1,l2)) -> *(prod(l1),prod(l2))

The set of usable rules consists of

  r2, r3, r9

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{323, x1 + 21, x2 + 10}
            *_A(x1,x2) = max{52, x1 + 9}
            |1|_A(x1) = x1 + 231
            |0|_A(x1) = max{169, x1 + 134}
            |#|_A = 62
            *#!_A(x1,x2) = 0
            +!_A(x1,x2) = max{311, x1 - 181, x2 + 121}
            *!_A(x1,x2) = max{6, x1 - 173, x2 + 1}
            |1|!_A(x1) = max{0, x1 - 11}
            |0|!_A(x1) = x1 + 7
            |#|!_A = 0
        
        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 + 9
              *_A(x1,x2) = x1 + x2 + 14
              |1|_A(x1) = x1 + 5
              |0|_A(x1) = x1 + 2
              |#|_A() = 1
              *#!_A(x1,x2) = x1
              +!_A(x1,x2) = x1 + x2 + 5
              *!_A(x1,x2) = x1 + x2 + 15
              |1|!_A(x1) = x1 + 4
              |0|!_A(x1) = 14
              |#|!_A() = 3
        
  
    partial status:
  
      pi(*#) = []
      pi(+) = [1, 2]
      pi(*) = [1, 2]
      pi(|1|) = [1]
      pi(|0|) = []
      pi(|#|) = []

The following rules are removed:

  p3, p4, p5, p6



-- SCC decomposition.

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

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

and R consists of:

r1: |0|(|#|()) -> |#|()
r2: +(x,|#|()) -> x
r3: +(|#|(),x) -> x
r4: +(|0|(x),|0|(y)) -> |0|(+(x,y))
r5: +(|0|(x),|1|(y)) -> |1|(+(x,y))
r6: +(|1|(x),|0|(y)) -> |1|(+(x,y))
r7: +(|1|(x),|1|(y)) -> |0|(+(+(x,y),|1|(|#|())))
r8: +(+(x,y),z) -> +(x,+(y,z))
r9: *(|#|(),x) -> |#|()
r10: *(|0|(x),y) -> |0|(*(x,y))
r11: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r12: *(*(x,y),z) -> *(x,*(y,z))
r13: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r14: app(nil(),l) -> l
r15: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r16: sum(nil()) -> |0|(|#|())
r17: sum(cons(x,l)) -> +(x,sum(l))
r18: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r19: prod(nil()) -> |1|(|#|())
r20: prod(cons(x,l)) -> *(x,prod(l))
r21: 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)) -> *#(x,z)
p2: *#(x,+(y,z)) -> *#(x,y)

and R consists of:

r1: |0|(|#|()) -> |#|()
r2: +(x,|#|()) -> x
r3: +(|#|(),x) -> x
r4: +(|0|(x),|0|(y)) -> |0|(+(x,y))
r5: +(|0|(x),|1|(y)) -> |1|(+(x,y))
r6: +(|1|(x),|0|(y)) -> |1|(+(x,y))
r7: +(|1|(x),|1|(y)) -> |0|(+(+(x,y),|1|(|#|())))
r8: +(+(x,y),z) -> +(x,+(y,z))
r9: *(|#|(),x) -> |#|()
r10: *(|0|(x),y) -> |0|(*(x,y))
r11: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r12: *(*(x,y),z) -> *(x,*(y,z))
r13: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r14: app(nil(),l) -> l
r15: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r16: sum(nil()) -> |0|(|#|())
r17: sum(cons(x,l)) -> +(x,sum(l))
r18: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r19: prod(nil()) -> |1|(|#|())
r20: prod(cons(x,l)) -> *(x,prod(l))
r21: 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{x1 - 37, x2 + 23}
            *#!_A(x1,x2) = max{73, x1 - 1}
            +!_A(x1,x2) = max{x1 + 11, x2 + 72}
        
        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) = x2
              +_A(x1,x2) = x1 + x2 + 2
              *#!_A(x1,x2) = x2
              +!_A(x1,x2) = 6
        
  
    partial status:
  
      pi(*#) = [1]
      pi(+) = []

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: +#(|0|(x),|0|(y)) -> +#(x,y)
p2: +#(+(x,y),z) -> +#(y,z)
p3: +#(+(x,y),z) -> +#(x,+(y,z))
p4: +#(|1|(x),|1|(y)) -> +#(x,y)
p5: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|()))
p6: +#(|0|(x),|1|(y)) -> +#(x,y)
p7: +#(|1|(x),|0|(y)) -> +#(x,y)

and R consists of:

r1: |0|(|#|()) -> |#|()
r2: +(x,|#|()) -> x
r3: +(|#|(),x) -> x
r4: +(|0|(x),|0|(y)) -> |0|(+(x,y))
r5: +(|0|(x),|1|(y)) -> |1|(+(x,y))
r6: +(|1|(x),|0|(y)) -> |1|(+(x,y))
r7: +(|1|(x),|1|(y)) -> |0|(+(+(x,y),|1|(|#|())))
r8: +(+(x,y),z) -> +(x,+(y,z))
r9: *(|#|(),x) -> |#|()
r10: *(|0|(x),y) -> |0|(*(x,y))
r11: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r12: *(*(x,y),z) -> *(x,*(y,z))
r13: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r14: app(nil(),l) -> l
r15: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r16: sum(nil()) -> |0|(|#|())
r17: sum(cons(x,l)) -> +(x,sum(l))
r18: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r19: prod(nil()) -> |1|(|#|())
r20: prod(cons(x,l)) -> *(x,prod(l))
r21: 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) = 0
            |0|_A(x1) = 25
            +_A(x1,x2) = max{x1 + 13, x2}
            |1|_A(x1) = 20
            |#|_A = 0
            +#!_A(x1,x2) = 0
            |0|!_A(x1) = 4
            +!_A(x1,x2) = 4
            |1|!_A(x1) = 1
            |#|!_A = 3
        
        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
              |0|_A(x1) = x1 + 9
              +_A(x1,x2) = x1 + x2 + 4
              |1|_A(x1) = x1 + 25
              |#|_A() = 12
              +#!_A(x1,x2) = x1 + x2
              |0|!_A(x1) = 59
              +!_A(x1,x2) = x1 + 51
              |1|!_A(x1) = 56
              |#|!_A() = 58
        
  
    partial status:
  
      pi(+#) = []
      pi(|0|) = []
      pi(+) = [1, 2]
      pi(|1|) = []
      pi(|#|) = []

The following rules are removed:

  p1, p2, p4, p5, p6, p7



-- 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: |0|(|#|()) -> |#|()
r2: +(x,|#|()) -> x
r3: +(|#|(),x) -> x
r4: +(|0|(x),|0|(y)) -> |0|(+(x,y))
r5: +(|0|(x),|1|(y)) -> |1|(+(x,y))
r6: +(|1|(x),|0|(y)) -> |1|(+(x,y))
r7: +(|1|(x),|1|(y)) -> |0|(+(+(x,y),|1|(|#|())))
r8: +(+(x,y),z) -> +(x,+(y,z))
r9: *(|#|(),x) -> |#|()
r10: *(|0|(x),y) -> |0|(*(x,y))
r11: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r12: *(*(x,y),z) -> *(x,*(y,z))
r13: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r14: app(nil(),l) -> l
r15: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r16: sum(nil()) -> |0|(|#|())
r17: sum(cons(x,l)) -> +(x,sum(l))
r18: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r19: prod(nil()) -> |1|(|#|())
r20: prod(cons(x,l)) -> *(x,prod(l))
r21: 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: |0|(|#|()) -> |#|()
r2: +(x,|#|()) -> x
r3: +(|#|(),x) -> x
r4: +(|0|(x),|0|(y)) -> |0|(+(x,y))
r5: +(|0|(x),|1|(y)) -> |1|(+(x,y))
r6: +(|1|(x),|0|(y)) -> |1|(+(x,y))
r7: +(|1|(x),|1|(y)) -> |0|(+(+(x,y),|1|(|#|())))
r8: +(+(x,y),z) -> +(x,+(y,z))
r9: *(|#|(),x) -> |#|()
r10: *(|0|(x),y) -> |0|(*(x,y))
r11: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r12: *(*(x,y),z) -> *(x,*(y,z))
r13: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r14: app(nil(),l) -> l
r15: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r16: sum(nil()) -> |0|(|#|())
r17: sum(cons(x,l)) -> +(x,sum(l))
r18: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r19: prod(nil()) -> |1|(|#|())
r20: prod(cons(x,l)) -> *(x,prod(l))
r21: 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 - 10}
            +_A(x1,x2) = max{11, x1, x2}
            |0|_A(x1) = max{12, x1 - 18}
            |#|_A = 5
            |1|_A(x1) = max{17, x1 - 18}
            +#!_A(x1,x2) = max{x1 + 83, x2 + 70}
            +!_A(x1,x2) = max{x1 + 80, x2 + 44}
            |0|!_A(x1) = x1 + 8
            |#|!_A = 11
            |1|!_A(x1) = 9
        
        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 + 5
              |0|_A(x1) = 7
              |#|_A() = 7
              |1|_A(x1) = 3
              +#!_A(x1,x2) = x1 + x2 + 3
              +!_A(x1,x2) = 7
              |0|!_A(x1) = 6
              |#|!_A() = 0
              |1|!_A(x1) = 2
        
  
    partial status:
  
      pi(+#) = [1, 2]
      pi(+) = [1, 2]
      pi(|0|) = []
      pi(|#|) = []
      pi(|1|) = []

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: |0|(|#|()) -> |#|()
r2: +(x,|#|()) -> x
r3: +(|#|(),x) -> x
r4: +(|0|(x),|0|(y)) -> |0|(+(x,y))
r5: +(|0|(x),|1|(y)) -> |1|(+(x,y))
r6: +(|1|(x),|0|(y)) -> |1|(+(x,y))
r7: +(|1|(x),|1|(y)) -> |0|(+(+(x,y),|1|(|#|())))
r8: +(+(x,y),z) -> +(x,+(y,z))
r9: *(|#|(),x) -> |#|()
r10: *(|0|(x),y) -> |0|(*(x,y))
r11: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r12: *(*(x,y),z) -> *(x,*(y,z))
r13: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r14: app(nil(),l) -> l
r15: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r16: sum(nil()) -> |0|(|#|())
r17: sum(cons(x,l)) -> +(x,sum(l))
r18: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r19: prod(nil()) -> |1|(|#|())
r20: prod(cons(x,l)) -> *(x,prod(l))
r21: 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{x1 + 227, x2 + 6}
            app#!_A(x1,x2) = max{0, x1 - 228}
            cons!_A(x1,x2) = max{3, x1 - 1, x2 - 6}
        
        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 + 5
              app#!_A(x1,x2) = x1
              cons!_A(x1,x2) = 14
        
  
    partial status:
  
      pi(app#) = [1]
      pi(cons) = [1, 2]

The following rules are removed:

  p1

Then no dependency pair remains.