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))
  sum(nil()) -> |0|(|#|())
  sum(cons(x,l)) -> +(x,sum(l))
  prod(nil()) -> |1|(|#|())
  prod(cons(x,l)) -> *(x,prod(l))


-- 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: sum#(nil()) -> |0|#(|#|())
p18: sum#(cons(x,l)) -> +#(x,sum(l))
p19: sum#(cons(x,l)) -> sum#(l)
p20: prod#(cons(x,l)) -> *#(x,prod(l))
p21: 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: sum(nil()) -> |0|(|#|())
r14: sum(cons(x,l)) -> +(x,sum(l))
r15: prod(nil()) -> |1|(|#|())
r16: prod(cons(x,l)) -> *(x,prod(l))

The estimated dependency graph contains the following SCCs:

  {p19}
  {p21}
  {p11, p14, p15, p16}
  {p2, p3, p4, p6, p7, p8, p9}


-- 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: |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: sum(nil()) -> |0|(|#|())
r14: sum(cons(x,l)) -> +(x,sum(l))
r15: prod(nil()) -> |1|(|#|())
r16: prod(cons(x,l)) -> *(x,prod(l))

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) = x1
              cons_A(x1,x2) = x1 + x2 + 3
              sum#!_A(x1) = x1
              cons!_A(x1,x2) = 6
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1) = x1
            cons_A(x1,x2) = max{x1 - 1, x2 + 3}
            sum#!_A(x1) = max{1, x1 - 1}
            cons!_A(x1,x2) = max{2, x1 - 1, x2 - 15}
        
  
    partial status:
  
      pi(sum#) = []
      pi(cons) = [2]

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: |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: sum(nil()) -> |0|(|#|())
r14: sum(cons(x,l)) -> +(x,sum(l))
r15: prod(nil()) -> |1|(|#|())
r16: prod(cons(x,l)) -> *(x,prod(l))

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
              cons_A(x1,x2) = x1 + x2 + 1
              prod#!_A(x1) = x1 + 15
              cons!_A(x1,x2) = 14
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            prod#_A(x1) = x1
            cons_A(x1,x2) = max{4, x1 - 80, x2}
            prod#!_A(x1) = max{2, x1 - 5}
            cons!_A(x1,x2) = max{1, x1 - 61, x2 - 79}
        
  
    partial status:
  
      pi(prod#) = []
      pi(cons) = [2]

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: *#(|1|(x),y) -> *#(x,y)
p4: *#(|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: sum(nil()) -> |0|(|#|())
r14: sum(cons(x,l)) -> +(x,sum(l))
r15: prod(nil()) -> |1|(|#|())
r16: prod(cons(x,l)) -> *(x,prod(l))

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              *#_A(x1,x2) = x1
              *_A(x1,x2) = x1 + x2 + 2
              |1|_A(x1) = x1 + 3
              |0|_A(x1) = x1 + 8
              +_A(x1,x2) = x1 + x2 + 2
              |#|_A() = 5
              *#!_A(x1,x2) = x1 + 1
              *!_A(x1,x2) = x1 + 4
              |1|!_A(x1) = x1
              |0|!_A(x1) = 8
              +!_A(x1,x2) = x1 + 1
              |#|!_A() = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            *#_A(x1,x2) = 0
            *_A(x1,x2) = max{x1 + 34, x2 + 2}
            |1|_A(x1) = max{21, x1 - 23}
            |0|_A(x1) = max{4, x1 - 33}
            +_A(x1,x2) = max{24, x1 + 8, x2 + 9}
            |#|_A = 34
            *#!_A(x1,x2) = max{28, x1 - 5}
            *!_A(x1,x2) = max{38, x1 - 66, x2 - 99}
            |1|!_A(x1) = max{27, x1 + 1}
            |0|!_A(x1) = max{26, x1 - 134}
            +!_A(x1,x2) = max{39, x1 - 22, x2 - 134}
            |#|!_A = 28
        
  
    partial status:
  
      pi(*#) = []
      pi(*) = []
      pi(|1|) = [1]
      pi(|0|) = [1]
      pi(+) = [1, 2]
      pi(|#|) = []

The following rules are removed:

  p1, p2, p3, p4

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: sum(nil()) -> |0|(|#|())
r14: sum(cons(x,l)) -> +(x,sum(l))
r15: prod(nil()) -> |1|(|#|())
r16: prod(cons(x,l)) -> *(x,prod(l))

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              +#_A(x1,x2) = 0
              |0|_A(x1) = x1 + 1
              +_A(x1,x2) = x1 + x2 + 2
              |1|_A(x1) = x1 + 7
              |#|_A() = 4
              +#!_A(x1,x2) = x1 + x2
              |0|!_A(x1) = 16
              +!_A(x1,x2) = 18
              |1|!_A(x1) = 15
              |#|!_A() = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            +#_A(x1,x2) = 0
            |0|_A(x1) = 22
            +_A(x1,x2) = max{19, x1 + 9, x2}
            |1|_A(x1) = 20
            |#|_A = 0
            +#!_A(x1,x2) = max{x1 + 70, x2 + 31}
            |0|!_A(x1) = max{3, x1 - 21}
            +!_A(x1,x2) = max{6, x1 - 18, x2 - 38}
            |1|!_A(x1) = x1 + 30
            |#|!_A = 1
        
  
    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: sum(nil()) -> |0|(|#|())
r14: sum(cons(x,l)) -> +(x,sum(l))
r15: prod(nil()) -> |1|(|#|())
r16: 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) -> +#(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: sum(nil()) -> |0|(|#|())
r14: sum(cons(x,l)) -> +(x,sum(l))
r15: prod(nil()) -> |1|(|#|())
r16: prod(cons(x,l)) -> *(x,prod(l))

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) = x1
              +_A(x1,x2) = x1 + x2
              |0|_A(x1) = 4
              |#|_A() = 4
              +#!_A(x1,x2) = x1 + 7
              +!_A(x1,x2) = x2 + 6
              |0|!_A(x1) = 4
              |#|!_A() = 2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            +#_A(x1,x2) = max{0, x1 - 1}
            +_A(x1,x2) = x1
            |0|_A(x1) = x1
            |#|_A = 4
            +#!_A(x1,x2) = 9
            +!_A(x1,x2) = max{x1 + 8, x2 + 3}
            |0|!_A(x1) = x1 + 8
            |#|!_A = 1
        
  
    partial status:
  
      pi(+#) = [1]
      pi(+) = [1, 2]
      pi(|0|) = []
      pi(|#|) = []

The following rules are removed:

  p1

Then no dependency pair remains.