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

  r1, r9, r10

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          sum#_A(x1) = x1
          cons_A(x1,x2) = x2 + 7
          |0|_A(x1) = 4
          |#|_A() = 3
          *_A(x1,x2) = x1 + 8
  
    base order 2:
  
      precedence:
      
        |#| > * > cons > sum# = |0|
      
  
    partial status:
  
      pi(sum#) = []
      pi(cons) = [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: 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

  r1, r9, r10

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          prod#_A(x1) = x1
          cons_A(x1,x2) = x2 + 7
          |0|_A(x1) = 4
          |#|_A() = 3
          *_A(x1,x2) = x1 + 8
  
    base order 2:
  
      precedence:
      
        |#| > * > cons > prod# = |0|
      
  
    partial status:
  
      pi(prod#) = []
      pi(cons) = [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: *#(*(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, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          *#_A(x1,x2) = x1 + 1
          *_A(x1,x2) = x1 + x2 + 8
          |1|_A(x1) = x1 + 7
          |0|_A(x1) = x1 + 2
          +_A(x1,x2) = x1 + x2 + 3
          |#|_A() = 1
  
    base order 2:
  
      precedence:
      
        + > * > |#| > |1| = |0| > *#
      
  
    partial status:
  
      pi(*#) = []
      pi(*) = [1, 2]
      pi(|1|) = [1]
      pi(|0|) = []
      pi(+) = []
      pi(|#|) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: *#(*(x,y),z) -> *#(x,*(y,z))
p2: *#(|1|(x),y) -> *#(x,y)
p3: *#(|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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: *#(*(x,y),z) -> *#(x,*(y,z))
p2: *#(|0|(x),y) -> *#(x,y)
p3: *#(|1|(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, r9

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          *#_A(x1,x2) = x1 + 1
          *_A(x1,x2) = x1 + 4
          |0|_A(x1) = x1 + 2
          |1|_A(x1) = x1 + 13
          +_A(x1,x2) = x1 + 5
          |#|_A() = 1
  
    base order 2:
  
      precedence:
      
        *# = * > + > |0| = |1| = |#|
      
  
    partial status:
  
      pi(*#) = [1]
      pi(*) = []
      pi(|0|) = [1]
      pi(|1|) = [1]
      pi(+) = []
      pi(|#|) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: *#(|0|(x),y) -> *#(x,y)
p2: *#(|1|(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 estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: *#(|0|(x),y) -> *#(x,y)
p2: *#(|1|(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

  r9

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          *#_A(x1,x2) = x1
          |0|_A(x1) = x1 + 2
          |1|_A(x1) = x1 + 4
          *_A(x1,x2) = 3
          |#|_A() = 1
  
    base order 2:
  
      precedence:
      
        *# > |0| = |1| = * > |#|
      
  
    partial status:
  
      pi(*#) = [1]
      pi(|0|) = [1]
      pi(|1|) = [1]
      pi(*) = []
      pi(|#|) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: *#(|1|(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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: *#(|1|(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

  r1, r4, r7, r9, r10

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          *#_A(x1,x2) = x1
          |1|_A(x1) = x1 + 5
          |0|_A(x1) = 2
          |#|_A() = 1
          +_A(x1,x2) = 4
          *_A(x1,x2) = 3
  
    base order 2:
  
      precedence:
      
        *# > |1| > * > + > |0| > |#|
      
  
    partial status:
  
      pi(*#) = [1]
      pi(|1|) = [1]
      pi(|0|) = []
      pi(|#|) = []
      pi(+) = []
      pi(*) = []

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: +#(|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, r9

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          +#_A(x1,x2) = x1 + x2 + 6
          |0|_A(x1) = x1 + 1
          +_A(x1,x2) = x1 + x2 + 2
          |1|_A(x1) = x1 + 5
          |#|_A() = 1
          *_A(x1,x2) = x1 + x2 + 6
  
    base order 2:
  
      precedence:
      
        +# > + > * > |0| > |1| > |#|
      
  
    partial status:
  
      pi(+#) = []
      pi(|0|) = []
      pi(+) = []
      pi(|1|) = [1]
      pi(|#|) = []
      pi(*) = []

The following rules are removed:

  p1



-- SCC decomposition.

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),|1|(y)) -> +#(x,y)
p4: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|()))
p5: +#(|0|(x),|1|(y)) -> +#(x,y)
p6: +#(|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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6}


-- Reduction pair.

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

p1: +#(+(x,y),z) -> +#(y,z)
p2: +#(|1|(x),|0|(y)) -> +#(x,y)
p3: +#(|0|(x),|1|(y)) -> +#(x,y)
p4: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|()))
p5: +#(|1|(x),|1|(y)) -> +#(x,y)
p6: +#(+(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, r3, r4, r5, r6, r7, r8, r9

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          +#_A(x1,x2) = x1 + x2 + 9
          +_A(x1,x2) = x1 + x2 + 1
          |1|_A(x1) = x1 + 8
          |0|_A(x1) = x1 + 5
          |#|_A() = 1
          *_A(x1,x2) = 4
  
    base order 2:
  
      precedence:
      
        +# > + > |1| = |0| > * > |#|
      
  
    partial status:
  
      pi(+#) = []
      pi(+) = []
      pi(|1|) = [1]
      pi(|0|) = [1]
      pi(|#|) = []
      pi(*) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: +#(|1|(x),|0|(y)) -> +#(x,y)
p2: +#(|0|(x),|1|(y)) -> +#(x,y)
p3: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|()))
p4: +#(|1|(x),|1|(y)) -> +#(x,y)
p5: +#(+(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, p2, p3, p4, p5}


-- Reduction pair.

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

p1: +#(|1|(x),|0|(y)) -> +#(x,y)
p2: +#(+(x,y),z) -> +#(x,+(y,z))
p3: +#(|1|(x),|1|(y)) -> +#(x,y)
p4: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|()))
p5: +#(|0|(x),|1|(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, r9

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          +#_A(x1,x2) = x1 + x2 + 11
          |1|_A(x1) = x1 + 10
          |0|_A(x1) = x1 + 6
          +_A(x1,x2) = x1 + x2 + 2
          |#|_A() = 1
          *_A(x1,x2) = 5
  
    base order 2:
  
      precedence:
      
        +# > |0| = + > * > |1| > |#|
      
  
    partial status:
  
      pi(+#) = []
      pi(|1|) = [1]
      pi(|0|) = [1]
      pi(+) = [1, 2]
      pi(|#|) = []
      pi(*) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: +#(+(x,y),z) -> +#(x,+(y,z))
p2: +#(|1|(x),|1|(y)) -> +#(x,y)
p3: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|()))
p4: +#(|0|(x),|1|(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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

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

p1: +#(+(x,y),z) -> +#(x,+(y,z))
p2: +#(|0|(x),|1|(y)) -> +#(x,y)
p3: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|()))
p4: +#(|1|(x),|1|(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, r9

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          +#_A(x1,x2) = x1 + x2 + 6
          +_A(x1,x2) = x1 + x2 + 2
          |0|_A(x1) = x1 + 1
          |1|_A(x1) = x1 + 5
          |#|_A() = 1
          *_A(x1,x2) = x1 + 3
  
    base order 2:
  
      precedence:
      
        +# > + > |1| > * > |0| > |#|
      
  
    partial status:
  
      pi(+#) = [1]
      pi(+) = []
      pi(|0|) = []
      pi(|1|) = [1]
      pi(|#|) = []
      pi(*) = [1]

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: +#(+(x,y),z) -> +#(x,+(y,z))
p2: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|()))
p3: +#(|1|(x),|1|(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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: +#(+(x,y),z) -> +#(x,+(y,z))
p2: +#(|1|(x),|1|(y)) -> +#(x,y)
p3: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|()))

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, r9

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          +#_A(x1,x2) = x1 + x2 + 1
          +_A(x1,x2) = x1 + x2 + 2
          |1|_A(x1) = x1 + 8
          |#|_A() = 1
          |0|_A(x1) = x1 + 4
          *_A(x1,x2) = x1 + 3
  
    base order 2:
  
      precedence:
      
        +# > + > |1| = |0| > * > |#|
      
  
    partial status:
  
      pi(+#) = []
      pi(+) = []
      pi(|1|) = [1]
      pi(|#|) = []
      pi(|0|) = []
      pi(*) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: +#(+(x,y),z) -> +#(x,+(y,z))
p2: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|()))

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, p2}


-- Reduction pair.

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

p1: +#(+(x,y),z) -> +#(x,+(y,z))
p2: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|()))

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, r9

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          +#_A(x1,x2) = x1 + x2 + 1
          +_A(x1,x2) = x1 + x2 + 2
          |1|_A(x1) = x1 + 5
          |#|_A() = 1
          |0|_A(x1) = x1 + 1
          *_A(x1,x2) = x1
  
    base order 2:
  
      precedence:
      
        * > |1| = |0| > +# = + > |#|
      
  
    partial status:
  
      pi(+#) = []
      pi(+) = []
      pi(|1|) = [1]
      pi(|#|) = []
      pi(|0|) = []
      pi(*) = [1]

The following rules are removed:

  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: |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, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          +#_A(x1,x2) = x1 + x2 + 1
          +_A(x1,x2) = x1 + x2 + 5
          |0|_A(x1) = 2
          |#|_A() = 1
          |1|_A(x1) = 4
          *_A(x1,x2) = 3
  
    base order 2:
  
      precedence:
      
        +# > + > |1| > * > |0| > |#|
      
  
    partial status:
  
      pi(+#) = [1]
      pi(+) = [1, 2]
      pi(|0|) = []
      pi(|#|) = []
      pi(|1|) = []
      pi(*) = []

The following rules are removed:

  p1

Then no dependency pair remains.