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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          sum#_A(x1) = x1 + 2
          app_A(x1,x2) = max{x1 + 1, x2 + 2}
          cons_A(x1,x2) = max{x1 + 2, x2 + 2}
    
      precedence:
      
        sum# = app = cons
      
      partial status:
    
        pi(sum#) = []
        pi(app) = [1, 2]
        pi(cons) = [2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          sum#_A(x1) = 4
          app_A(x1,x2) = max{x1 + 1, x2 + 3}
          cons_A(x1,x2) = max{3, x2 + 1}
    
      precedence:
      
        sum# > app = cons
      
      partial status:
    
        pi(sum#) = []
        pi(app) = [2]
        pi(cons) = [2]
    

The next rules are strictly ordered:

  p1, p2, p3

We remove them from the problem.  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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          prod#_A(x1) = x1 + 2
          app_A(x1,x2) = max{x1 + 1, x2 + 2}
          cons_A(x1,x2) = max{x1 + 2, x2 + 2}
    
      precedence:
      
        prod# = app = cons
      
      partial status:
    
        pi(prod#) = []
        pi(app) = [1, 2]
        pi(cons) = [2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          prod#_A(x1) = 4
          app_A(x1,x2) = max{x1 + 1, x2 + 3}
          cons_A(x1,x2) = max{3, x2 + 1}
    
      precedence:
      
        prod# > app = cons
      
      partial status:
    
        pi(prod#) = []
        pi(app) = [2]
        pi(cons) = [2]
    

The next rules are strictly ordered:

  p1, p2, p3

We remove them from the problem.  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

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          *#_A(x1,x2) = x1 + 5
          +_A(x1,x2) = max{8, x1, x2}
          *_A(x1,x2) = max{x1 + 6, x2}
          |1|_A(x1) = max{2, x1}
          |0|_A(x1) = max{3, x1}
          |#|_A = 1
    
      precedence:
      
        *# > * > + = |1| = |0| = |#|
      
      partial status:
    
        pi(*#) = []
        pi(+) = [1, 2]
        pi(*) = [1, 2]
        pi(|1|) = []
        pi(|0|) = []
        pi(|#|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          *#_A(x1,x2) = 0
          +_A(x1,x2) = max{9, x2}
          *_A(x1,x2) = x1 + 9
          |1|_A(x1) = 9
          |0|_A(x1) = 8
          |#|_A = 13
    
      precedence:
      
        + = * > *# = |#| > |1| = |0|
      
      partial status:
    
        pi(*#) = []
        pi(+) = []
        pi(*) = []
        pi(|1|) = []
        pi(|0|) = []
        pi(|#|) = []
    

The next rules are strictly ordered:

  p4

We remove them from the problem.

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

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          *#_A(x1,x2) = max{3, x1 + 1, x2 + 1}
          +_A(x1,x2) = max{x1, x2}
          |0|_A(x1) = max{1, x1}
          |1|_A(x1) = max{2, x1 + 1}
          *_A(x1,x2) = max{2, x2}
    
      precedence:
      
        *# = + = |0| = |1| = *
      
      partial status:
    
        pi(*#) = [1, 2]
        pi(+) = [1, 2]
        pi(|0|) = [1]
        pi(|1|) = [1]
        pi(*) = [2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          *#_A(x1,x2) = max{x1 + 1, x2 + 1}
          +_A(x1,x2) = max{x1, x2}
          |0|_A(x1) = x1
          |1|_A(x1) = x1
          *_A(x1,x2) = x2
    
      precedence:
      
        *# = + = |0| = |1| = *
      
      partial status:
    
        pi(*#) = [1, 2]
        pi(+) = [1, 2]
        pi(|0|) = [1]
        pi(|1|) = [1]
        pi(*) = [2]
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5

We remove them from the problem.  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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            +#_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,1),(1,1)) x2 + (6,18)
            |0|_A(x1) = ((1,0),(1,1)) x1 + (1,2)
            +_A(x1,x2) = x1 + x2 + (2,3)
            |1|_A(x1) = ((1,0),(1,1)) x1 + (5,17)
            |#|_A() = (1,1)
    
      precedence:
      
        +# = + > |0| > |1| = |#|
      
      partial status:
    
        pi(+#) = [1]
        pi(|0|) = [1]
        pi(+) = [1, 2]
        pi(|1|) = [1]
        pi(|#|) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            +#_A(x1,x2) = x1
            |0|_A(x1) = x1 + (2,1)
            +_A(x1,x2) = x1
            |1|_A(x1) = x1 + (2,1)
            |#|_A() = (1,1)
    
      precedence:
      
        + = |1| = |#| > |0| > +#
      
      partial status:
    
        pi(+#) = [1]
        pi(|0|) = [1]
        pi(+) = [1]
        pi(|1|) = [1]
        pi(|#|) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p4

We remove them from the problem.

-- SCC decomposition.

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

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

  {p1, p2, p3}


-- Reduction pair.

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

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

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            +#_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,1),(0,1)) x2 + (3,0)
            |1|_A(x1) = ((1,0),(1,1)) x1 + (2,3)
            +_A(x1,x2) = x1 + x2 + (0,1)
            |#|_A() = (0,0)
            |0|_A(x1) = ((1,0),(1,1)) x1 + (1,0)
    
      precedence:
      
        + > +# = |1| = |#| = |0|
      
      partial status:
    
        pi(+#) = [2]
        pi(|1|) = []
        pi(+) = [1, 2]
        pi(|#|) = []
        pi(|0|) = [1]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            +#_A(x1,x2) = (1,1)
            |1|_A(x1) = (5,3)
            +_A(x1,x2) = ((1,1),(0,0)) x2 + (4,4)
            |#|_A() = (2,2)
            |0|_A(x1) = (3,2)
    
      precedence:
      
        +# = |1| = + = |0| > |#|
      
      partial status:
    
        pi(+#) = []
        pi(|1|) = []
        pi(+) = []
        pi(|#|) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

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

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = max{2, x1 + 1, x2 + 1}
          |0|_A(x1) = max{1, x1}
          |1|_A(x1) = max{1, x1}
    
      precedence:
      
        +# = |0| = |1|
      
      partial status:
    
        pi(+#) = [1, 2]
        pi(|0|) = [1]
        pi(|1|) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = max{x1 + 1, x2 - 1}
          |0|_A(x1) = x1
          |1|_A(x1) = x1
    
      precedence:
      
        +# = |0| = |1|
      
      partial status:
    
        pi(+#) = [1]
        pi(|0|) = [1]
        pi(|1|) = [1]
    

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.  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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = x1 + 1
          cons_A(x1,x2) = max{x1, x2 + 1}
    
      precedence:
      
        app# = cons
      
      partial status:
    
        pi(app#) = [1]
        pi(cons) = [1, 2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = max{0, x1 - 1}
          cons_A(x1,x2) = max{x1, x2}
    
      precedence:
      
        app# = cons
      
      partial status:
    
        pi(app#) = []
        pi(cons) = [1, 2]
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.