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:

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

The next rules are strictly ordered:

  p1

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#(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:

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

The next rules are strictly ordered:

  p1

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) -> *#(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

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

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 + 4
          *_A(x1,x2) = max{x1 + 5, x2}
          |1|_A(x1) = max{3, x1}
          |0|_A(x1) = max{2, x1}
          |#|_A = 0
          +_A(x1,x2) = max{1, x1, x2}
    
      precedence:
      
        *# > * = |#| > |0| = + > |1|
      
      partial status:
    
        pi(*#) = []
        pi(*) = [1, 2]
        pi(|1|) = []
        pi(|0|) = []
        pi(|#|) = []
        pi(+) = [1, 2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          *#_A(x1,x2) = 0
          *_A(x1,x2) = max{15, x1 + 11, x2}
          |1|_A(x1) = 16
          |0|_A(x1) = 16
          |#|_A = 0
          +_A(x1,x2) = max{17, x1 + 9, x2}
    
      precedence:
      
        |1| > *# = * = |#| > + > |0|
      
      partial status:
    
        pi(*#) = []
        pi(*) = [1, 2]
        pi(|1|) = []
        pi(|0|) = []
        pi(|#|) = []
        pi(+) = [2]
    

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

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

  (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}
          *_A(x1,x2) = max{x1, x2}
          |0|_A(x1) = max{1, x1}
          |1|_A(x1) = max{1, x1}
    
      precedence:
      
        *# = * = |0| = |1|
      
      partial status:
    
        pi(*#) = [1]
        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{0, x1 - 2}
          *_A(x1,x2) = max{x1 + 3, x2 + 1}
          |0|_A(x1) = max{3, x1 + 1}
          |1|_A(x1) = max{3, x1 + 1}
    
      precedence:
      
        *# = * = |0| = |1|
      
      partial status:
    
        pi(*#) = []
        pi(*) = [2]
        pi(|0|) = [1]
        pi(|1|) = [1]
    

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: +#(|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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            +#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (6,4)
            |0|_A(x1) = ((1,0),(1,1)) x1 + (1,3)
            +_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (2,10)
            |1|_A(x1) = ((1,0),(0,0)) x1 + (5,2)
            |#|_A() = (1,1)
    
      precedence:
      
        +# = |0| = + = |1| = |#|
      
      partial status:
    
        pi(+#) = []
        pi(|0|) = []
        pi(+) = []
        pi(|1|) = []
        pi(|#|) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            +#_A(x1,x2) = (2,5)
            |0|_A(x1) = ((1,0),(1,0)) x1 + (0,4)
            +_A(x1,x2) = ((1,0),(1,1)) x1
            |1|_A(x1) = (1,2)
            |#|_A() = (0,1)
    
      precedence:
      
        +# > |1| = |#| > |0| = +
      
      partial status:
    
        pi(+#) = []
        pi(|0|) = []
        pi(+) = [1]
        pi(|1|) = []
        pi(|#|) = []
    

The next rules are strictly ordered:

  p2, p6

We remove them from the problem.

-- SCC decomposition.

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

p1: +#(|0|(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: +#(|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}


-- Reduction pair.

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

p1: +#(|0|(x),|0|(y)) -> +#(x,y)
p2: +#(|1|(x),|0|(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 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: lexicographic order
          interpretations:
            +#_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (6,12)
            |0|_A(x1) = ((1,0),(1,1)) x1 + (1,2)
            |1|_A(x1) = x1 + (5,11)
            +_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (2,1)
            |#|_A() = (1,0)
    
      precedence:
      
        +# = |0| = |1| = + = |#|
      
      partial status:
    
        pi(+#) = []
        pi(|0|) = []
        pi(|1|) = []
        pi(+) = []
        pi(|#|) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            +#_A(x1,x2) = x1 + (2,2)
            |0|_A(x1) = ((1,0),(0,0)) x1 + (12,6)
            |1|_A(x1) = x1 + (7,4)
            +_A(x1,x2) = ((1,0),(0,0)) x1 + (4,1)
            |#|_A() = (1,0)
    
      precedence:
      
        |0| = + > |1| = |#| > +#
      
      partial status:
    
        pi(+#) = [1]
        pi(|0|) = []
        pi(|1|) = []
        pi(+) = []
        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: +#(+(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

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{x1 + 24, x2 + 12}
          +_A(x1,x2) = max{11, x1 + 6, x2}
          |0|_A(x1) = 5
          |#|_A = 0
          |1|_A(x1) = 2
    
      precedence:
      
        +# = + > |0| = |#| = |1|
      
      partial status:
    
        pi(+#) = [1, 2]
        pi(+) = [1, 2]
        pi(|0|) = []
        pi(|#|) = []
        pi(|1|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = max{x1 + 10, x2 + 14}
          +_A(x1,x2) = max{13, x1 + 4, x2 + 5}
          |0|_A(x1) = 21
          |#|_A = 14
          |1|_A(x1) = 11
    
      precedence:
      
        +# = + = |0| = |#| = |1|
      
      partial status:
    
        pi(+#) = [1]
        pi(+) = [1, 2]
        pi(|0|) = []
        pi(|#|) = []
        pi(|1|) = []
    

The next rules are strictly ordered:

  p1

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