YES

We show the termination of the TRS R:

  minus(x,|0|()) -> x
  minus(s(x),s(y)) -> minus(x,y)
  quot(|0|(),s(y)) -> |0|()
  quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
  le(|0|(),y) -> true()
  le(s(x),|0|()) -> false()
  le(s(x),s(y)) -> le(x,y)
  app(nil(),y) -> y
  app(add(n,x),y) -> add(n,app(x,y))
  low(n,nil()) -> nil()
  low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
  if_low(true(),n,add(m,x)) -> add(m,low(n,x))
  if_low(false(),n,add(m,x)) -> low(n,x)
  high(n,nil()) -> nil()
  high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
  if_high(true(),n,add(m,x)) -> high(n,x)
  if_high(false(),n,add(m,x)) -> add(m,high(n,x))
  quicksort(nil()) -> nil()
  quicksort(add(n,x)) -> app(quicksort(low(n,x)),add(n,quicksort(high(n,x))))

-- SCC decomposition.

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

p1: minus#(s(x),s(y)) -> minus#(x,y)
p2: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))
p3: quot#(s(x),s(y)) -> minus#(x,y)
p4: le#(s(x),s(y)) -> le#(x,y)
p5: app#(add(n,x),y) -> app#(x,y)
p6: low#(n,add(m,x)) -> if_low#(le(m,n),n,add(m,x))
p7: low#(n,add(m,x)) -> le#(m,n)
p8: if_low#(true(),n,add(m,x)) -> low#(n,x)
p9: if_low#(false(),n,add(m,x)) -> low#(n,x)
p10: high#(n,add(m,x)) -> if_high#(le(m,n),n,add(m,x))
p11: high#(n,add(m,x)) -> le#(m,n)
p12: if_high#(true(),n,add(m,x)) -> high#(n,x)
p13: if_high#(false(),n,add(m,x)) -> high#(n,x)
p14: quicksort#(add(n,x)) -> app#(quicksort(low(n,x)),add(n,quicksort(high(n,x))))
p15: quicksort#(add(n,x)) -> quicksort#(low(n,x))
p16: quicksort#(add(n,x)) -> low#(n,x)
p17: quicksort#(add(n,x)) -> quicksort#(high(n,x))
p18: quicksort#(add(n,x)) -> high#(n,x)

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: le(|0|(),y) -> true()
r6: le(s(x),|0|()) -> false()
r7: le(s(x),s(y)) -> le(x,y)
r8: app(nil(),y) -> y
r9: app(add(n,x),y) -> add(n,app(x,y))
r10: low(n,nil()) -> nil()
r11: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r12: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r13: if_low(false(),n,add(m,x)) -> low(n,x)
r14: high(n,nil()) -> nil()
r15: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r16: if_high(true(),n,add(m,x)) -> high(n,x)
r17: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r18: quicksort(nil()) -> nil()
r19: quicksort(add(n,x)) -> app(quicksort(low(n,x)),add(n,quicksort(high(n,x))))

The estimated dependency graph contains the following SCCs:

  {p2}
  {p1}
  {p15, p17}
  {p10, p12, p13}
  {p6, p8, p9}
  {p4}
  {p5}


-- Reduction pair.

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

p1: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: le(|0|(),y) -> true()
r6: le(s(x),|0|()) -> false()
r7: le(s(x),s(y)) -> le(x,y)
r8: app(nil(),y) -> y
r9: app(add(n,x),y) -> add(n,app(x,y))
r10: low(n,nil()) -> nil()
r11: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r12: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r13: if_low(false(),n,add(m,x)) -> low(n,x)
r14: high(n,nil()) -> nil()
r15: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r16: if_high(true(),n,add(m,x)) -> high(n,x)
r17: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r18: quicksort(nil()) -> nil()
r19: quicksort(add(n,x)) -> app(quicksort(low(n,x)),add(n,quicksort(high(n,x))))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

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

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: minus#(s(x),s(y)) -> minus#(x,y)

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: le(|0|(),y) -> true()
r6: le(s(x),|0|()) -> false()
r7: le(s(x),s(y)) -> le(x,y)
r8: app(nil(),y) -> y
r9: app(add(n,x),y) -> add(n,app(x,y))
r10: low(n,nil()) -> nil()
r11: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r12: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r13: if_low(false(),n,add(m,x)) -> low(n,x)
r14: high(n,nil()) -> nil()
r15: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r16: if_high(true(),n,add(m,x)) -> high(n,x)
r17: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r18: quicksort(nil()) -> nil()
r19: quicksort(add(n,x)) -> app(quicksort(low(n,x)),add(n,quicksort(high(n,x))))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

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: quicksort#(add(n,x)) -> quicksort#(high(n,x))
p2: quicksort#(add(n,x)) -> quicksort#(low(n,x))

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: le(|0|(),y) -> true()
r6: le(s(x),|0|()) -> false()
r7: le(s(x),s(y)) -> le(x,y)
r8: app(nil(),y) -> y
r9: app(add(n,x),y) -> add(n,app(x,y))
r10: low(n,nil()) -> nil()
r11: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r12: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r13: if_low(false(),n,add(m,x)) -> low(n,x)
r14: high(n,nil()) -> nil()
r15: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r16: if_high(true(),n,add(m,x)) -> high(n,x)
r17: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r18: quicksort(nil()) -> nil()
r19: quicksort(add(n,x)) -> app(quicksort(low(n,x)),add(n,quicksort(high(n,x))))

The set of usable rules consists of

  r5, r6, r7, r10, r11, r12, r13, r14, r15, r16, r17

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            quicksort#_A(x1) = ((1,0),(0,0)) x1
            add_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (3,5)
            high_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(1,0)) x2 + (0,3)
            low_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (2,0)
            le_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (0,9)
            |0|_A() = (7,1)
            true_A() = (6,4)
            s_A(x1) = x1 + (5,0)
            false_A() = (4,2)
            if_low_A(x1,x2,x3) = ((1,0),(1,0)) x2 + ((1,0),(1,0)) x3 + (2,3)
            if_high_A(x1,x2,x3) = ((0,0),(1,0)) x1 + x3
            nil_A() = (1,1)
    
      precedence:
      
        quicksort# = add = high = low = le = |0| = true = s = false = if_low = if_high = nil
      
      partial status:
    
        pi(quicksort#) = []
        pi(add) = []
        pi(high) = []
        pi(low) = []
        pi(le) = []
        pi(|0|) = []
        pi(true) = []
        pi(s) = []
        pi(false) = []
        pi(if_low) = []
        pi(if_high) = []
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            quicksort#_A(x1) = (0,0)
            add_A(x1,x2) = (2,3)
            high_A(x1,x2) = (2,1)
            low_A(x1,x2) = (3,4)
            le_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (5,5)
            |0|_A() = (0,1)
            true_A() = (4,0)
            s_A(x1) = x1
            false_A() = (4,2)
            if_low_A(x1,x2,x3) = (3,4)
            if_high_A(x1,x2,x3) = ((1,0),(1,1)) x3 + (0,1)
            nil_A() = (1,5)
    
      precedence:
      
        nil > add = high > if_high > le > low = |0| = true = false = if_low > s > quicksort#
      
      partial status:
    
        pi(quicksort#) = []
        pi(add) = []
        pi(high) = []
        pi(low) = []
        pi(le) = []
        pi(|0|) = []
        pi(true) = []
        pi(s) = [1]
        pi(false) = []
        pi(if_low) = []
        pi(if_high) = [3]
        pi(nil) = []
    

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: if_high#(false(),n,add(m,x)) -> high#(n,x)
p2: high#(n,add(m,x)) -> if_high#(le(m,n),n,add(m,x))
p3: if_high#(true(),n,add(m,x)) -> high#(n,x)

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: le(|0|(),y) -> true()
r6: le(s(x),|0|()) -> false()
r7: le(s(x),s(y)) -> le(x,y)
r8: app(nil(),y) -> y
r9: app(add(n,x),y) -> add(n,app(x,y))
r10: low(n,nil()) -> nil()
r11: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r12: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r13: if_low(false(),n,add(m,x)) -> low(n,x)
r14: high(n,nil()) -> nil()
r15: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r16: if_high(true(),n,add(m,x)) -> high(n,x)
r17: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r18: quicksort(nil()) -> nil()
r19: quicksort(add(n,x)) -> app(quicksort(low(n,x)),add(n,quicksort(high(n,x))))

The set of usable rules consists of

  r5, r6, r7

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            if_high#_A(x1,x2,x3) = ((1,0),(1,1)) x2 + ((1,0),(1,1)) x3 + (1,1)
            false_A() = (3,1)
            add_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (3,2)
            high#_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,0)) x2 + (2,4)
            le_A(x1,x2) = ((1,0),(1,0)) x1 + (2,1)
            true_A() = (3,5)
            |0|_A() = (5,6)
            s_A(x1) = x1 + (4,2)
    
      precedence:
      
        le > true = |0| > false = s > if_high# = add = high#
      
      partial status:
    
        pi(if_high#) = [2, 3]
        pi(false) = []
        pi(add) = []
        pi(high#) = []
        pi(le) = []
        pi(true) = []
        pi(|0|) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            if_high#_A(x1,x2,x3) = x2 + x3 + (1,1)
            false_A() = (1,3)
            add_A(x1,x2) = (0,0)
            high#_A(x1,x2) = (0,0)
            le_A(x1,x2) = (2,2)
            true_A() = (1,1)
            |0|_A() = (2,4)
            s_A(x1) = (3,4)
    
      precedence:
      
        s > le > |0| > if_high# = add = high# > false > true
      
      partial status:
    
        pi(if_high#) = [2, 3]
        pi(false) = []
        pi(add) = []
        pi(high#) = []
        pi(le) = []
        pi(true) = []
        pi(|0|) = []
        pi(s) = []
    

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: if_low#(false(),n,add(m,x)) -> low#(n,x)
p2: low#(n,add(m,x)) -> if_low#(le(m,n),n,add(m,x))
p3: if_low#(true(),n,add(m,x)) -> low#(n,x)

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: le(|0|(),y) -> true()
r6: le(s(x),|0|()) -> false()
r7: le(s(x),s(y)) -> le(x,y)
r8: app(nil(),y) -> y
r9: app(add(n,x),y) -> add(n,app(x,y))
r10: low(n,nil()) -> nil()
r11: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r12: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r13: if_low(false(),n,add(m,x)) -> low(n,x)
r14: high(n,nil()) -> nil()
r15: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r16: if_high(true(),n,add(m,x)) -> high(n,x)
r17: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r18: quicksort(nil()) -> nil()
r19: quicksort(add(n,x)) -> app(quicksort(low(n,x)),add(n,quicksort(high(n,x))))

The set of usable rules consists of

  r5, r6, r7

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            if_low#_A(x1,x2,x3) = ((1,0),(1,1)) x2 + x3 + (2,3)
            false_A() = (5,1)
            add_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (3,1)
            low#_A(x1,x2) = x1 + x2 + (4,2)
            le_A(x1,x2) = ((0,0),(1,0)) x1 + (6,4)
            true_A() = (1,3)
            |0|_A() = (6,4)
            s_A(x1) = x1 + (7,5)
    
      precedence:
      
        if_low# = false = add = low# = le = true = |0| = s
      
      partial status:
    
        pi(if_low#) = []
        pi(false) = []
        pi(add) = []
        pi(low#) = []
        pi(le) = []
        pi(true) = []
        pi(|0|) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            if_low#_A(x1,x2,x3) = x2 + x3 + (2,3)
            false_A() = (0,0)
            add_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (2,1)
            low#_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,1)) x2 + (3,0)
            le_A(x1,x2) = (0,0)
            true_A() = (3,1)
            |0|_A() = (4,2)
            s_A(x1) = (1,1)
    
      precedence:
      
        add = low# = |0| > if_low# = true = s > false = le
      
      partial status:
    
        pi(if_low#) = [2, 3]
        pi(false) = []
        pi(add) = []
        pi(low#) = [2]
        pi(le) = []
        pi(true) = []
        pi(|0|) = []
        pi(s) = []
    

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: le#(s(x),s(y)) -> le#(x,y)

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: le(|0|(),y) -> true()
r6: le(s(x),|0|()) -> false()
r7: le(s(x),s(y)) -> le(x,y)
r8: app(nil(),y) -> y
r9: app(add(n,x),y) -> add(n,app(x,y))
r10: low(n,nil()) -> nil()
r11: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r12: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r13: if_low(false(),n,add(m,x)) -> low(n,x)
r14: high(n,nil()) -> nil()
r15: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r16: if_high(true(),n,add(m,x)) -> high(n,x)
r17: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r18: quicksort(nil()) -> nil()
r19: quicksort(add(n,x)) -> app(quicksort(low(n,x)),add(n,quicksort(high(n,x))))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

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: app#(add(n,x),y) -> app#(x,y)

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: le(|0|(),y) -> true()
r6: le(s(x),|0|()) -> false()
r7: le(s(x),s(y)) -> le(x,y)
r8: app(nil(),y) -> y
r9: app(add(n,x),y) -> add(n,app(x,y))
r10: low(n,nil()) -> nil()
r11: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r12: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r13: if_low(false(),n,add(m,x)) -> low(n,x)
r14: high(n,nil()) -> nil()
r15: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r16: if_high(true(),n,add(m,x)) -> high(n,x)
r17: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r18: quicksort(nil()) -> nil()
r19: quicksort(add(n,x)) -> app(quicksort(low(n,x)),add(n,quicksort(high(n,x))))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (2,2)
            add_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (1,1)
    
      precedence:
      
        app# = add
      
      partial status:
    
        pi(app#) = []
        pi(add) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = (1,1)
            add_A(x1,x2) = x1 + (2,2)
    
      precedence:
      
        app# = add
      
      partial status:
    
        pi(app#) = []
        pi(add) = [1]
    

The next rules are strictly ordered:

  p1

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