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:

  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:
              quot#_A(x1,x2) = x1
              s_A(x1) = x1 + 3
              minus_A(x1,x2) = x1 + 1
              |0|_A() = 1
              quot#!_A(x1,x2) = x2 + 3
              s!_A(x1) = 5
              minus!_A(x1,x2) = 4
              |0|!_A() = 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            quot#_A(x1,x2) = max{0, x1 - 51, x2 - 82}
            s_A(x1) = max{80, x1 + 49}
            minus_A(x1,x2) = max{50, x1 + 17}
            |0|_A = 80
            quot#!_A(x1,x2) = max{30, x1 - 18}
            s!_A(x1) = 27
            minus!_A(x1,x2) = max{29, x1 - 50}
            |0|!_A = 26
        
  
    partial status:
  
      pi(quot#) = [1, 2]
      pi(s) = [1]
      pi(minus) = [1]
      pi(|0|) = []

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: 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:

  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:
              minus#_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 4
              minus#!_A(x1,x2) = x1 + x2
              s!_A(x1) = 6
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            minus#_A(x1,x2) = 0
            s_A(x1) = max{11, x1 - 2}
            minus#!_A(x1,x2) = max{10, x2 - 1}
            s!_A(x1) = 0
        
  
    partial status:
  
      pi(minus#) = []
      pi(s) = []

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: 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:

  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:
              quicksort#_A(x1) = 0
              add_A(x1,x2) = x1 + x2 + 6
              high_A(x1,x2) = x2 + 1
              low_A(x1,x2) = x2 + 1
              le_A(x1,x2) = x1 + x2 + 1
              |0|_A() = 2
              true_A() = 1
              s_A(x1) = x1 + 3
              false_A() = 1
              nil_A() = 2
              if_low_A(x1,x2,x3) = x3 + 1
              if_high_A(x1,x2,x3) = x3 + 1
              quicksort#!_A(x1) = x1
              add!_A(x1,x2) = x2 + 6
              high!_A(x1,x2) = x2 + 3
              low!_A(x1,x2) = x2 + 5
              le!_A(x1,x2) = x1 + 8
              |0|!_A() = 1
              true!_A() = 0
              s!_A(x1) = 3
              false!_A() = 2
              nil!_A() = 0
              if_low!_A(x1,x2,x3) = x3 + 2
              if_high!_A(x1,x2,x3) = x3 + 2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            quicksort#_A(x1) = max{11, x1 - 6}
            add_A(x1,x2) = max{x1 + 5, x2}
            high_A(x1,x2) = max{x1 - 14, x2}
            low_A(x1,x2) = max{17, x1 - 14, x2}
            le_A(x1,x2) = max{2, x1 - 1, x2 + 1}
            |0|_A = 13
            true_A = 12
            s_A(x1) = max{3, x1}
            false_A = 14
            nil_A = 13
            if_low_A(x1,x2,x3) = max{17, x1 - 15, x2 - 14, x3}
            if_high_A(x1,x2,x3) = max{1, x1 - 15, x2 - 14, x3}
            quicksort#!_A(x1) = x1 + 15
            add!_A(x1,x2) = 3
            high!_A(x1,x2) = max{x1 + 21, x2 + 14}
            low!_A(x1,x2) = 2
            le!_A(x1,x2) = 13
            |0|!_A = 16
            true!_A = 11
            s!_A(x1) = x1 + 12
            false!_A = 4
            nil!_A = 1
            if_low!_A(x1,x2,x3) = 1
            if_high!_A(x1,x2,x3) = max{x1 - 3, x2 + 4, x3 + 4}
        
  
    partial status:
  
      pi(quicksort#) = [1]
      pi(add) = [2]
      pi(high) = [2]
      pi(low) = [2]
      pi(le) = [1, 2]
      pi(|0|) = []
      pi(true) = []
      pi(s) = [1]
      pi(false) = []
      pi(nil) = []
      pi(if_low) = [3]
      pi(if_high) = [3]

The following rules are removed:

  p1, p2

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

  (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:
              if_high#_A(x1,x2,x3) = x3
              false_A() = 5
              add_A(x1,x2) = x1 + x2 + 2
              high#_A(x1,x2) = x2 + 2
              le_A(x1,x2) = x1 + x2 + 1
              true_A() = 4
              if_high#!_A(x1,x2,x3) = 3
              false!_A() = 2
              add!_A(x1,x2) = 4
              high#!_A(x1,x2) = 3
              le!_A(x1,x2) = 1
              true!_A() = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_high#_A(x1,x2,x3) = max{1, x2 - 1}
            false_A = 10
            add_A(x1,x2) = max{x1 + 76, x2 + 15}
            high#_A(x1,x2) = max{1, x1 - 1}
            le_A(x1,x2) = max{10, x1 - 196, x2 - 218}
            true_A = 10
            if_high#!_A(x1,x2,x3) = max{85, x3 + 9}
            false!_A = 205
            add!_A(x1,x2) = max{204, x1 + 22, x2 + 17}
            high#!_A(x1,x2) = max{23, x2 + 10}
            le!_A(x1,x2) = max{1, x2 - 233}
            true!_A = 5
        
  
    partial status:
  
      pi(if_high#) = []
      pi(false) = []
      pi(add) = []
      pi(high#) = []
      pi(le) = []
      pi(true) = []

The following rules are removed:

  p1, p2, p3

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:

  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:
              if_low#_A(x1,x2,x3) = x3
              false_A() = 1
              add_A(x1,x2) = x1 + x2 + 8
              low#_A(x1,x2) = x2 + 8
              le_A(x1,x2) = x1 + x2 + 1
              true_A() = 1
              |0|_A() = 3
              s_A(x1) = x1 + 1
              if_low#!_A(x1,x2,x3) = x2 + x3 + 6
              false!_A() = 2
              add!_A(x1,x2) = x2 + 6
              low#!_A(x1,x2) = x1 + x2 + 7
              le!_A(x1,x2) = x1 + 5
              true!_A() = 0
              |0|!_A() = 3
              s!_A(x1) = x1 + 4
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_low#_A(x1,x2,x3) = max{33, x1 - 28, x2 - 27, x3 - 59}
            false_A = 29
            add_A(x1,x2) = max{x1 + 19, x2 + 58}
            low#_A(x1,x2) = max{33, x1 - 27, x2 - 59}
            le_A(x1,x2) = max{29, x1 - 73}
            true_A = 60
            |0|_A = 133
            s_A(x1) = max{72, x1 + 71}
            if_low#!_A(x1,x2,x3) = max{20, x1 - 1, x3 - 56}
            false!_A = 5
            add!_A(x1,x2) = max{7, x2 - 22}
            low#!_A(x1,x2) = x2 + 2
            le!_A(x1,x2) = 61
            true!_A = 5
            |0|!_A = 66
            s!_A(x1) = max{6, x1 - 3}
        
  
    partial status:
  
      pi(if_low#) = []
      pi(false) = []
      pi(add) = []
      pi(low#) = []
      pi(le) = []
      pi(true) = []
      pi(|0|) = []
      pi(s) = []

The following rules are removed:

  p1, p2, p3

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:

  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:
              le#_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 1
              le#!_A(x1,x2) = x1 + x2 + 3
              s!_A(x1) = 2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            le#_A(x1,x2) = 0
            s_A(x1) = max{2, x1 - 5}
            le#!_A(x1,x2) = max{21, x2 + 19}
            s!_A(x1) = x1 + 18
        
  
    partial status:
  
      pi(le#) = []
      pi(s) = []

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: 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:

  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:
              app#_A(x1,x2) = x1
              add_A(x1,x2) = x1 + x2 + 7
              app#!_A(x1,x2) = 13
              add!_A(x1,x2) = x1 + 13
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = 0
            add_A(x1,x2) = max{x1 + 266, x2 + 31}
            app#!_A(x1,x2) = max{159, x1 - 1}
            add!_A(x1,x2) = max{159, x1 + 28, x2 + 24}
        
  
    partial status:
  
      pi(app#) = []
      pi(add) = [2]

The following rules are removed:

  p1

Then no dependency pair remains.