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, r10, r18

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!):
        
          max/plus interpretation on natural numbers:
          
            quot#_A(x1,x2) = 0
            s_A(x1) = x1 + 11
            minus_A(x1,x2) = max{x1 + 11, x2 - 12}
            |0|_A = 6
            low_A(x1,x2) = max{40, x1 - 17, x2 - 17}
            nil_A = 16
            quicksort_A(x1) = max{23, x1 - 25}
            quot#!_A(x1,x2) = max{x1 + 26, x2 + 3}
            s!_A(x1) = max{7, x1 - 2}
            minus!_A(x1,x2) = max{2, x1 - 12}
            |0|!_A = 6
            low!_A(x1,x2) = max{2, x1 - 84, x2 - 171}
            nil!_A = 0
            quicksort!_A(x1) = max{6, x1 - 41}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            quot#_A(x1,x2) = max{0, x1 - 10, x2 - 11}
            s_A(x1) = x1 + 9
            minus_A(x1,x2) = max{9, x1 + 3, x2 + 6}
            |0|_A = 8
            low_A(x1,x2) = max{39, x1 - 13, x2 - 13}
            nil_A = 8
            quicksort_A(x1) = max{25, x1 - 27}
            quot#!_A(x1,x2) = max{1, x1 - 4, x2 - 7}
            s!_A(x1) = max{6, x1 - 162}
            minus!_A(x1,x2) = max{4, x1 - 5, x2 - 157}
            |0|!_A = 2
            low!_A(x1,x2) = max{0, x1 - 78, x2 - 122}
            nil!_A = 0
            quicksort!_A(x1) = max{0, x1 - 40}
        
  
    partial status:
  
      pi(quot#) = [1]
      pi(s) = [1]
      pi(minus) = [1]
      pi(|0|) = []
      pi(low) = []
      pi(nil) = []
      pi(quicksort) = []

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!):
        
          max/plus interpretation on natural numbers:
          
            minus#_A(x1,x2) = max{0, x1 - 3, x2 - 3}
            s_A(x1) = max{11, x1 + 2}
            minus#!_A(x1,x2) = max{8, x1 - 12, x2 - 17}
            s!_A(x1) = max{7, x1 - 16}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            minus#_A(x1,x2) = max{0, x1 - 3, x2 - 3}
            s_A(x1) = max{11, x1 + 2}
            minus#!_A(x1,x2) = max{1, x1 - 12, x2 - 12}
            s!_A(x1) = max{5, x1 - 2}
        
  
    partial status:
  
      pi(minus#) = [1]
      pi(s) = [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: 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

  r3, 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!):
        
          max/plus interpretation on natural numbers:
          
            quicksort#_A(x1) = 0
            add_A(x1,x2) = 69
            high_A(x1,x2) = x1 + 68
            low_A(x1,x2) = x1 + 77
            quot_A(x1,x2) = max{24, x1 - 1, x2 - 4}
            |0|_A = 13
            s_A(x1) = max{2, x1}
            le_A(x1,x2) = max{67, x2 + 55}
            true_A = 67
            false_A = 68
            nil_A = 61
            if_low_A(x1,x2,x3) = max{x2 + 77, x3 - 29}
            if_high_A(x1,x2,x3) = max{x1 + 1, x2 + 68}
            quicksort#!_A(x1) = 2
            add!_A(x1,x2) = max{15, x2 - 40}
            high!_A(x1,x2) = x1 + 29
            low!_A(x1,x2) = x1 + 139
            quot!_A(x1,x2) = max{x1 + 22, x2 + 18}
            |0|!_A = 17
            s!_A(x1) = max{12, x1 + 1}
            le!_A(x1,x2) = 14
            true!_A = 16
            false!_A = 13
            nil!_A = 28
            if_low!_A(x1,x2,x3) = max{x2 + 139, x3 - 68}
            if_high!_A(x1,x2,x3) = max{x1 - 54, x2 + 29, x3 - 53}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            quicksort#_A(x1) = 0
            add_A(x1,x2) = max{x1 + 23, x2 + 21}
            high_A(x1,x2) = max{2, x2}
            low_A(x1,x2) = max{21, x2 + 19}
            quot_A(x1,x2) = max{x1 + 2, x2 - 1}
            |0|_A = 2
            s_A(x1) = x1 + 2
            le_A(x1,x2) = max{34, x2 + 1}
            true_A = 16
            false_A = 18
            nil_A = 0
            if_low_A(x1,x2,x3) = max{41, x3 + 19}
            if_high_A(x1,x2,x3) = max{22, x3}
            quicksort#!_A(x1) = max{9, x1 - 7}
            add!_A(x1,x2) = 0
            high!_A(x1,x2) = max{x1 + 12, x2 + 15}
            low!_A(x1,x2) = max{8, x2 - 5}
            quot!_A(x1,x2) = 1
            |0|!_A = 2
            s!_A(x1) = max{3, x1 - 4}
            le!_A(x1,x2) = max{37, x1 - 5, x2 + 13}
            true!_A = 3
            false!_A = 37
            nil!_A = 0
            if_low!_A(x1,x2,x3) = max{15, x3 - 16}
            if_high!_A(x1,x2,x3) = max{x1 - 2, x2 + 12, x3 - 6}
        
  
    partial status:
  
      pi(quicksort#) = []
      pi(add) = [1, 2]
      pi(high) = [2]
      pi(low) = [2]
      pi(quot) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(le) = [1]
      pi(true) = []
      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

  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!):
        
          max/plus interpretation on natural numbers:
          
            if_high#_A(x1,x2,x3) = max{0, x1 - 1, x2 - 3, x3 - 26}
            false_A = 0
            add_A(x1,x2) = max{24, x1 + 17, x2 + 17}
            high#_A(x1,x2) = max{0, x1 - 3, x2 - 26}
            le_A(x1,x2) = max{0, x1 - 49, x2 - 49}
            true_A = 0
            |0|_A = 33
            s_A(x1) = max{48, x1 + 47}
            if_high#!_A(x1,x2,x3) = max{17, x3 - 14}
            false!_A = 14
            add!_A(x1,x2) = max{16, x1 - 4, x2 - 25}
            high#!_A(x1,x2) = max{2, x2 - 6}
            le!_A(x1,x2) = max{12, x1 - 34, x2 - 49}
            true!_A = 1
            |0|!_A = 15
            s!_A(x1) = max{17, x1 - 99}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_high#_A(x1,x2,x3) = max{0, x1 - 1, x2 - 483, x3 - 16}
            false_A = 0
            add_A(x1,x2) = max{10, x1 - 284, x2 + 4}
            high#_A(x1,x2) = max{0, x1 - 483, x2 - 12}
            le_A(x1,x2) = max{0, x1 - 492, x2 - 482}
            true_A = 0
            |0|_A = 481
            s_A(x1) = max{491, x1 + 11}
            if_high#!_A(x1,x2,x3) = max{289, x3 + 285}
            false!_A = 5
            add!_A(x1,x2) = max{6, x1 - 495, x2 - 143}
            high#!_A(x1,x2) = x2 + 286
            le!_A(x1,x2) = max{3, x1 - 458, x2 - 492}
            true!_A = 9
            |0|!_A = 4
            s!_A(x1) = max{6, x1 - 458}
        
  
    partial status:
  
      pi(if_high#) = [2]
      pi(false) = []
      pi(add) = [1]
      pi(high#) = [1]
      pi(le) = []
      pi(true) = []
      pi(|0|) = []
      pi(s) = [1]

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, r10

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!):
        
          max/plus interpretation on natural numbers:
          
            if_low#_A(x1,x2,x3) = max{0, x1 - 1, x2 - 3, x3 - 10}
            false_A = 0
            add_A(x1,x2) = max{8, x1 + 6, x2 + 6}
            low#_A(x1,x2) = max{0, x1 - 3, x2 - 10}
            le_A(x1,x2) = max{0, x1 - 15, x2 - 15}
            true_A = 0
            |0|_A = 5
            s_A(x1) = max{14, x1 + 4}
            low_A(x1,x2) = max{50, x1 + 15}
            nil_A = 16
            if_low#!_A(x1,x2,x3) = 0
            false!_A = 21
            add!_A(x1,x2) = max{0, x2 - 4}
            low#!_A(x1,x2) = max{0, x2 - 5}
            le!_A(x1,x2) = max{24, x2 - 28}
            true!_A = 23
            |0|!_A = 22
            s!_A(x1) = max{23, x1 - 29}
            low!_A(x1,x2) = max{26, x1 + 3, x2 - 17}
            nil!_A = 2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_low#_A(x1,x2,x3) = max{0, x1 - 1, x2 - 3, x3 - 36}
            false_A = 0
            add_A(x1,x2) = max{34, x1 - 111, x2 + 7}
            low#_A(x1,x2) = max{0, x1 - 3, x2 - 36}
            le_A(x1,x2) = max{0, x1 - 146, x2 - 14}
            true_A = 0
            |0|_A = 13
            s_A(x1) = max{16, x1 + 13}
            low_A(x1,x2) = max{110, x1 - 111}
            nil_A = 7
            if_low#!_A(x1,x2,x3) = max{0, x1 - 1, x2 - 1, x3 - 36}
            false!_A = 2
            add!_A(x1,x2) = max{x1 + 2, x2 + 192}
            low#!_A(x1,x2) = max{1, x2 - 37}
            le!_A(x1,x2) = max{25, x1 - 120, x2 - 136}
            true!_A = 5
            |0|!_A = 6
            s!_A(x1) = x1 + 24
            low!_A(x1,x2) = max{193, x1 - 54, x2 - 37}
            nil!_A = 0
        
  
    partial status:
  
      pi(if_low#) = [3]
      pi(false) = []
      pi(add) = [2]
      pi(low#) = [2]
      pi(le) = []
      pi(true) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(low) = [2]
      pi(nil) = []

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!):
        
          max/plus interpretation on natural numbers:
          
            le#_A(x1,x2) = max{0, x1 - 2, x2 - 2}
            s_A(x1) = max{7, x1 + 1}
            le#!_A(x1,x2) = max{45, x1 - 8, x2 - 10}
            s!_A(x1) = max{44, x1 - 9}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            le#_A(x1,x2) = max{0, x1 - 6, x2 - 6}
            s_A(x1) = max{8, x1 + 5}
            le#!_A(x1,x2) = max{5, x1 - 88, x2 - 9}
            s!_A(x1) = max{4, x1 - 87}
        
  
    partial status:
  
      pi(le#) = [1]
      pi(s) = [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: 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!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 2}
            add_A(x1,x2) = max{4, x1 - 46, x2 + 1}
            app#!_A(x1,x2) = max{3, x1 - 5, x2 - 1}
            add!_A(x1,x2) = max{2, x1 - 104, x2 - 61}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 7}
            add_A(x1,x2) = max{14, x1 - 85, x2 + 6}
            app#!_A(x1,x2) = max{7, x1 - 131, x2 - 1}
            add!_A(x1,x2) = max{6, x1 - 220, x2 - 130}
        
  
    partial status:
  
      pi(app#) = [1]
      pi(add) = [2]

The following rules are removed:

  p1

Then no dependency pair remains.