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, r5

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        quot#_A(x1,x2) = max{x1 + 7, x2 - 1}
        s_A(x1) = max{12, x1 + 6}
        minus_A(x1,x2) = max{10, x1 + 5}
        |0|_A = 8
        le_A(x1,x2) = max{24, x1 - 13, x2 - 13}
        true_A = 7
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          quot#_A(x1,x2) = max{0, x1 - 5, x2 - 5}
          s_A(x1) = max{3, x1 + 1}
          minus_A(x1,x2) = max{4, x1 + 1, x2 - 2}
          |0|_A = 3
          le_A(x1,x2) = max{11, x1 - 4, x2 - 4}
          true_A = 0
          quot#!_A(x1,x2) = x2 + 1
          s!_A(x1) = max{0, x1 - 193}
          minus!_A(x1,x2) = max{0, x1 - 193}
          |0|!_A = 2
          le!_A(x1,x2) = max{2, x1 - 192, x2 - 4}
          true!_A = 1
  
    partial status:
  
      pi(quot#) = [1]
      pi(s) = [1]
      pi(minus) = []
      pi(|0|) = []
      pi(le) = []
      pi(true) = []

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:
  
      max/plus interpretation on natural numbers:
      
        minus#_A(x1,x2) = 0
        s_A(x1) = max{3, x1 - 16}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          minus#_A(x1,x2) = max{0, x1 - 12, x2 - 12}
          s_A(x1) = x1 + 11
          minus#!_A(x1,x2) = max{7, x1 - 3, x2 - 128}
          s!_A(x1) = max{6, x1 - 127}
  
    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:
  
      max/plus interpretation on natural numbers:
      
        quicksort#_A(x1) = max{95, x1 + 78}
        add_A(x1,x2) = max{x1 + 58, x2 + 24}
        high_A(x1,x2) = max{16, x2 + 7}
        low_A(x1,x2) = max{23, x2}
        le_A(x1,x2) = max{3, x1 - 1, x2 - 18}
        |0|_A = 21
        true_A = 19
        s_A(x1) = x1 + 6
        false_A = 4
        nil_A = 0
        if_low_A(x1,x2,x3) = max{46, x3}
        if_high_A(x1,x2,x3) = max{65, x3 + 7}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          quicksort#_A(x1) = max{0, x1 - 20}
          add_A(x1,x2) = max{20, x1 - 5, x2}
          high_A(x1,x2) = x2
          low_A(x1,x2) = max{19, x1 - 38, x2}
          le_A(x1,x2) = 7
          |0|_A = 35
          true_A = 1
          s_A(x1) = max{15, x1 - 13}
          false_A = 1
          nil_A = 27
          if_low_A(x1,x2,x3) = max{20, x1 - 13, x2 - 38, x3}
          if_high_A(x1,x2,x3) = max{1, x1 - 12, x3}
          quicksort#!_A(x1) = 0
          add!_A(x1,x2) = x1 + 4
          high!_A(x1,x2) = max{57, x1 + 2, x2 + 35}
          low!_A(x1,x2) = max{52, x1 - 37, x2 + 30}
          le!_A(x1,x2) = max{58, x1 - 36, x2 - 36}
          |0|!_A = 60
          true!_A = 59
          s!_A(x1) = 57
          false!_A = 1
          nil!_A = 58
          if_low!_A(x1,x2,x3) = max{31, x1 - 13, x2 - 38, x3 + 4}
          if_high!_A(x1,x2,x3) = max{56, x1 - 2, x3 + 9}
  
    partial status:
  
      pi(quicksort#) = []
      pi(add) = [1]
      pi(high) = [2]
      pi(low) = []
      pi(le) = []
      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

  r5, r6, r7

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        if_high#_A(x1,x2,x3) = 1
        false_A = 4
        add_A(x1,x2) = max{6, x1 - 24, x2 - 35}
        high#_A(x1,x2) = 1
        le_A(x1,x2) = max{44, x1 - 22, x2 - 6}
        true_A = 2
        |0|_A = 5
        s_A(x1) = max{25, x1 + 2}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          if_high#_A(x1,x2,x3) = max{0, x1 - 3, x2 - 161, x3 - 28}
          false_A = 2
          add_A(x1,x2) = max{x1 - 94, x2 + 26}
          high#_A(x1,x2) = max{0, x1 - 161, x2 - 28}
          le_A(x1,x2) = max{2, x1 - 158, x2 - 158}
          true_A = 2
          |0|_A = 149
          s_A(x1) = max{157, x1 + 18}
          if_high#!_A(x1,x2,x3) = max{146, x3 + 95}
          false!_A = 5
          add!_A(x1,x2) = max{3, x1 - 306, x2 - 200}
          high#!_A(x1,x2) = max{122, x2 + 121}
          le!_A(x1,x2) = max{6, x1 - 150, x2 - 143}
          true!_A = 2
          |0|!_A = 6
          s!_A(x1) = max{7, x1 - 143}
  
    partial status:
  
      pi(if_high#) = []
      pi(false) = []
      pi(add) = []
      pi(high#) = []
      pi(le) = []
      pi(true) = []
      pi(|0|) = []
      pi(s) = [1]

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: if_high#(false(),n,add(m,x)) -> high#(n,x)
p2: 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 estimated dependency graph contains the following SCCs:

  (no SCCs)

-- 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:
  
      max/plus interpretation on natural numbers:
      
        if_low#_A(x1,x2,x3) = 4
        false_A = 7
        add_A(x1,x2) = max{x1 + 21, x2 - 24}
        low#_A(x1,x2) = 4
        le_A(x1,x2) = max{46, x1 - 10, x2 - 10}
        true_A = 5
        |0|_A = 8
        s_A(x1) = x1 + 9
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          if_low#_A(x1,x2,x3) = max{0, x1 - 20, x2 - 3, x3 - 16}
          false_A = 0
          add_A(x1,x2) = max{14, x1 - 72, x2 + 2}
          low#_A(x1,x2) = max{0, x1 - 3, x2 - 16}
          le_A(x1,x2) = max{19, x1 - 117, x2 - 97}
          true_A = 19
          |0|_A = 4
          s_A(x1) = x1 + 96
          if_low#!_A(x1,x2,x3) = max{88, x3 + 73}
          false!_A = 1
          add!_A(x1,x2) = max{5, x1 - 49, x2 - 56}
          low#!_A(x1,x2) = max{88, x2 + 75}
          le!_A(x1,x2) = max{90, x1 - 5, x2 - 97}
          true!_A = 90
          |0|!_A = 91
          s!_A(x1) = max{94, x1 - 4}
  
    partial status:
  
      pi(if_low#) = []
      pi(false) = []
      pi(add) = [1]
      pi(low#) = []
      pi(le) = []
      pi(true) = []
      pi(|0|) = []
      pi(s) = [1]

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: if_low#(false(),n,add(m,x)) -> low#(n,x)
p2: 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 estimated dependency graph contains the following SCCs:

  (no SCCs)

-- 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:
  
      max/plus interpretation on natural numbers:
      
        le#_A(x1,x2) = 1
        s_A(x1) = max{5, x1 - 12}
  
    base order 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{18, x1 + 5}
          le#!_A(x1,x2) = max{5, x1 - 6, x2 - 384}
          s!_A(x1) = max{4, x1 - 385}
  
    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:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = x2 + 7
        add_A(x1,x2) = max{5, x1 - 22, x2 - 23}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = max{0, x1 - 3}
          add_A(x1,x2) = max{14, x1 - 61, x2 + 2}
          app#!_A(x1,x2) = max{10, x1 - 3}
          add!_A(x1,x2) = max{9, x1 - 189, x2 - 220}
  
    partial status:
  
      pi(app#) = []
      pi(add) = []

The following rules are removed:

  p1

Then no dependency pair remains.