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, r3, r14, r18

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          quot#_A(x1,x2) = x1 + 1
          s_A(x1) = x1 + 4
          minus_A(x1,x2) = x1 + 1
          |0|_A() = 2
          quot_A(x1,x2) = 3
          high_A(x1,x2) = x1 + x2 + 3
          nil_A() = 1
          quicksort_A(x1) = x1 + 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          quot#_A(x1,x2) = max{51, x1 + 34, x2 - 35}
          s_A(x1) = x1 + 34
          minus_A(x1,x2) = max{17, x1 + 16}
          |0|_A = 0
          quot_A(x1,x2) = max{17, x1}
          high_A(x1,x2) = max{40, x1 - 125, x2 - 42}
          nil_A = 9
          quicksort_A(x1) = max{39, x1 - 10}
          quot#!_A(x1,x2) = max{67, x1 + 34}
          s!_A(x1) = x1 + 52
          minus!_A(x1,x2) = max{66, x1 - 35, x2 - 35}
          |0|!_A = 53
          quot!_A(x1,x2) = max{49, x1 + 34}
          high!_A(x1,x2) = max{50, x1 - 36, x2 - 8}
          nil!_A = 0
          quicksort!_A(x1) = max{0, x1 - 58}
  
    partial status:
  
      pi(quot#) = []
      pi(s) = []
      pi(minus) = [1]
      pi(|0|) = []
      pi(quot) = []
      pi(high) = [2]
      pi(nil) = []
      pi(quicksort) = [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: 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

  r14

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          minus#_A(x1,x2) = 0
          s_A(x1) = 8
          high_A(x1,x2) = x1 + x2 + 4
          nil_A() = 1
  
    base order 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{67, x1 + 2}
          high_A(x1,x2) = max{57, x1 - 48, x2 - 47}
          nil_A = 30
          minus#!_A(x1,x2) = max{1, x1 - 3, x2 - 68}
          s!_A(x1) = max{70, x1 + 68}
          high!_A(x1,x2) = max{92, x1 - 1, x2 - 31}
          nil!_A = 91
  
    partial status:
  
      pi(minus#) = []
      pi(s) = []
      pi(high) = [1]
      pi(nil) = []

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, r18

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          quicksort#_A(x1) = x1 + 3
          add_A(x1,x2) = x2 + 2
          high_A(x1,x2) = x2
          low_A(x1,x2) = x2 + 1
          le_A(x1,x2) = 5
          |0|_A() = 5
          true_A() = 4
          s_A(x1) = x1 + 6
          false_A() = 0
          nil_A() = 1
          if_low_A(x1,x2,x3) = x3 + 1
          if_high_A(x1,x2,x3) = x3
          quicksort_A(x1) = 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          quicksort#_A(x1) = 0
          add_A(x1,x2) = max{9, x1 - 15, x2}
          high_A(x1,x2) = max{x1 + 8, x2}
          low_A(x1,x2) = max{16, x1 - 15, x2}
          le_A(x1,x2) = max{0, x1 - 4, x2 - 7}
          |0|_A = 3
          true_A = 0
          s_A(x1) = max{3, x1 + 1}
          false_A = 0
          nil_A = 9
          if_low_A(x1,x2,x3) = max{16, x2 - 15, x3}
          if_high_A(x1,x2,x3) = max{x2 + 8, x3}
          quicksort_A(x1) = x1 + 7
          quicksort#!_A(x1) = max{1, x1 - 10}
          add!_A(x1,x2) = 6
          high!_A(x1,x2) = max{2, x1 - 26, x2 - 10}
          low!_A(x1,x2) = max{x1 + 5, x2 + 22}
          le!_A(x1,x2) = max{8, x2 - 6}
          |0|!_A = 8
          true!_A = 8
          s!_A(x1) = max{8, x1 - 5}
          false!_A = 5
          nil!_A = 1
          if_low!_A(x1,x2,x3) = max{x2 + 5, x3 + 22}
          if_high!_A(x1,x2,x3) = max{0, x2 - 27, x3 - 36}
          quicksort!_A(x1) = max{3, x1 - 13}
  
    partial status:
  
      pi(quicksort#) = [1]
      pi(add) = []
      pi(high) = [2]
      pi(low) = []
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(s) = [1]
      pi(false) = []
      pi(nil) = []
      pi(if_low) = []
      pi(if_high) = [3]
      pi(quicksort) = []

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

  r3, r5, r10, r14, r18

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          if_high#_A(x1,x2,x3) = x3
          false_A() = 8
          add_A(x1,x2) = x2 + 2
          high#_A(x1,x2) = x2 + 2
          le_A(x1,x2) = 7
          true_A() = 3
          quot_A(x1,x2) = 10
          |0|_A() = 9
          s_A(x1) = x1 + 11
          low_A(x1,x2) = x1 + x2 + 4
          nil_A() = 1
          high_A(x1,x2) = x1 + x2 + 9
          quicksort_A(x1) = x1 + 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          if_high#_A(x1,x2,x3) = max{1, x2 - 2, x3 - 15}
          false_A = 0
          add_A(x1,x2) = max{14, x1 + 1, x2 + 6}
          high#_A(x1,x2) = max{1, x1 - 2, x2 - 15}
          le_A(x1,x2) = max{x1, x2}
          true_A = 0
          quot_A(x1,x2) = max{1, x1 - 246, x2 - 407}
          |0|_A = 0
          s_A(x1) = max{163, x1 + 162}
          low_A(x1,x2) = max{67, x1 - 7}
          nil_A = 1
          high_A(x1,x2) = max{40, x1 - 10, x2 - 16}
          quicksort_A(x1) = max{24, x1 - 41}
          if_high#!_A(x1,x2,x3) = max{73, x2 + 72, x3 + 71}
          false!_A = 74
          add!_A(x1,x2) = max{72, x2 - 160}
          high#!_A(x1,x2) = max{71, x2 - 161}
          le!_A(x1,x2) = max{73, x1 - 164}
          true!_A = 1
          quot!_A(x1,x2) = max{1, x1 - 84, x2 - 86}
          |0|!_A = 75
          s!_A(x1) = max{245, x1 + 243}
          low!_A(x1,x2) = max{73, x1 - 162, x2 - 165}
          nil!_A = 1
          high!_A(x1,x2) = max{71, x2 - 167}
          quicksort!_A(x1) = max{0, x1 - 143}
  
    partial status:
  
      pi(if_high#) = []
      pi(false) = []
      pi(add) = [2]
      pi(high#) = []
      pi(le) = []
      pi(true) = []
      pi(quot) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(low) = [2]
      pi(nil) = []
      pi(high) = []
      pi(quicksort) = []

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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          if_low#_A(x1,x2,x3) = 1
          false_A() = 2
          add_A(x1,x2) = x2 + 5
          low#_A(x1,x2) = 1
          le_A(x1,x2) = 13
          true_A() = 2
          |0|_A() = 3
          s_A(x1) = x1 + 14
  
    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 - 1, x2 - 3, x3 - 35}
          false_A = 0
          add_A(x1,x2) = max{33, x1 + 28, x2 + 28}
          low#_A(x1,x2) = max{0, x1 - 3, x2 - 35}
          le_A(x1,x2) = max{0, x1 - 16, x2 - 16}
          true_A = 0
          |0|_A = 2
          s_A(x1) = x1 + 15
          if_low#!_A(x1,x2,x3) = max{12, x3 - 47}
          false!_A = 1
          add!_A(x1,x2) = max{5, x2 - 20}
          low#!_A(x1,x2) = max{4, x2 - 21}
          le!_A(x1,x2) = max{6, x1 - 3, x2 - 15}
          true!_A = 1
          |0|!_A = 2
          s!_A(x1) = x1 + 15
  
    partial status:
  
      pi(if_low#) = []
      pi(false) = []
      pi(add) = [2]
      pi(low#) = []
      pi(le) = []
      pi(true) = []
      pi(|0|) = []
      pi(s) = [1]

The following rules are removed:

  p1, p3



-- SCC decomposition.

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

p1: low#(n,add(m,x)) -> if_low#(le(m,n),n,add(m,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

  r14

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          le#_A(x1,x2) = 0
          s_A(x1) = 8
          high_A(x1,x2) = x1 + x2 + 4
          nil_A() = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          le#_A(x1,x2) = max{0, x1 - 89, x2 - 89}
          s_A(x1) = max{129, x1 + 88}
          high_A(x1,x2) = max{100, x1 - 82, x2 - 77}
          nil_A = 74
          le#!_A(x1,x2) = max{1, x1 - 89, x2 - 89}
          s!_A(x1) = max{222, x1 + 220}
          high!_A(x1,x2) = max{302, x1 - 1, x2 - 75}
          nil!_A = 301
  
    partial status:
  
      pi(le#) = []
      pi(s) = []
      pi(high) = [1]
      pi(nil) = []

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

  r5, r6

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          app#_A(x1,x2) = x1
          add_A(x1,x2) = x2 + 3
          le_A(x1,x2) = x2 + 2
          |0|_A() = 4
          true_A() = 1
          s_A(x1) = x1 + 3
          false_A() = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = max{0, x1 - 8}
          add_A(x1,x2) = max{x1 - 64, x2 + 7}
          le_A(x1,x2) = max{86, x1 - 2, x2 - 2}
          |0|_A = 1
          true_A = 0
          s_A(x1) = max{1, x1 - 216}
          false_A = 0
          app#!_A(x1,x2) = max{68, x1 - 32}
          add!_A(x1,x2) = max{x1 + 69, x2 - 31}
          le!_A(x1,x2) = max{214, x2 - 2}
          |0|!_A = 216
          true!_A = 1
          s!_A(x1) = x1 + 214
          false!_A = 214
  
    partial status:
  
      pi(app#) = [1]
      pi(add) = []
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(s) = [1]
      pi(false) = []

The following rules are removed:

  p1

Then no dependency pair remains.