YES
We show the termination of the TRS R:

  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))
  head(add(n,x)) -> n
  tail(add(n,x)) -> x
  isempty(nil()) -> true()
  isempty(add(n,x)) -> false()
  quicksort(x) -> if_qs(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
  if_qs(true(),x,n,y) -> nil()
  if_qs(false(),x,n,y) -> app(quicksort(x),add(n,quicksort(y)))


-- SCC decomposition.

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

p1: le#(s(x),s(y)) -> le#(x,y)
p2: app#(add(n,x),y) -> app#(x,y)
p3: low#(n,add(m,x)) -> if_low#(le(m,n),n,add(m,x))
p4: low#(n,add(m,x)) -> le#(m,n)
p5: if_low#(true(),n,add(m,x)) -> low#(n,x)
p6: if_low#(false(),n,add(m,x)) -> low#(n,x)
p7: high#(n,add(m,x)) -> if_high#(le(m,n),n,add(m,x))
p8: high#(n,add(m,x)) -> le#(m,n)
p9: if_high#(true(),n,add(m,x)) -> high#(n,x)
p10: if_high#(false(),n,add(m,x)) -> high#(n,x)
p11: quicksort#(x) -> if_qs#(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
p12: quicksort#(x) -> isempty#(x)
p13: quicksort#(x) -> low#(head(x),tail(x))
p14: quicksort#(x) -> head#(x)
p15: quicksort#(x) -> tail#(x)
p16: quicksort#(x) -> high#(head(x),tail(x))
p17: if_qs#(false(),x,n,y) -> app#(quicksort(x),add(n,quicksort(y)))
p18: if_qs#(false(),x,n,y) -> quicksort#(x)
p19: if_qs#(false(),x,n,y) -> quicksort#(y)

and R consists of:

r1: le(|0|(),y) -> true()
r2: le(s(x),|0|()) -> false()
r3: le(s(x),s(y)) -> le(x,y)
r4: app(nil(),y) -> y
r5: app(add(n,x),y) -> add(n,app(x,y))
r6: low(n,nil()) -> nil()
r7: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r8: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r9: if_low(false(),n,add(m,x)) -> low(n,x)
r10: high(n,nil()) -> nil()
r11: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r12: if_high(true(),n,add(m,x)) -> high(n,x)
r13: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r14: head(add(n,x)) -> n
r15: tail(add(n,x)) -> x
r16: isempty(nil()) -> true()
r17: isempty(add(n,x)) -> false()
r18: quicksort(x) -> if_qs(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
r19: if_qs(true(),x,n,y) -> nil()
r20: if_qs(false(),x,n,y) -> app(quicksort(x),add(n,quicksort(y)))

The estimated dependency graph contains the following SCCs:

  {p11, p18, p19}
  {p7, p9, p10}
  {p3, p5, p6}
  {p1}
  {p2}


-- Reduction pair.

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

p1: if_qs#(false(),x,n,y) -> quicksort#(y)
p2: quicksort#(x) -> if_qs#(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
p3: if_qs#(false(),x,n,y) -> quicksort#(x)

and R consists of:

r1: le(|0|(),y) -> true()
r2: le(s(x),|0|()) -> false()
r3: le(s(x),s(y)) -> le(x,y)
r4: app(nil(),y) -> y
r5: app(add(n,x),y) -> add(n,app(x,y))
r6: low(n,nil()) -> nil()
r7: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r8: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r9: if_low(false(),n,add(m,x)) -> low(n,x)
r10: high(n,nil()) -> nil()
r11: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r12: if_high(true(),n,add(m,x)) -> high(n,x)
r13: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r14: head(add(n,x)) -> n
r15: tail(add(n,x)) -> x
r16: isempty(nil()) -> true()
r17: isempty(add(n,x)) -> false()
r18: quicksort(x) -> if_qs(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
r19: if_qs(true(),x,n,y) -> nil()
r20: if_qs(false(),x,n,y) -> app(quicksort(x),add(n,quicksort(y)))

The set of usable rules consists of

  r1, r2, r3, r6, r7, r8, r9, r10, r11, r12, r13, r15, r16, r17, r19

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        if_qs#_A(x1,x2,x3,x4) = max{x1 + 2, x2 + 28, x4 + 37}
        false_A = 110
        quicksort#_A(x1) = max{111, x1 + 27}
        isempty_A(x1) = x1 + 24
        low_A(x1,x2) = max{46, x2 + 1}
        head_A(x1) = 51
        tail_A(x1) = max{47, x1 - 42}
        high_A(x1,x2) = max{49, x2 + 2}
        le_A(x1,x2) = max{108, x1 + 2}
        |0|_A = 108
        true_A = 109
        s_A(x1) = x1 + 111
        nil_A = 86
        add_A(x1,x2) = max{88, x1 - 15, x2 + 43}
        if_low_A(x1,x2,x3) = max{45, x1 - 21, x3 + 1}
        if_high_A(x1,x2,x3) = max{90, x1 - 18, x3 + 2}
        if_qs_A(x1,x2,x3,x4) = max{88, x1 - 62, x2 - 10, x3 - 1, x4 - 89}
  
    base order 2:
  
      precedence:
      
        if_qs > low = high = |0| = s = if_high > le = if_low > false = add > quicksort# > if_qs# = isempty = head = tail > true > nil
      
  
    partial status:
  
      pi(if_qs#) = []
      pi(false) = []
      pi(quicksort#) = []
      pi(isempty) = []
      pi(low) = [2]
      pi(head) = []
      pi(tail) = []
      pi(high) = [2]
      pi(le) = [1]
      pi(|0|) = []
      pi(true) = []
      pi(s) = [1]
      pi(nil) = []
      pi(add) = []
      pi(if_low) = []
      pi(if_high) = [3]
      pi(if_qs) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: quicksort#(x) -> if_qs#(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
p2: if_qs#(false(),x,n,y) -> quicksort#(x)

and R consists of:

r1: le(|0|(),y) -> true()
r2: le(s(x),|0|()) -> false()
r3: le(s(x),s(y)) -> le(x,y)
r4: app(nil(),y) -> y
r5: app(add(n,x),y) -> add(n,app(x,y))
r6: low(n,nil()) -> nil()
r7: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r8: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r9: if_low(false(),n,add(m,x)) -> low(n,x)
r10: high(n,nil()) -> nil()
r11: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r12: if_high(true(),n,add(m,x)) -> high(n,x)
r13: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r14: head(add(n,x)) -> n
r15: tail(add(n,x)) -> x
r16: isempty(nil()) -> true()
r17: isempty(add(n,x)) -> false()
r18: quicksort(x) -> if_qs(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
r19: if_qs(true(),x,n,y) -> nil()
r20: if_qs(false(),x,n,y) -> app(quicksort(x),add(n,quicksort(y)))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: quicksort#(x) -> if_qs#(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
p2: if_qs#(false(),x,n,y) -> quicksort#(x)

and R consists of:

r1: le(|0|(),y) -> true()
r2: le(s(x),|0|()) -> false()
r3: le(s(x),s(y)) -> le(x,y)
r4: app(nil(),y) -> y
r5: app(add(n,x),y) -> add(n,app(x,y))
r6: low(n,nil()) -> nil()
r7: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r8: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r9: if_low(false(),n,add(m,x)) -> low(n,x)
r10: high(n,nil()) -> nil()
r11: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r12: if_high(true(),n,add(m,x)) -> high(n,x)
r13: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r14: head(add(n,x)) -> n
r15: tail(add(n,x)) -> x
r16: isempty(nil()) -> true()
r17: isempty(add(n,x)) -> false()
r18: quicksort(x) -> if_qs(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
r19: if_qs(true(),x,n,y) -> nil()
r20: if_qs(false(),x,n,y) -> app(quicksort(x),add(n,quicksort(y)))

The set of usable rules consists of

  r6, r7, r8, r9, r15, r16, r17, r19

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        quicksort#_A(x1) = max{53, x1 + 32}
        if_qs#_A(x1,x2,x3,x4) = max{x1 + 18, x2 + 33}
        isempty_A(x1) = max{13, x1 - 3}
        low_A(x1,x2) = max{19, x2}
        head_A(x1) = 24
        tail_A(x1) = max{0, x1 - 2}
        high_A(x1,x2) = max{45, x1 + 19}
        false_A = 36
        nil_A = 18
        add_A(x1,x2) = max{x1 + 44, x2 + 3}
        if_low_A(x1,x2,x3) = max{21, x3}
        le_A(x1,x2) = max{x1 - 2, x2 + 13}
        true_A = 14
        if_qs_A(x1,x2,x3,x4) = max{37, x1 - 1, x3 - 25}
  
    base order 2:
  
      precedence:
      
        tail = high = if_qs > low = head = add = if_low = le > quicksort# > if_qs# = false > isempty > true > nil
      
  
    partial status:
  
      pi(quicksort#) = [1]
      pi(if_qs#) = []
      pi(isempty) = []
      pi(low) = []
      pi(head) = []
      pi(tail) = []
      pi(high) = []
      pi(false) = []
      pi(nil) = []
      pi(add) = []
      pi(if_low) = []
      pi(le) = []
      pi(true) = []
      pi(if_qs) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: if_qs#(false(),x,n,y) -> quicksort#(x)

and R consists of:

r1: le(|0|(),y) -> true()
r2: le(s(x),|0|()) -> false()
r3: le(s(x),s(y)) -> le(x,y)
r4: app(nil(),y) -> y
r5: app(add(n,x),y) -> add(n,app(x,y))
r6: low(n,nil()) -> nil()
r7: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r8: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r9: if_low(false(),n,add(m,x)) -> low(n,x)
r10: high(n,nil()) -> nil()
r11: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r12: if_high(true(),n,add(m,x)) -> high(n,x)
r13: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r14: head(add(n,x)) -> n
r15: tail(add(n,x)) -> x
r16: isempty(nil()) -> true()
r17: isempty(add(n,x)) -> false()
r18: quicksort(x) -> if_qs(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
r19: if_qs(true(),x,n,y) -> nil()
r20: if_qs(false(),x,n,y) -> app(quicksort(x),add(n,quicksort(y)))

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_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: le(|0|(),y) -> true()
r2: le(s(x),|0|()) -> false()
r3: le(s(x),s(y)) -> le(x,y)
r4: app(nil(),y) -> y
r5: app(add(n,x),y) -> add(n,app(x,y))
r6: low(n,nil()) -> nil()
r7: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r8: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r9: if_low(false(),n,add(m,x)) -> low(n,x)
r10: high(n,nil()) -> nil()
r11: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r12: if_high(true(),n,add(m,x)) -> high(n,x)
r13: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r14: head(add(n,x)) -> n
r15: tail(add(n,x)) -> x
r16: isempty(nil()) -> true()
r17: isempty(add(n,x)) -> false()
r18: quicksort(x) -> if_qs(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
r19: if_qs(true(),x,n,y) -> nil()
r20: if_qs(false(),x,n,y) -> app(quicksort(x),add(n,quicksort(y)))

The set of usable rules consists of

  r1, r2, r3, r6, r10, r17, r19

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        if_high#_A(x1,x2,x3) = max{65, x1 + 60, x2 + 62, x3 + 43}
        false_A = 74
        add_A(x1,x2) = max{x1 + 59, x2 + 22}
        high#_A(x1,x2) = max{64, x1 + 62, x2 + 44}
        le_A(x1,x2) = max{3, x1, x2 + 1}
        true_A = 27
        |0|_A = 75
        s_A(x1) = x1
        low_A(x1,x2) = max{x1 + 6, x2 - 42}
        nil_A = 5
        high_A(x1,x2) = max{23, x1 - 23, x2 - 58}
        isempty_A(x1) = max{75, x1 - 78}
        if_qs_A(x1,x2,x3,x4) = max{x1 + 8, x2 - 12, x3 + 19, x4 + 2}
  
    base order 2:
  
      precedence:
      
        if_qs > if_high# = high# > add = low = isempty > |0| = high > false > le > true = s > nil
      
  
    partial status:
  
      pi(if_high#) = [3]
      pi(false) = []
      pi(add) = []
      pi(high#) = [2]
      pi(le) = [2]
      pi(true) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(low) = [1]
      pi(nil) = []
      pi(high) = []
      pi(isempty) = []
      pi(if_qs) = []

The following rules are removed:

  p3



-- 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: high#(n,add(m,x)) -> if_high#(le(m,n),n,add(m,x))

and R consists of:

r1: le(|0|(),y) -> true()
r2: le(s(x),|0|()) -> false()
r3: le(s(x),s(y)) -> le(x,y)
r4: app(nil(),y) -> y
r5: app(add(n,x),y) -> add(n,app(x,y))
r6: low(n,nil()) -> nil()
r7: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r8: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r9: if_low(false(),n,add(m,x)) -> low(n,x)
r10: high(n,nil()) -> nil()
r11: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r12: if_high(true(),n,add(m,x)) -> high(n,x)
r13: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r14: head(add(n,x)) -> n
r15: tail(add(n,x)) -> x
r16: isempty(nil()) -> true()
r17: isempty(add(n,x)) -> false()
r18: quicksort(x) -> if_qs(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
r19: if_qs(true(),x,n,y) -> nil()
r20: if_qs(false(),x,n,y) -> app(quicksort(x),add(n,quicksort(y)))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- 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))

and R consists of:

r1: le(|0|(),y) -> true()
r2: le(s(x),|0|()) -> false()
r3: le(s(x),s(y)) -> le(x,y)
r4: app(nil(),y) -> y
r5: app(add(n,x),y) -> add(n,app(x,y))
r6: low(n,nil()) -> nil()
r7: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r8: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r9: if_low(false(),n,add(m,x)) -> low(n,x)
r10: high(n,nil()) -> nil()
r11: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r12: if_high(true(),n,add(m,x)) -> high(n,x)
r13: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r14: head(add(n,x)) -> n
r15: tail(add(n,x)) -> x
r16: isempty(nil()) -> true()
r17: isempty(add(n,x)) -> false()
r18: quicksort(x) -> if_qs(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
r19: if_qs(true(),x,n,y) -> nil()
r20: if_qs(false(),x,n,y) -> app(quicksort(x),add(n,quicksort(y)))

The set of usable rules consists of

  r1, r2, r3, r6, r10, r17, r19

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        if_high#_A(x1,x2,x3) = max{26, x1, x3 - 22}
        false_A = 0
        add_A(x1,x2) = max{x1 + 32, x2 + 25}
        high#_A(x1,x2) = max{1, x2 - 5}
        le_A(x1,x2) = max{0, x1 - 1}
        |0|_A = 13
        true_A = 0
        s_A(x1) = x1 + 13
        low_A(x1,x2) = max{10, x1 - 13, x2}
        nil_A = 0
        high_A(x1,x2) = max{32, x1 + 20, x2 - 23}
        isempty_A(x1) = max{0, x1 - 24}
        if_qs_A(x1,x2,x3,x4) = max{26, x1 - 15, x2 - 11, x3 - 2, x4 - 33}
  
    base order 2:
  
      precedence:
      
        if_qs > if_high# = le > add = |0| = s = low = isempty > high# = high > false = true > nil
      
  
    partial status:
  
      pi(if_high#) = [1]
      pi(false) = []
      pi(add) = [2]
      pi(high#) = []
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(s) = []
      pi(low) = [2]
      pi(nil) = []
      pi(high) = []
      pi(isempty) = []
      pi(if_qs) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: high#(n,add(m,x)) -> if_high#(le(m,n),n,add(m,x))

and R consists of:

r1: le(|0|(),y) -> true()
r2: le(s(x),|0|()) -> false()
r3: le(s(x),s(y)) -> le(x,y)
r4: app(nil(),y) -> y
r5: app(add(n,x),y) -> add(n,app(x,y))
r6: low(n,nil()) -> nil()
r7: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r8: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r9: if_low(false(),n,add(m,x)) -> low(n,x)
r10: high(n,nil()) -> nil()
r11: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r12: if_high(true(),n,add(m,x)) -> high(n,x)
r13: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r14: head(add(n,x)) -> n
r15: tail(add(n,x)) -> x
r16: isempty(nil()) -> true()
r17: isempty(add(n,x)) -> false()
r18: quicksort(x) -> if_qs(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
r19: if_qs(true(),x,n,y) -> nil()
r20: if_qs(false(),x,n,y) -> app(quicksort(x),add(n,quicksort(y)))

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: le(|0|(),y) -> true()
r2: le(s(x),|0|()) -> false()
r3: le(s(x),s(y)) -> le(x,y)
r4: app(nil(),y) -> y
r5: app(add(n,x),y) -> add(n,app(x,y))
r6: low(n,nil()) -> nil()
r7: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r8: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r9: if_low(false(),n,add(m,x)) -> low(n,x)
r10: high(n,nil()) -> nil()
r11: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r12: if_high(true(),n,add(m,x)) -> high(n,x)
r13: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r14: head(add(n,x)) -> n
r15: tail(add(n,x)) -> x
r16: isempty(nil()) -> true()
r17: isempty(add(n,x)) -> false()
r18: quicksort(x) -> if_qs(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
r19: if_qs(true(),x,n,y) -> nil()
r20: if_qs(false(),x,n,y) -> app(quicksort(x),add(n,quicksort(y)))

The set of usable rules consists of

  r1, r2, r3, r6, r10, r17, r19

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        if_low#_A(x1,x2,x3) = max{x1 - 6, x2 + 42, x3 + 17}
        false_A = 45
        add_A(x1,x2) = max{x1 - 5, x2 + 10}
        low#_A(x1,x2) = max{x1 + 42, x2 + 26}
        le_A(x1,x2) = x2 + 48
        true_A = 47
        |0|_A = 46
        s_A(x1) = x1 + 49
        low_A(x1,x2) = max{1, x1 - 21, x2 - 19}
        nil_A = 0
        high_A(x1,x2) = max{43, x1 - 26, x2 + 4}
        isempty_A(x1) = max{56, x1 - 11}
        if_qs_A(x1,x2,x3,x4) = max{3, x1 - 57, x2 - 45, x3 - 53, x4 - 45}
  
    base order 2:
  
      precedence:
      
        if_qs > add > high > |0| > low = isempty > if_low# = low# = true > le = s = nil > false
      
  
    partial status:
  
      pi(if_low#) = [3]
      pi(false) = []
      pi(add) = []
      pi(low#) = [2]
      pi(le) = []
      pi(true) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(low) = []
      pi(nil) = []
      pi(high) = []
      pi(isempty) = []
      pi(if_qs) = []

The following rules are removed:

  p1



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

and R consists of:

r1: le(|0|(),y) -> true()
r2: le(s(x),|0|()) -> false()
r3: le(s(x),s(y)) -> le(x,y)
r4: app(nil(),y) -> y
r5: app(add(n,x),y) -> add(n,app(x,y))
r6: low(n,nil()) -> nil()
r7: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r8: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r9: if_low(false(),n,add(m,x)) -> low(n,x)
r10: high(n,nil()) -> nil()
r11: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r12: if_high(true(),n,add(m,x)) -> high(n,x)
r13: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r14: head(add(n,x)) -> n
r15: tail(add(n,x)) -> x
r16: isempty(nil()) -> true()
r17: isempty(add(n,x)) -> false()
r18: quicksort(x) -> if_qs(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
r19: if_qs(true(),x,n,y) -> nil()
r20: if_qs(false(),x,n,y) -> app(quicksort(x),add(n,quicksort(y)))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

and R consists of:

r1: le(|0|(),y) -> true()
r2: le(s(x),|0|()) -> false()
r3: le(s(x),s(y)) -> le(x,y)
r4: app(nil(),y) -> y
r5: app(add(n,x),y) -> add(n,app(x,y))
r6: low(n,nil()) -> nil()
r7: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r8: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r9: if_low(false(),n,add(m,x)) -> low(n,x)
r10: high(n,nil()) -> nil()
r11: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r12: if_high(true(),n,add(m,x)) -> high(n,x)
r13: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r14: head(add(n,x)) -> n
r15: tail(add(n,x)) -> x
r16: isempty(nil()) -> true()
r17: isempty(add(n,x)) -> false()
r18: quicksort(x) -> if_qs(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
r19: if_qs(true(),x,n,y) -> nil()
r20: if_qs(false(),x,n,y) -> app(quicksort(x),add(n,quicksort(y)))

The set of usable rules consists of

  r1, r2, r3, r6, r10, r19

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        low#_A(x1,x2) = max{x1 + 37, x2 + 11}
        add_A(x1,x2) = max{13, x1 + 6, x2 + 9}
        if_low#_A(x1,x2,x3) = max{x1 - 1, x2 + 37, x3 + 3}
        le_A(x1,x2) = x2 + 36
        true_A = 35
        |0|_A = 14
        s_A(x1) = max{14, x1 + 12}
        false_A = 13
        low_A(x1,x2) = 11
        nil_A = 0
        high_A(x1,x2) = max{13, x2}
        if_qs_A(x1,x2,x3,x4) = max{14, x1 - 36, x2 - 23, x3 - 52, x4 - 50}
  
    base order 2:
  
      precedence:
      
        if_qs > add = |0| > low# = if_low# = s = high > le = true = false = low > nil
      
  
    partial status:
  
      pi(low#) = [2]
      pi(add) = [2]
      pi(if_low#) = [3]
      pi(le) = []
      pi(true) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(false) = []
      pi(low) = []
      pi(nil) = []
      pi(high) = [2]
      pi(if_qs) = []

The following rules are removed:

  p2



-- 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: le(|0|(),y) -> true()
r2: le(s(x),|0|()) -> false()
r3: le(s(x),s(y)) -> le(x,y)
r4: app(nil(),y) -> y
r5: app(add(n,x),y) -> add(n,app(x,y))
r6: low(n,nil()) -> nil()
r7: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r8: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r9: if_low(false(),n,add(m,x)) -> low(n,x)
r10: high(n,nil()) -> nil()
r11: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r12: if_high(true(),n,add(m,x)) -> high(n,x)
r13: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r14: head(add(n,x)) -> n
r15: tail(add(n,x)) -> x
r16: isempty(nil()) -> true()
r17: isempty(add(n,x)) -> false()
r18: quicksort(x) -> if_qs(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
r19: if_qs(true(),x,n,y) -> nil()
r20: if_qs(false(),x,n,y) -> app(quicksort(x),add(n,quicksort(y)))

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: le(|0|(),y) -> true()
r2: le(s(x),|0|()) -> false()
r3: le(s(x),s(y)) -> le(x,y)
r4: app(nil(),y) -> y
r5: app(add(n,x),y) -> add(n,app(x,y))
r6: low(n,nil()) -> nil()
r7: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r8: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r9: if_low(false(),n,add(m,x)) -> low(n,x)
r10: high(n,nil()) -> nil()
r11: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r12: if_high(true(),n,add(m,x)) -> high(n,x)
r13: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r14: head(add(n,x)) -> n
r15: tail(add(n,x)) -> x
r16: isempty(nil()) -> true()
r17: isempty(add(n,x)) -> false()
r18: quicksort(x) -> if_qs(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
r19: if_qs(true(),x,n,y) -> nil()
r20: if_qs(false(),x,n,y) -> app(quicksort(x),add(n,quicksort(y)))

The set of usable rules consists of

  r1, r2, r6, r10, r17, r19

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        le#_A(x1,x2) = max{x1 - 1, x2 + 6}
        s_A(x1) = x1 + 5
        le_A(x1,x2) = max{10, x1 - 6, x2 - 24}
        |0|_A = 5
        true_A = 4
        false_A = 4
        low_A(x1,x2) = max{8, x1 - 23, x2 - 51}
        nil_A = 0
        high_A(x1,x2) = max{5, x1 - 46, x2 - 46}
        isempty_A(x1) = max{5, x1 - 49}
        add_A(x1,x2) = max{7, x1 - 12, x2 - 21}
        if_qs_A(x1,x2,x3,x4) = max{21, x1 - 54, x2 - 34, x3 - 11, x4 - 11}
  
    base order 2:
  
      precedence:
      
        if_qs > le > le# = s = |0| = low = isempty = add > true > nil = high > false
      
  
    partial status:
  
      pi(le#) = [2]
      pi(s) = [1]
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(false) = []
      pi(low) = []
      pi(nil) = []
      pi(high) = []
      pi(isempty) = []
      pi(add) = []
      pi(if_qs) = []

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: le(|0|(),y) -> true()
r2: le(s(x),|0|()) -> false()
r3: le(s(x),s(y)) -> le(x,y)
r4: app(nil(),y) -> y
r5: app(add(n,x),y) -> add(n,app(x,y))
r6: low(n,nil()) -> nil()
r7: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r8: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r9: if_low(false(),n,add(m,x)) -> low(n,x)
r10: high(n,nil()) -> nil()
r11: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r12: if_high(true(),n,add(m,x)) -> high(n,x)
r13: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r14: head(add(n,x)) -> n
r15: tail(add(n,x)) -> x
r16: isempty(nil()) -> true()
r17: isempty(add(n,x)) -> false()
r18: quicksort(x) -> if_qs(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
r19: if_qs(true(),x,n,y) -> nil()
r20: if_qs(false(),x,n,y) -> app(quicksort(x),add(n,quicksort(y)))

The set of usable rules consists of

  r1, r2, r6, r10, r17, r19

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{27, x1 + 14, x2 + 1}
        add_A(x1,x2) = max{13, x1 - 13, x2 + 5}
        le_A(x1,x2) = max{8, x1 - 43, x2 - 27}
        |0|_A = 8
        true_A = 7
        s_A(x1) = max{7, x1 - 44}
        false_A = 6
        low_A(x1,x2) = max{9, x1 - 26, x2 - 29}
        nil_A = 0
        high_A(x1,x2) = max{11, x1 - 12, x2 - 15}
        isempty_A(x1) = max{8, x1 - 35}
        if_qs_A(x1,x2,x3,x4) = max{16, x1 - 11, x2 - 34, x3 - 12, x4 - 12}
  
    base order 2:
  
      precedence:
      
        if_qs > app# = add > high > low > le > |0| = s > false > isempty > true > nil
      
  
    partial status:
  
      pi(app#) = [1]
      pi(add) = [2]
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(s) = []
      pi(false) = []
      pi(low) = []
      pi(nil) = []
      pi(high) = []
      pi(isempty) = []
      pi(if_qs) = []

The following rules are removed:

  p1

Then no dependency pair remains.