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(cons(N,L),Y) -> cons(N,app(L,Y))
  low(N,nil()) -> nil()
  low(N,cons(M,L)) -> iflow(le(M,N),N,cons(M,L))
  iflow(true(),N,cons(M,L)) -> cons(M,low(N,L))
  iflow(false(),N,cons(M,L)) -> low(N,L)
  high(N,nil()) -> nil()
  high(N,cons(M,L)) -> ifhigh(le(M,N),N,cons(M,L))
  ifhigh(true(),N,cons(M,L)) -> high(N,L)
  ifhigh(false(),N,cons(M,L)) -> cons(M,high(N,L))
  quicksort(nil()) -> nil()
  quicksort(cons(N,L)) -> app(quicksort(low(N,L)),cons(N,quicksort(high(N,L))))

-- SCC decomposition.

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

p1: le#(s(X),s(Y)) -> le#(X,Y)
p2: app#(cons(N,L),Y) -> app#(L,Y)
p3: low#(N,cons(M,L)) -> iflow#(le(M,N),N,cons(M,L))
p4: low#(N,cons(M,L)) -> le#(M,N)
p5: iflow#(true(),N,cons(M,L)) -> low#(N,L)
p6: iflow#(false(),N,cons(M,L)) -> low#(N,L)
p7: high#(N,cons(M,L)) -> ifhigh#(le(M,N),N,cons(M,L))
p8: high#(N,cons(M,L)) -> le#(M,N)
p9: ifhigh#(true(),N,cons(M,L)) -> high#(N,L)
p10: ifhigh#(false(),N,cons(M,L)) -> high#(N,L)
p11: quicksort#(cons(N,L)) -> app#(quicksort(low(N,L)),cons(N,quicksort(high(N,L))))
p12: quicksort#(cons(N,L)) -> quicksort#(low(N,L))
p13: quicksort#(cons(N,L)) -> low#(N,L)
p14: quicksort#(cons(N,L)) -> quicksort#(high(N,L))
p15: quicksort#(cons(N,L)) -> high#(N,L)

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(cons(N,L),Y) -> cons(N,app(L,Y))
r6: low(N,nil()) -> nil()
r7: low(N,cons(M,L)) -> iflow(le(M,N),N,cons(M,L))
r8: iflow(true(),N,cons(M,L)) -> cons(M,low(N,L))
r9: iflow(false(),N,cons(M,L)) -> low(N,L)
r10: high(N,nil()) -> nil()
r11: high(N,cons(M,L)) -> ifhigh(le(M,N),N,cons(M,L))
r12: ifhigh(true(),N,cons(M,L)) -> high(N,L)
r13: ifhigh(false(),N,cons(M,L)) -> cons(M,high(N,L))
r14: quicksort(nil()) -> nil()
r15: quicksort(cons(N,L)) -> app(quicksort(low(N,L)),cons(N,quicksort(high(N,L))))

The estimated dependency graph contains the following SCCs:

  {p12, p14}
  {p7, p9, p10}
  {p3, p5, p6}
  {p1}
  {p2}


-- Reduction pair.

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

p1: quicksort#(cons(N,L)) -> quicksort#(high(N,L))
p2: quicksort#(cons(N,L)) -> quicksort#(low(N,L))

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(cons(N,L),Y) -> cons(N,app(L,Y))
r6: low(N,nil()) -> nil()
r7: low(N,cons(M,L)) -> iflow(le(M,N),N,cons(M,L))
r8: iflow(true(),N,cons(M,L)) -> cons(M,low(N,L))
r9: iflow(false(),N,cons(M,L)) -> low(N,L)
r10: high(N,nil()) -> nil()
r11: high(N,cons(M,L)) -> ifhigh(le(M,N),N,cons(M,L))
r12: ifhigh(true(),N,cons(M,L)) -> high(N,L)
r13: ifhigh(false(),N,cons(M,L)) -> cons(M,high(N,L))
r14: quicksort(nil()) -> nil()
r15: quicksort(cons(N,L)) -> app(quicksort(low(N,L)),cons(N,quicksort(high(N,L))))

The set of usable rules consists of

  r1, r2, r3, r6, r7, r8, r9, r10, r11, r12, r13

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        quicksort#_A(x1) = max{11, x1 + 4}
        cons_A(x1,x2) = max{x1 + 14, x2 + 10}
        high_A(x1,x2) = x2 + 6
        low_A(x1,x2) = max{9, x2 + 2}
        le_A(x1,x2) = x1 + 36
        |0|_A = 27
        true_A = 40
        s_A(x1) = max{28, x1}
        false_A = 28
        iflow_A(x1,x2,x3) = max{x1 - 20, x3 + 2}
        ifhigh_A(x1,x2,x3) = max{x1 - 16, x3 + 6}
        nil_A = 2
  
    precedence:
    
      cons = high = low = le = |0| = true = s = false = iflow = ifhigh = nil > quicksort#
    
    partial status:
  
      pi(quicksort#) = [1]
      pi(cons) = [1]
      pi(high) = [2]
      pi(low) = [2]
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(s) = [1]
      pi(false) = []
      pi(iflow) = [3]
      pi(ifhigh) = [3]
      pi(nil) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: quicksort#(cons(N,L)) -> quicksort#(low(N,L))

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(cons(N,L),Y) -> cons(N,app(L,Y))
r6: low(N,nil()) -> nil()
r7: low(N,cons(M,L)) -> iflow(le(M,N),N,cons(M,L))
r8: iflow(true(),N,cons(M,L)) -> cons(M,low(N,L))
r9: iflow(false(),N,cons(M,L)) -> low(N,L)
r10: high(N,nil()) -> nil()
r11: high(N,cons(M,L)) -> ifhigh(le(M,N),N,cons(M,L))
r12: ifhigh(true(),N,cons(M,L)) -> high(N,L)
r13: ifhigh(false(),N,cons(M,L)) -> cons(M,high(N,L))
r14: quicksort(nil()) -> nil()
r15: quicksort(cons(N,L)) -> app(quicksort(low(N,L)),cons(N,quicksort(high(N,L))))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: quicksort#(cons(N,L)) -> quicksort#(low(N,L))

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(cons(N,L),Y) -> cons(N,app(L,Y))
r6: low(N,nil()) -> nil()
r7: low(N,cons(M,L)) -> iflow(le(M,N),N,cons(M,L))
r8: iflow(true(),N,cons(M,L)) -> cons(M,low(N,L))
r9: iflow(false(),N,cons(M,L)) -> low(N,L)
r10: high(N,nil()) -> nil()
r11: high(N,cons(M,L)) -> ifhigh(le(M,N),N,cons(M,L))
r12: ifhigh(true(),N,cons(M,L)) -> high(N,L)
r13: ifhigh(false(),N,cons(M,L)) -> cons(M,high(N,L))
r14: quicksort(nil()) -> nil()
r15: quicksort(cons(N,L)) -> app(quicksort(low(N,L)),cons(N,quicksort(high(N,L))))

The set of usable rules consists of

  r1, r2, r3, r6, r7, r8, r9

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        quicksort#_A(x1) = max{10, x1 + 6}
        cons_A(x1,x2) = x2 + 10
        low_A(x1,x2) = x2 + 2
        le_A(x1,x2) = max{x1 + 14, x2 + 23}
        |0|_A = 23
        true_A = 22
        s_A(x1) = max{2, x1}
        false_A = 1
        iflow_A(x1,x2,x3) = x3 + 2
        nil_A = 8
  
    precedence:
    
      cons = low = le = |0| = true = iflow = nil > quicksort# > s = false
    
    partial status:
  
      pi(quicksort#) = [1]
      pi(cons) = []
      pi(low) = []
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(s) = []
      pi(false) = []
      pi(iflow) = []
      pi(nil) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

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

p1: ifhigh#(false(),N,cons(M,L)) -> high#(N,L)
p2: high#(N,cons(M,L)) -> ifhigh#(le(M,N),N,cons(M,L))
p3: ifhigh#(true(),N,cons(M,L)) -> high#(N,L)

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(cons(N,L),Y) -> cons(N,app(L,Y))
r6: low(N,nil()) -> nil()
r7: low(N,cons(M,L)) -> iflow(le(M,N),N,cons(M,L))
r8: iflow(true(),N,cons(M,L)) -> cons(M,low(N,L))
r9: iflow(false(),N,cons(M,L)) -> low(N,L)
r10: high(N,nil()) -> nil()
r11: high(N,cons(M,L)) -> ifhigh(le(M,N),N,cons(M,L))
r12: ifhigh(true(),N,cons(M,L)) -> high(N,L)
r13: ifhigh(false(),N,cons(M,L)) -> cons(M,high(N,L))
r14: quicksort(nil()) -> nil()
r15: quicksort(cons(N,L)) -> app(quicksort(low(N,L)),cons(N,quicksort(high(N,L))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        ifhigh#_A(x1,x2,x3) = max{x2 + 2, x3}
        false_A = 1
        cons_A(x1,x2) = max{x1 + 2, x2 + 2}
        high#_A(x1,x2) = max{x1 + 2, x2}
        le_A(x1,x2) = 2
        true_A = 1
        |0|_A = 0
        s_A(x1) = max{6, x1 + 2}
  
    precedence:
    
      ifhigh# = false = cons = high# = le = true = |0| = s
    
    partial status:
  
      pi(ifhigh#) = [3]
      pi(false) = []
      pi(cons) = []
      pi(high#) = [2]
      pi(le) = []
      pi(true) = []
      pi(|0|) = []
      pi(s) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: high#(N,cons(M,L)) -> ifhigh#(le(M,N),N,cons(M,L))
p2: ifhigh#(true(),N,cons(M,L)) -> high#(N,L)

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(cons(N,L),Y) -> cons(N,app(L,Y))
r6: low(N,nil()) -> nil()
r7: low(N,cons(M,L)) -> iflow(le(M,N),N,cons(M,L))
r8: iflow(true(),N,cons(M,L)) -> cons(M,low(N,L))
r9: iflow(false(),N,cons(M,L)) -> low(N,L)
r10: high(N,nil()) -> nil()
r11: high(N,cons(M,L)) -> ifhigh(le(M,N),N,cons(M,L))
r12: ifhigh(true(),N,cons(M,L)) -> high(N,L)
r13: ifhigh(false(),N,cons(M,L)) -> cons(M,high(N,L))
r14: quicksort(nil()) -> nil()
r15: quicksort(cons(N,L)) -> app(quicksort(low(N,L)),cons(N,quicksort(high(N,L))))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: high#(N,cons(M,L)) -> ifhigh#(le(M,N),N,cons(M,L))
p2: ifhigh#(true(),N,cons(M,L)) -> high#(N,L)

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(cons(N,L),Y) -> cons(N,app(L,Y))
r6: low(N,nil()) -> nil()
r7: low(N,cons(M,L)) -> iflow(le(M,N),N,cons(M,L))
r8: iflow(true(),N,cons(M,L)) -> cons(M,low(N,L))
r9: iflow(false(),N,cons(M,L)) -> low(N,L)
r10: high(N,nil()) -> nil()
r11: high(N,cons(M,L)) -> ifhigh(le(M,N),N,cons(M,L))
r12: ifhigh(true(),N,cons(M,L)) -> high(N,L)
r13: ifhigh(false(),N,cons(M,L)) -> cons(M,high(N,L))
r14: quicksort(nil()) -> nil()
r15: quicksort(cons(N,L)) -> app(quicksort(low(N,L)),cons(N,quicksort(high(N,L))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        high#_A(x1,x2) = max{x1 + 9, x2 + 5}
        cons_A(x1,x2) = max{10, x1 + 1, x2 + 4}
        ifhigh#_A(x1,x2,x3) = max{x1 + 1, x2 + 9, x3 + 1}
        le_A(x1,x2) = 4
        true_A = 3
        |0|_A = 0
        s_A(x1) = max{8, x1 + 2}
        false_A = 1
  
    precedence:
    
      high# = cons = ifhigh# = le = true = |0| = s = false
    
    partial status:
  
      pi(high#) = [2]
      pi(cons) = [2]
      pi(ifhigh#) = [3]
      pi(le) = []
      pi(true) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(false) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: high#(N,cons(M,L)) -> ifhigh#(le(M,N),N,cons(M,L))

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(cons(N,L),Y) -> cons(N,app(L,Y))
r6: low(N,nil()) -> nil()
r7: low(N,cons(M,L)) -> iflow(le(M,N),N,cons(M,L))
r8: iflow(true(),N,cons(M,L)) -> cons(M,low(N,L))
r9: iflow(false(),N,cons(M,L)) -> low(N,L)
r10: high(N,nil()) -> nil()
r11: high(N,cons(M,L)) -> ifhigh(le(M,N),N,cons(M,L))
r12: ifhigh(true(),N,cons(M,L)) -> high(N,L)
r13: ifhigh(false(),N,cons(M,L)) -> cons(M,high(N,L))
r14: quicksort(nil()) -> nil()
r15: quicksort(cons(N,L)) -> app(quicksort(low(N,L)),cons(N,quicksort(high(N,L))))

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- Reduction pair.

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

p1: iflow#(false(),N,cons(M,L)) -> low#(N,L)
p2: low#(N,cons(M,L)) -> iflow#(le(M,N),N,cons(M,L))
p3: iflow#(true(),N,cons(M,L)) -> low#(N,L)

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(cons(N,L),Y) -> cons(N,app(L,Y))
r6: low(N,nil()) -> nil()
r7: low(N,cons(M,L)) -> iflow(le(M,N),N,cons(M,L))
r8: iflow(true(),N,cons(M,L)) -> cons(M,low(N,L))
r9: iflow(false(),N,cons(M,L)) -> low(N,L)
r10: high(N,nil()) -> nil()
r11: high(N,cons(M,L)) -> ifhigh(le(M,N),N,cons(M,L))
r12: ifhigh(true(),N,cons(M,L)) -> high(N,L)
r13: ifhigh(false(),N,cons(M,L)) -> cons(M,high(N,L))
r14: quicksort(nil()) -> nil()
r15: quicksort(cons(N,L)) -> app(quicksort(low(N,L)),cons(N,quicksort(high(N,L))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        iflow#_A(x1,x2,x3) = max{x2 + 2, x3}
        false_A = 1
        cons_A(x1,x2) = max{x1 + 2, x2 + 2}
        low#_A(x1,x2) = max{x1 + 2, x2}
        le_A(x1,x2) = 2
        true_A = 1
        |0|_A = 0
        s_A(x1) = max{6, x1 + 2}
  
    precedence:
    
      iflow# = false = cons = low# = le = true = |0| = s
    
    partial status:
  
      pi(iflow#) = [3]
      pi(false) = []
      pi(cons) = []
      pi(low#) = [2]
      pi(le) = []
      pi(true) = []
      pi(|0|) = []
      pi(s) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: low#(N,cons(M,L)) -> iflow#(le(M,N),N,cons(M,L))
p2: iflow#(true(),N,cons(M,L)) -> low#(N,L)

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(cons(N,L),Y) -> cons(N,app(L,Y))
r6: low(N,nil()) -> nil()
r7: low(N,cons(M,L)) -> iflow(le(M,N),N,cons(M,L))
r8: iflow(true(),N,cons(M,L)) -> cons(M,low(N,L))
r9: iflow(false(),N,cons(M,L)) -> low(N,L)
r10: high(N,nil()) -> nil()
r11: high(N,cons(M,L)) -> ifhigh(le(M,N),N,cons(M,L))
r12: ifhigh(true(),N,cons(M,L)) -> high(N,L)
r13: ifhigh(false(),N,cons(M,L)) -> cons(M,high(N,L))
r14: quicksort(nil()) -> nil()
r15: quicksort(cons(N,L)) -> app(quicksort(low(N,L)),cons(N,quicksort(high(N,L))))

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,cons(M,L)) -> iflow#(le(M,N),N,cons(M,L))
p2: iflow#(true(),N,cons(M,L)) -> low#(N,L)

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(cons(N,L),Y) -> cons(N,app(L,Y))
r6: low(N,nil()) -> nil()
r7: low(N,cons(M,L)) -> iflow(le(M,N),N,cons(M,L))
r8: iflow(true(),N,cons(M,L)) -> cons(M,low(N,L))
r9: iflow(false(),N,cons(M,L)) -> low(N,L)
r10: high(N,nil()) -> nil()
r11: high(N,cons(M,L)) -> ifhigh(le(M,N),N,cons(M,L))
r12: ifhigh(true(),N,cons(M,L)) -> high(N,L)
r13: ifhigh(false(),N,cons(M,L)) -> cons(M,high(N,L))
r14: quicksort(nil()) -> nil()
r15: quicksort(cons(N,L)) -> app(quicksort(low(N,L)),cons(N,quicksort(high(N,L))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        low#_A(x1,x2) = max{x1 + 9, x2 + 5}
        cons_A(x1,x2) = max{10, x1 + 1, x2 + 4}
        iflow#_A(x1,x2,x3) = max{x1 + 1, x2 + 9, x3 + 1}
        le_A(x1,x2) = 4
        true_A = 3
        |0|_A = 0
        s_A(x1) = max{8, x1 + 2}
        false_A = 1
  
    precedence:
    
      low# = cons = iflow# = le = true = |0| = s = false
    
    partial status:
  
      pi(low#) = [2]
      pi(cons) = [2]
      pi(iflow#) = [3]
      pi(le) = []
      pi(true) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(false) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: low#(N,cons(M,L)) -> iflow#(le(M,N),N,cons(M,L))

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(cons(N,L),Y) -> cons(N,app(L,Y))
r6: low(N,nil()) -> nil()
r7: low(N,cons(M,L)) -> iflow(le(M,N),N,cons(M,L))
r8: iflow(true(),N,cons(M,L)) -> cons(M,low(N,L))
r9: iflow(false(),N,cons(M,L)) -> low(N,L)
r10: high(N,nil()) -> nil()
r11: high(N,cons(M,L)) -> ifhigh(le(M,N),N,cons(M,L))
r12: ifhigh(true(),N,cons(M,L)) -> high(N,L)
r13: ifhigh(false(),N,cons(M,L)) -> cons(M,high(N,L))
r14: quicksort(nil()) -> nil()
r15: quicksort(cons(N,L)) -> app(quicksort(low(N,L)),cons(N,quicksort(high(N,L))))

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(cons(N,L),Y) -> cons(N,app(L,Y))
r6: low(N,nil()) -> nil()
r7: low(N,cons(M,L)) -> iflow(le(M,N),N,cons(M,L))
r8: iflow(true(),N,cons(M,L)) -> cons(M,low(N,L))
r9: iflow(false(),N,cons(M,L)) -> low(N,L)
r10: high(N,nil()) -> nil()
r11: high(N,cons(M,L)) -> ifhigh(le(M,N),N,cons(M,L))
r12: ifhigh(true(),N,cons(M,L)) -> high(N,L)
r13: ifhigh(false(),N,cons(M,L)) -> cons(M,high(N,L))
r14: quicksort(nil()) -> nil()
r15: quicksort(cons(N,L)) -> app(quicksort(low(N,L)),cons(N,quicksort(high(N,L))))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        le#_A(x1,x2) = max{0, x1 - 2, x2 - 2}
        s_A(x1) = max{3, x1 + 1}
  
    precedence:
    
      le# = s
    
    partial status:
  
      pi(le#) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

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

p1: app#(cons(N,L),Y) -> app#(L,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(cons(N,L),Y) -> cons(N,app(L,Y))
r6: low(N,nil()) -> nil()
r7: low(N,cons(M,L)) -> iflow(le(M,N),N,cons(M,L))
r8: iflow(true(),N,cons(M,L)) -> cons(M,low(N,L))
r9: iflow(false(),N,cons(M,L)) -> low(N,L)
r10: high(N,nil()) -> nil()
r11: high(N,cons(M,L)) -> ifhigh(le(M,N),N,cons(M,L))
r12: ifhigh(true(),N,cons(M,L)) -> high(N,L)
r13: ifhigh(false(),N,cons(M,L)) -> cons(M,high(N,L))
r14: quicksort(nil()) -> nil()
r15: quicksort(cons(N,L)) -> app(quicksort(low(N,L)),cons(N,quicksort(high(N,L))))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        app#_A(x1,x2) = max{0, x1 - 2}
        cons_A(x1,x2) = max{x1 + 3, x2 + 1}
  
    precedence:
    
      app# = cons
    
    partial status:
  
      pi(app#) = []
      pi(cons) = [2]

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.